Displaying 1-10 of 68 results found.
Final digits of A005165(2n) for large n, read from right.
+20
2
9, 1, 0, 7, 7, 5, 5, 3, 6, 1, 6, 1, 2, 6, 3, 9, 3, 5, 6, 4, 4, 1, 2, 8, 8, 5, 7, 6, 3, 2, 6, 8, 6, 5, 5, 0, 1, 6, 0, 4, 8, 5, 9, 1, 6, 9, 4, 6, 3, 4, 0, 7, 3, 8, 1, 7, 6, 8, 1, 3, 1, 6, 8, 8, 1, 1, 9, 8, 9, 2, 8, 8, 8, 6, 0, 2, 9, 5, 3, 9, 9, 6, 6, 3, 2, 9, 9, 3, 6, 3, 4, 2, 4, 7, 9, 0, 7, 7, 4, 5, 4, 6, 4, 8, 6
EXAMPLE
A005165(20)=2317323290554617019; A005165(30)=256705485669535347568006115577019.
Final digits of A005165(2n+1) for large n, read from right.
+20
1
1, 8, 9, 2, 2, 4, 4, 6, 3, 8, 3, 8, 7, 3, 6, 0, 6, 4, 3, 5, 5, 8, 7, 1, 1, 4, 2, 3, 6, 7, 3, 1, 3, 4, 4, 9, 8, 3, 9, 5, 1, 4, 0, 8, 3, 0, 5, 3, 6, 5, 9, 2, 6, 1, 8, 2, 3, 1, 8, 6, 8, 3, 1, 1, 8, 8, 0, 1, 0, 7, 1, 1, 1, 3, 9, 7, 0, 4, 6, 0, 0, 3, 3, 6, 7, 0, 0, 6, 3, 6, 5, 7, 5, 2, 0, 9, 2, 2, 5, 4, 5, 3, 5, 1, 3
EXAMPLE
A005165(21)=48773618881154822981; A005165(31)=7966133168508387470157556764422981.
0, 1, 2, 7, 26, 127, 746, 5167, 41066, 368047, 3669866, 40284847, 482671466, 6267305647, 87660962666, 1313941673647, 21010450850666, 357001369769647, 6423384156578666, 122002101778601647, 2439325392333218666, 51212944273488041647, 1126440053169940898666
FORMULA
a(n) = Sum_{i=0..floor((n-1)/2)} (n-2*i)!.
D-finite with recurrence a(n) -n*a(n-1) -a(n-2) +n*a(n-3)=0.
Numbers n such that A005165(n) is a semiprime.
+20
0
9, 11, 12, 17, 18, 23, 26, 31, 62, 64, 67, 76, 78, 84, 87, 88, 123, 125, 131
EXAMPLE
a(1) = 9 so A005165(9) = 326981 = 79 * 4139 is a semiprime.
Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.
(Formerly M0692 N0256)
+10
6165
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155
COMMENTS
D. E. Knuth writes: "Before Fibonacci wrote his work, the sequence F_{n} had already been discussed by Indian scholars, who had long been interested in rhythmic patterns that are formed from one-beat and two-beat notes. The number of such rhythms having n beats altogether is F_{n+1}; therefore both Gopāla (before 1135) and Hemachandra (c. 1150) mentioned the numbers 1, 2, 3, 5, 8, 13, 21, ... explicitly." (TAOCP Vol. 1, 2nd ed.) - Peter Luschny, Jan 11 2015 In keeping with historical accounts (see the references by P. Singh and S. Kak), the generalized Fibonacci sequence a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, ... can also be described as the Gopala-Hemachandra numbers H(n) = H(n-1) + H(n-2), with F(n) = H(n) for a = b = 1, and Lucas sequence L(n) = H(n) for a = 2, b = 1. - Lekraj Beedassy, Jan 11 2015 Susantha Goonatilake writes: "[T]his sequence was well known in South Asia and used in the metrical sciences. Its development is attributed in part to Pingala (200 BC), later being associated with Virahanka (circa 700 AD), Gopala (circa 1135), and Hemachandra (circa 1150)—all of whom lived and worked prior to Fibonacci." (Toward a Global Science: Mining Civilizational Knowledge, p. 126) - Russ Cox, Sep 08 2021 Also sometimes called Hemachandra numbers.
Also sometimes called Lamé's sequence.
For a photograph of "Fibonacci"'s 1202 book, see the Leonardo of Pisa link below.
F(n+2) = number of binary sequences of length n that have no consecutive 0's.
F(n+2) = number of subsets of {1,2,...,n} that contain no consecutive integers.
F(n+1) = number of tilings of a 2 X n rectangle by 2 X 1 dominoes.
F(n+1) = number of matchings (i.e., Hosoya index) in a path graph on n vertices: F(5)=5 because the matchings of the path graph on the vertices A, B, C, D are the empty set, {AB}, {BC}, {CD} and {AB, CD}. - Emeric Deutsch, Jun 18 2001 F(n) = number of compositions of n+1 with no part equal to 1. [Cayley, Grimaldi]
Positive terms are the solutions to z = 2*x*y^4 + (x^2)*y^3 - 2*(x^3)*y^2 - y^5 - (x^4)*y + 2*y for x,y >= 0 (Ribenboim, page 193). When x=F(n), y=F(n + 1) and z > 0 then z=F(n + 1).
For Fibonacci search see Knuth, Vol. 3; Horowitz and Sahni; etc.
F(n+1) is the number of perfect matchings in ladder graph L_n = P_2 X P_n. - Sharon Sela (sharonsela(AT)hotmail.com), May 19 2002
F(n+1) = number of (3412,132)-, (3412,213)- and (3412,321)-avoiding involutions in S_n.
This is also the Horadam sequence (0,1,1,1). - Ross La Haye, Aug 18 2003 An INVERT transform of A019590. INVERT([1,1,2,3,5,8,...]) gives A000129. INVERT([1,2,3,5,8,13,21,...]) gives A028859. - Antti Karttunen, Dec 12 2003 Number of meaningful differential operations of the k-th order on the space R^3. - Branko Malesevic, Mar 02 2004 F(n) = number of compositions of n-1 with no part greater than 2. Example: F(4) = 3 because we have 3 = 1+1+1 = 1+2 = 2+1.
F(n) = number of compositions of n into odd parts; e.g., F(6) counts 1+1+1+1+1+1, 1+1+1+3, 1+1+3+1, 1+3+1+1, 1+5, 3+1+1+1, 3+3, 5+1. - Clark Kimberling, Jun 22 2004 F(n) = number of binary words of length n beginning with 0 and having all runlengths odd; e.g., F(6) counts 010101, 010111, 010001, 011101, 011111, 000101, 000111, 000001. - Clark Kimberling, Jun 22 2004 The number of sequences (s(0),s(1),...,s(n)) such that 0 < s(i) < 5, |s(i)-s(i-1)|=1 and s(0)=1 is F(n+1); e.g., F(5+1) = 8 corresponds to 121212, 121232, 121234, 123212, 123232, 123234, 123432, 123434. - Clark Kimberling, Jun 22 2004 [corrected by Neven Juric, Jan 09 2009] Likewise F(6+1) = 13 corresponds to these thirteen sequences with seven numbers: 1212121, 1212123, 1212321, 1212323, 1212343, 1232121, 1232123, 1232321, 1232323, 1232343, 1234321, 1234323, 1234343. - Neven Juric, Jan 09 2008
A relationship between F(n) and the Mandelbrot set is discussed in the link "Le nombre d'or dans l'ensemble de Mandelbrot" (in French). - Gerald McGarvey, Sep 19 2004 For n > 0, the continued fraction for F(2n-1)*phi = [F(2n); L(2n-1), L(2n-1), L(2n-1), ...] and the continued fraction for F(2n)*phi = [F(2n+1)-1; 1, L(2n)-2, 1, L(2n)-2, ...]. Also true: F(2n)*phi = [F(2n+1); -L(2n), L(2n), -L(2n), L(2n), ...] where L(i) is the i-th Lucas number ( A000204). - Clark Kimberling, Nov 28 2004 [corrected by Hieronymus Fischer, Oct 20 2010] For any nonzero number k, the continued fraction [4,4,...,4,k], which is n 4's and a single k, equals (F(3n) + k*F(3n+3))/(F(3n-3) + k*F(3n)). - Greg Dresden, Aug 07 2019 F(n+1) (for n >= 1) = number of permutations p of 1,2,3,...,n such that |k-p(k)| <= 1 for k=1,2,...,n. (For <= 2 and <= 3, see A002524 and A002526.) - Clark Kimberling, Nov 28 2004 The ratios F(n+1)/F(n) for n > 0 are the convergents to the simple continued fraction expansion of the golden section. - Jonathan Sondow, Dec 19 2004 Lengths of successive words (starting with a) under the substitution: {a -> ab, b -> a}. - Jeroen F.J. Laros, Jan 22 2005 The Fibonacci sequence, like any additive sequence, naturally tends to be geometric with common ratio not a rational power of 10; consequently, for a sufficiently large number of terms, Benford's law of first significant digit (i.e., first digit 1 <= d <= 9 occurring with probability log_10(d+1) - log_10(d)) holds. - Lekraj Beedassy, Apr 29 2005 (See Brown-Duncan, 1970. - N. J. A. Sloane, Feb 12 2017) F(n+2) = Sum_{k=0..n} binomial(floor((n+k)/2),k), row sums of A046854. - Paul Barry, Mar 11 2003 Number of order ideals of the "zig-zag" poset. See vol. 1, ch. 3, prob. 23 of Stanley. - Mitch Harris, Dec 27 2005 F(n+1)/F(n) is also the Farey fraction sequence (see A097545 for explanation) for the golden ratio, which is the only number whose Farey fractions and continued fractions are the same. - Joshua Zucker, May 08 2006 a(n+2) is the number of paths through 2 plates of glass with n reflections (reflections occurring at plate/plate or plate/air interfaces). Cf. A006356- A006359. - Mitch Harris, Jul 06 2006 F(n+1) equals the number of downsets (i.e., decreasing subsets) of an n-element fence, i.e., an ordered set of height 1 on {1,2,...,n} with 1 > 2 < 3 > 4 < ... n and no other comparabilities. Alternatively, F(n+1) equals the number of subsets A of {1,2,...,n} with the property that, if an odd k is in A, then the adjacent elements of {1,2,...,n} belong to A, i.e., both k - 1 and k + 1 are in A (provided they are in {1,2,...,n}). - Brian Davey, Aug 25 2006 Number of Kekulé structures in polyphenanthrenes. See the paper by Lukovits and Janezic for details. - Parthasarathy Nambi, Aug 22 2006 Inverse: With phi = (sqrt(5) + 1)/2, round(log_phi(sqrt((sqrt(5) a(n) + sqrt(5 a(n)^2 - 4))(sqrt(5) a(n) + sqrt(5 a(n)^2 + 4)))/2)) = n for n >= 3, obtained by rounding the arithmetic mean of the inverses given in A001519 and A001906. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007 A result of Jacobi from 1848 states that every symmetric matrix over a p.i.d. is congruent to a triple-diagonal matrix. Consider the maximal number T(n) of summands in the determinant of an n X n triple-diagonal matrix. This is the same as the number of summands in such a determinant in which the main-, sub- and superdiagonal elements are all nonzero. By expanding on the first row we see that the sequence of T(n)'s is the Fibonacci sequence without the initial stammer on the 1's. - Larry Gerstein (gerstein(AT)math.ucsb.edu), Mar 30 2007
Suppose psi=log(phi). We get the representation F(n)=(2/sqrt(5))*sinh(n*psi) if n is even; F(n)=(2/sqrt(5))*cosh(n*psi) if n is odd. There is a similar representation for Lucas numbers ( A000032). Many Fibonacci formulas now easily follow from appropriate sinh and cosh formulas. For example: the de Moivre theorem (cosh(x)+sinh(x))^m = cosh(mx)+sinh(mx) produces L(n)^2 + 5F(n)^2 = 2L(2n) and L(n)F(n) = F(2n) (setting x=n*psi and m=2). - Hieronymus Fischer, Apr 18 2007 Inverse: floor(log_phi(sqrt(5)*F(n)) + 1/2) = n, for n > 1. Also for n > 0, floor((1/2)*log_phi(5*F(n)*F(n+1))) = n. Extension valid for integer n, except n=0,-1: floor((1/2)*sign(F(n)*F(n+1))*log_phi|5*F(n)*F(n+1)|) = n (where sign(x) = sign of x). - Hieronymus Fischer, May 02 2007 F(n+2) = the number of Khalimsky-continuous functions with a two-point codomain. - Shiva Samieinia (shiva(AT)math.su.se), Oct 04 2007
This is a_1(n) in the Doroslovacki reference.
The sequence of first differences, F(n+1)-F(n), is essentially the same sequence: 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... - Colm Mulcahy, Mar 03 2008 Except for the initial term, the numerator of the convergents to the recursion x = 1/(x+1). - Cino Hilliard, Sep 15 2008 F(n) is the number of possible binary sequences of length n that obey the sequential construction rule: if last symbol is 0, add the complement (1); else add 0 or 1. Here 0,1 are metasymbols for any 2-valued symbol set. This rule has obvious similarities to JFJ Laros's rule, but is based on addition rather than substitution and creates a tree rather than a single sequence. - Ross Drewe, Oct 05 2008 F(n) = Product_{k=1..(n-1)/2} (1 + 4*cos^2 k*Pi/n), where terms = roots to the Fibonacci product polynomials, A152063. - Gary W. Adamson, Nov 22 2008 Output of Kasteleyn's formula for the number of perfect matchings of an m X n grid specializes to the Fibonacci sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009 (F(n),F(n+4)) satisfies the Diophantine equation: X^2 + Y^2 - 7XY = 9*(-1)^n. - Mohamed Bouhamida, Sep 06 2009 (F(n),F(n+2)) satisfies the Diophantine equation: X^2 + Y^2 - 3XY = (-1)^n. - Mohamed Bouhamida, Sep 08 2009 Difference between number of closed walks of length n+1 from a node on a pentagon and number of walks of length n+1 between two adjacent nodes on a pentagon. - Henry Bottomley, Feb 10 2010 F(n+1) = number of Motzkin paths of length n having exactly one weak ascent. A Motzkin path of length n is a lattice path from (0,0) to (n,0) consisting of U=(1,1), D=(1,-1) and H=(1,0) steps and never going below the x-axis. A weak ascent in a Motzkin path is a maximal sequence of consecutive U and H steps. Example: a(5)=5 because we have (HHHH), (HHU)D, (HUH)D, (UHH)D, and (UU)DD (the unique weak ascent is shown between parentheses; see A114690). - Emeric Deutsch, Mar 11 2010 From the Pinter and Ziegler reference's abstract: authors "show that essentially the Fibonacci sequence is the unique binary recurrence which contains infinitely many three-term arithmetic progressions. A criterion for general linear recurrences having infinitely many three-term arithmetic progressions is also given." - Jonathan Vos Post, May 22 2010 F(n+1) = number of paths of length n starting at initial node on the path graph P_4. - Johannes W. Meijer, May 27 2010 F(k) = number of cyclotomic polynomials in denominator of generating function for number of ways to place k nonattacking queens on an n X n board. - Vaclav Kotesovec, Jun 07 2010 As n->oo, (a(n)/a(n-1) - a(n-1)/a(n)) tends to 1.0. Example: a(12)/a(11) - a(11)/a(12) = 144/89 - 89/144 = 0.99992197.... - Gary W. Adamson, Jul 16 2010 Fibonacci numbers are those numbers m such that m*phi is closer to an integer than k*phi for all k, 1 <= k < m. More formally: a(0)=0, a(1)=1, a(2)=1, a(n+1) = minimal m > a(n) such that m*phi is closer to an integer than a(n)*phi.
For all numbers 1 <= k < F(n), the inequality |k*phi-round(k*phi)| > |F(n)*phi-round(F(n)*phi)| holds.
F(n)*phi - round(F(n)*phi) = -((-phi)^(-n)), for n > 1.
Fract(1/2 + F(n)*phi) = 1/2 -(-phi)^(-n), for n > 1.
Fract(F(n)*phi) = (1/2)*(1 + (-1)^n) - (-phi)^(-n), n > 1.
Inverse: n = -log_phi |1/2 - fract(1/2 + F(n)*phi)|.
(End)
F(n+k)^2 - F(n)^2 = F(k)*F(2n+k), for even k. - Gary Detlefs, Dec 04 2010 F(n+k)^2 + F(n)^2 = F(k)*F(2n+k), for odd k. - Gary Detlefs, Dec 04 2010 The number of free permutations of [n].
The number of permutations of [n] for which s_k in supp(w) implies s_{k+-1} not in supp(w).
The number of permutations of [n] in which every decomposition into length(w) reflections is actually composed of simple reflections. (End)
The sequence F(n+1)^(1/n) is increasing. The sequence F(n+2)^(1/n) is decreasing. - Thomas Ordowski, Apr 19 2012 Two conjectures: For n > 1, F(n+2)^2 mod F(n+1)^2 = F(n)*F(n+1) - (-1)^n. For n > 0, (F(2n) + F(2n+2))^2 = F(4n+3) + Sum_{k = 2..2n} F(2k). - Alex Ratushnyak, May 06 2012 Proof of Ratushnyak's first conjecture: For n > 1, F(n+2)^2 - F(n)*F(n+1) + (-1)^n = 2*F(n+1)^2.
Consider: F(n+2)^2 - F(n)*F(n+1) - 2*F(n+1)^2
= F(n+2)^2 - F(n+1)^2 - F(n+1)^2 - F(n)*F(n+1)
= (F(n+2) + F(n+1))*(F(n+2) - F(n+1)) - F(n+1)*(F(n+1) + F(n))
= F(n+3)*F(n) - F(n+1)*F(n+2) = -(-1)^n.
Proof of second conjecture: L(n) stands for Lucas number sequence from A000032. Consider the fact that
L(2n+1)^2 = L(4n+2) - 2
(F(2n) + F(2n+2))^2 = F(4n+1) + F(4n+3) - 2
(F(2n) + F(2n+2))^2 = (Sum_{k = 2..2n} F(2k)) + F(4n+3).
(End)
The relationship: INVERT transform of (1,1,0,0,0,...) = (1, 2, 3, 5, 8, ...), while the INVERT transform of (1,0,1,0,1,0,1,...) = (1, 1, 2, 3, 5, 8, ...) is equivalent to: The numbers of compositions using parts 1 and 2 is equivalent to the numbers of compositions using parts == 1 (mod 2) (i.e., the odd integers). Generally, the numbers of compositions using parts 1 and k is equivalent to the numbers of compositions of (n+1) using parts 1 mod k. Cf. A000930 for k = 3 and A003269 for k = 4. Example: for k = 2, n = 4 we have the compositions (22; 211, 121; 112; 1111) = 5; but using parts 1 and 3 we have for n = 5: (311, 131, 113, 11111, 5) = 5. - Gary W. Adamson, Jul 05 2012 The sequence F(n) is the binomial transformation of the alternating sequence (-1)^(n-1)*F(n), whereas the sequence F(n+1) is the binomial transformation of the alternating sequence (-1)^n*F(n-1). Both of these facts follow easily from the equalities a(n;1)=F(n+1) and b(n;1)=F(n) where a(n;d) and b(n;d) are so-called "delta-Fibonacci" numbers as defined in comments to A014445 (see also the papers of Witula et al.). - Roman Witula, Jul 24 2012 F(n) is the number of different (n-1)-digit binary numbers such that all substrings of length > 1 have at least one digit equal to 1. Example: for n = 5 there are 8 binary numbers with n - 1 = 4 digits (1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111), only the F(n) = 5 numbers 1010, 1011, 1101, 1110 and 1111 have the desired property. - Hieronymus Fischer, Nov 30 2012 For positive n, F(n+1) equals the determinant of the n X n tridiagonal matrix with 1's along the main diagonal, i's along the superdiagonal and along the subdiagonal where i = sqrt(-1). Example: Det([1,i,0,0; i,1,i,0; 0,i,1,i; 0,0,i,1]) = F(4+1) = 5. - Philippe Deléham, Feb 24 2013 For n >= 1, number of compositions of n where there is a drop between every second pair of parts, starting with the first and second part; see example. Also, a(n+1) is the number of compositions where there is a drop between every second pair of parts, starting with the second and third part; see example. - Joerg Arndt, May 21 2013 [see the Hopkins/Tangboonduangjit reference for a proof, see also the Checa reference for alternative proofs and statistics] For n >= 4, F(n-1) is the number of simple permutations in the geometric grid class given in A226433. - Jay Pantone, Sep 08 2013 a(n) are the pentagon (not pentagonal) numbers because the algebraic degree 2 number rho(5) = 2*cos(Pi/5) = phi (golden section), the length ratio diagonal/side in a pentagon, has minimal polynomial C(5,x) = x^2 - x - 1 (see A187360, n=5), hence rho(5)^n = a(n-1)*1 + a(n)*rho(5), n >= 0, in the power basis of the algebraic number field Q(rho(5)). One needs a(-1) = 1 here. See also the P. Steinbach reference under A049310. - Wolfdieter Lang, Oct 01 2013 Define F(-n) to be F(n) for n odd and -F(n) for n even. Then for all n and k, F(n+2k)^2 - F(n)^2 = F(n+k)*( F(n+3k) - F(n-k) ). - Charlie Marion, Dec 20 2013 ( F(n), F(n+2k) ) satisfies the Diophantine equation: X^2 + Y^2 - L(2k)*X*Y = F(4k)^2*(-1)^n. This generalizes Bouhamida's comments dated Sep 06 2009 and Sep 08 2009. - Charlie Marion, Jan 07 2014 For any prime p there is an infinite periodic subsequence within F(n) divisible by p, that begins at index n = 0 with value 0, and its first nonzero term at n = A001602(i), and period k = A001602(i). Also see A236479. - Richard R. Forberg, Jan 26 2014 Range of row n of the circular Pascal array of order 5. - Shaun V. Ault, May 30 2014 [orig. Kicey-Klimko 2011, and observations by Glen Whitehead; more general work found in Ault-Kicey 2014] Nonnegative range of the quintic polynomial 2*y - y^5 + 2*x*y^4 + x^2*y^3 - 2*x^3*y^2 - x^4*y with x, y >= 0, see Jones 1975. - Charles R Greathouse IV, Jun 01 2014 The expression round(1/(F(k+1)/F(n) + F(k)/F(n+1))), for n > 0, yields a Fibonacci sequence with k-1 leading zeros (with rounding 0.5 to 0). - Richard R. Forberg, Aug 04 2014 Conjecture: For n > 0, F(n) is the number of all admissible residue classes for which specific finite subsequences of the Collatz 3n + 1 function consists of n+2 terms. This has been verified for 0 < n < 51. For details see Links. - Mike Winkler, Oct 03 2014 a(4)=3 and a(6)=8 are the only Fibonacci numbers that are of the form prime+1. - Emmanuel Vantieghem, Oct 02 2014 a(1)=1=a(2), a(3)=2 are the only Fibonacci numbers that are of the form prime-1. - Emmanuel Vantieghem, Jun 07 2015 Any consecutive pair (m, k) of the Fibonacci sequence a(n) illustrates a fair equivalence between m miles and k kilometers. For instance, 8 miles ~ 13 km; 13 miles ~ 21 km. - Lekraj Beedassy, Oct 06 2014 a(n+1) counts closed walks on K_2, containing one loop on the other vertex. Equivalently the (1,1)_entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1; 1,1). - David Neil McGrath, Oct 29 2014 a(n-1) counts closed walks on the graph G(1-vertex;l-loop,2-loop). - David Neil McGrath, Nov 26 2014 Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x) = [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x). Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)]. Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078). BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)]. Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x). Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1]. (End)
F(n+1) equals the number of binary words of length n avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015 We prove Conjecture 1 of Rashid listed in the Formula section.
We use the following notation: F(n)= A000045(n), the Fibonacci numbers, and L(n) = A000032(n), the Lucas numbers. The fundamental Fibonacci-Lucas recursion asserts that G(n) = G(n-1) + G(n-2), with "L" or "F" replacing "G". We need the following prerequisites which we label (A), (B), (C), (D). The prerequisites are formulas in the Koshy book listed in the References section. (A) F(m-1) + F(m+1) = L(m) (Koshy, p. 97, #32), (B) L(2m) + 2*(-1)^m = L(m)^2 (Koshy p. 97, #41), (C) F(m+k)*F(m-k) = (-1)^n*F(k)^2 (Koshy, p. 113, #24, Tagiuri's identity), and (D) F(n)^2 + F(n+1)^2 = F(2n+1) (Koshy, p. 97, #30).
We must also prove (E), L(n+2)*F(n-1) = F(2n+1)+2*(-1)^n. To prove (E), first note that by (A), proof of (E) is equivalent to proving that F(n+1)*F(n-1) + F(n+3)*F(n-1) = F(2n+1) + 2*(-1)^n. But by (C) with k=1, we have F(n+1)*F(n-1) = F(n)^2 + (-1)^n. Applying (C) again with k=2 and m=n+1, we have F(n+3)*F(n-1) = F(n+1) + (-1)^n. Adding these two applications of (C) together and using (D) we have F(n+1)*F(n-1) + F(n+3)*F(n-1) = F(n)^2 + F(n+1)^2 + 2*(-1)^n = F(2n+1) + 2(-1)^n, completing the proof of (E).
We now prove Conjecture 1. By (A) and the Fibonacci-Lucas recursion, we have F(2n+1) + F(2n+2) + F(2n+3) + F(2n+4) = (F(2n+1) + F(2n+3)) + (F(2n+2) + F(2n+4)) = L(2n+2) +L(2n+3) = L(2n+4). But then by (B), with m=2n+4, we have sqrt(L(2n+4) + 2(-1)^n) = L(n+2). Finally by (E), we have L(n+2)*F(n-1) = F(2n+1) + 2*(-1)^n. Dividing both sides by F(n-1), we have (F(2n+1) + 2*(-1)^n)/F(n-1) = L(n+2) = sqrt(F(2n+1) + F(2n+2) + F(2n+3) + F(2n+4) + 2(-1)^n), as required.
(End)
In Fibonacci's Liber Abaci the rabbit problem appears in the translation of L. E. Sigler on pp. 404-405, and a remark [27] on p. 637. - Wolfdieter Lang, Apr 17 2015 a(n) counts partially ordered partitions of (n-1) into parts 1,2,3 where only the order of adjacent 1's and 2's are unimportant. (See example.) - David Neil McGrath, Jul 27 2015 F(n) divides F(n*k). Proved by Marjorie Bicknell and Verner E Hoggatt Jr. - Juhani Heino, Aug 24 2015 F(n) is the number of UDU-equivalence classes of ballot paths of length n. Two ballot paths of length n with steps U = (1,1), D = (1,-1) are UDU-equivalent whenever the positions of UDU are the same in both paths. - Kostas Manes, Aug 25 2015 Cassini's identity F(2n+1) * F(2n+3) = F(2n+2)^2 + 1 is the basis for a geometrical paradox (or dissection fallacy) in A262342. - Jonathan Sondow, Oct 23 2015 For n >= 4, F(n) is the number of up-down words on alphabet {1,2,3} of length n-2. - Ran Pan, Nov 23 2015 F(n+2) is the number of terms in p(n), where p(n)/q(n) is the n-th convergent of the formal infinite continued fraction [a(0),a(1),...]; e.g., p(3) = a(0)a(1)a(2)a(3) + a(0)a(1) + a(0)a(3) + a(2)a(3) + 1 has F(5) terms. Also, F(n+1) is the number of terms in q(n). - Clark Kimberling, Dec 23 2015 F(n+1) (for n >= 1) is the permanent of an n X n matrix M with M(i,j)=1 if |i-j| <= 1 and 0 otherwise. - Dmitry Efimov, Jan 08 2016 A trapezoid has three sides of lengths in order F(n), F(n+2), F(n). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*F(n+1). For a trapezoid with lengths of sides in order F(n+2), F(n), F(n+2), the fourth side will be F(n+3). - J. M. Bergot, Mar 17 2016 (1) Join two triangles with lengths of sides L(n), F(n+3), L(n+2) and F(n+2), L(n+1), L(n+2) (where L(n)= A000032(n)) along the common side of length L(n+2) to create an irregular quadrilateral. Its area is approximately 5*F(2*n-1) - (F(2*n-7) - F(2*n-13))/5. (2) Join two triangles with lengths of sides L(n), F(n+2), F(n+3) and L(n+1), F(n+1), F(n+3) along the common side F(n+3) to form an irregular quadrilateral. Its area is approximately 4*F(2*n-1) - 2*(F(2*n-7) + F(2*n-18)). - J. M. Bergot, Apr 06 2016 Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*.
Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2, x}, g(3) = {3, 2x, x+1, x^2}, etc.
Let T(r) be the tree obtained by substituting r for x.
If a positive integer N is not a square and r = sqrt(N), then the number of (not necessarily distinct) integers in g(n) is A000045(n), for n >= 1. See A274142. (End) Consider the partitions of n, with all summands initially listed in nonincreasing order. Freeze all the 1's in place and then allow all the other summands to change their order, without displacing any of the 1's. The resulting number of arrangements is a(n+1). - Gregory L. Simay, Jun 14 2016 Limit of the matrix power M^k shown in A163733, Sep 14 2016, as k->infinity results in a single column vector equal to the Fibonacci sequence. - Gary W. Adamson, Sep 19 2016 F(n) and Lucas numbers L(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 10 2017 Also the number of independent vertex sets and vertex covers in the (n-2)-path graph. - Eric W. Weisstein, Sep 22 2017 Shifted numbers of {UD, DU, FD, DF}-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are P-equivalent iff the positions of pattern P are identical in these paths. - Sergey Kirgizov, Apr 08 2018 For n > 0, F(n) = the number of Markov equivalence classes with skeleton the path on n nodes. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018 For n >= 2, also: number of terms in A032858 (every other base-3 digit is strictly smaller than its neighbors) with n-2 digits in base 3. - M. F. Hasler, Oct 05 2018 F(n+1) is the number of fixed points of the Foata transformation on S_n. - Kevin Long, Oct 17 2018 F(n+2) is the dimension of the Hecke algebra of type A_n with independent parameters (0,1,0,1,...) or (1,0,1,0,...). See Corollary 1.5 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019 The sequence is the second INVERT transform of (1, -1, 2, -3, 5, -8, 13, ...) and is the first sequence in an infinite set of successive INVERT transforms generated from (1, 0, 1, 0, 1, ...). Refer to the array shown in A073133. - Gary W. Adamson, Jul 16 2019 F(n*k)/F(k) = Sum_{i=0..n-1; j=0..n-1; i+2*j=n-1} (-1)^(j*(k-1))*L(k)^i*((i+j)!/(i!*j!)).
F((2*m+1)*k)/F(k) = Sum_{i=0..m-1} (-1)^(i*k)*L((2*m-2*i)*k) + (-1)^(m*k).
F(2*m*k)/F(k) = Sum_{i=0..m-1} (-1)^(i*k)*L((2*m-2*i-1)*k).
F(m+s)*F(n+r) - F(m+r)*F(n+s) = (-1)^(n+s)*F(m-n)*F(r-s).
F(m+r)*F(n+s) + F(m+s)*F(n+r) = (2*L(m+n+r+s) - (-1)^(n+s)*L(m-n)*L(r-s))/5.
L(m+r)*L(n+s) - 5*F(m+s)*F(n+r) = (-1)^(n+s)*L(m-n)*L(r-s).
L(m+r)*L(n+s) + 5*F(m+s)*F(n+r) = 2*L(m+n+r+s) + (-1)^(n+s)*5*F(m-n)*F(r-s).
L(m+r)*L(n+s) - L(m+s)*L(n+r) = (-1)^(n+s)*5*F(m-n)*F(r-s). (End)
F(n+1) is the number of permutations in S_n whose principal order ideals in the weak order are Boolean lattices. - Bridget Tenner, Jan 16 2020 F(n+1) is the number of permutations w in S_n that form Boolean intervals [s, w] in the weak order for every simple reflection s in the support of w. - Bridget Tenner, Jan 16 2020 F(n+1) is the number of subsets of {1,2,.,.,n} in which all differences between successive elements of subsets are odd. For example, for n = 6, F(7) = 13 and the 13 subsets are {6}, {1,6}, {3,6}, {5,6}, {2,3,6}, {2,5,6}, {4,5,6}, {1,2,3,6}, {1,2,5,6}, {1,4,5,6}, {3,4,5,6}, {2,3,4,5,6}, {1,2,3,4,5,6}. For even differences between elements see Comment in A016116. - Enrique Navarrete, Jul 01 2020 F(n) is the number of subsets of {1,2,...,n} in which the smallest element of the subset equals the size of the subset (this type of subset is sometimes called extraordinary). For example, F(6) = 8 and the subsets are {1}, {2,3}, {2,4}, {2,5}, {3,4,5}, {2,6}, {3,4,6}, {3,5,6}. It is easy to see that these subsets follow the Fibonacci recursion F(n) = F(n-1) + F(n-2) since we get F(n) such subsets by keeping all F(n-1) subsets from the previous stage (in the example, the F(5)=5 subsets that don't include 6), and by adding one to all elements and appending an additional element n to each subset in F(n-2) subsets (in the example, by applying this to the F(4)=3 subsets {1}, {2,3}, {2,4} we obtain {2,6}, {3,4,6}, {3,5,6}). - Enrique Navarrete, Sep 28 2020 Named "série de Fibonacci" by Lucas (1877) after the Italian mathematician Fibonacci (Leonardo Bonacci, c. 1170 - c. 1240/50). In 1876 he named the sequence "série de Lamé" after the French mathematician Gabriel Lamé (1795 - 1870). - Amiram Eldar, Apr 16 2021 F(n) is the number of edge coverings of the path with n edges. - M. Farrokhi D. G., Sep 30 2021 LCM(F(m), F(n)) is a Fibonacci number if and only if either F(m) divides F(n) or F(n) divides F(m). - M. Farrokhi D. G., Sep 30 2021 Every nonunit positive rational number has at most one representation as the quotient of two Fibonacci numbers. - M. Farrokhi D. G., Sep 30 2021 The infinite sum F(n)/10^(n-1) for all natural numbers n is equal to 100/89. More generally, the sum of F(n)/(k^(n-1)) for all natural numbers n is equal to k^2/(k^2-k-1). Jonatan Djurachkovitch, Dec 31 2023 For n >= 1, number of compositions (c(1),c(2),...,c(k)) of n where c(1), c(3), c(5), ... are 1. To obtain such compositions K(n) of length n increase all parts c(2) by one in all of K(n-1) and prepend two parts 1 in all of K(n-2). - Joerg Arndt, Jan 05 2024 Cohn (1964) proved that a(12) = 12^2 is the only square in the sequence greater than a(1) = 1. - M. F. Hasler, Dec 18 2024 Product_{i=n-2..n+2} F(i) = F(n)^5 - F(n). For example, (F(4)F(5)F(6)F(7)F(8))=(3 * 5 * 8 * 13 * 21) = 8^5 - 8. - Jules Beauchamp, Apr 28 2025 If n >= 3, then for any choice of n not necessarily distinct real numbers in the interval [1..F(n)-1], there are three that are lengths of the sides of a triangle. This is not true for [1..F(n)], as the choice of F(1),F(2),..,F(n) demonstrates. - Veselin Jungic, Oct 27 2025 If n*(great) denotes n repetitions of "great", then F(n+4) is the number of n*(great)-grandparents a honeybee queen or worker has, and F(n+3) is the number of n*(great)-grandparents a honeybee drone has. This is because males hatch from unfertilized eggs and females hatch from fertilized ones. - Johann Peters, Apr 15 2026 2*F(n)*F(n+1), F(n-1)*F(n+2), and F(2*n+1) form a Pythagorean triple (see Gelca and Andreescu). - Stefano Spezia, Aug 27 2026
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LINKS
Paula M. M. C. Catarino, and Anabela Borges, On Leonardo numbers, Acta Mathematica Universitatis Comenianae (2019), 1-12. J. H. E. Cohn, On Square Fibonacci Numbers, Journal of the London Mathematical Society, Volume s1-39, Issue 1, 1964, Pages 537-540, doi:10.1112/jlms/s1-39.1.537. G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255. MR1990179C. J. Glasby, S. P. Glasby, and F. Pleijel, Worms by number, Proc. Roy. Soc. B, Proc. Biol. Sci. 275 (1647) (2008) 2071-2076. Brady Haran and Colm Mulcahy, Little Fibs, Numberphile video (2016). Eric Lehman, F. Thomson Leighton, and Albert R. Meyer, Mathematics for Computer Science, Massachusetts Institute of Technology, 2015. See pp. 147, 156, 158, 181, 635-638, 874-877. B C Manjunath, Fibonacci Tala, recorded musical performance and synchronized player (2018). Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)] Eric Weisstein's World of Mathematics, Clique. Eric Weisstein's World of Mathematics, Matching. Eric Weisstein's World of Mathematics, Path Graph.
FORMULA
G.f.: x / (1 - x - x^2).
G.f.: Sum_{n>=0} x^n * Product_{k=1..n} (k + x)/(1 + k*x). - Paul D. Hanna, Oct 26 2013 F(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)).
Alternatively, F(n) = ((1/2+sqrt(5)/2)^n - (1/2-sqrt(5)/2)^n)/sqrt(5).
F(n) = F(n-1) + F(n-2) = -(-1)^n F(-n).
F(n) = round(phi^n/sqrt(5)).
F(n+1) = Sum_{j=0..floor(n/2)} binomial(n-j, j).
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Jan 03 2017 E.g.f.: (2/sqrt(5))*exp(x/2)*sinh(sqrt(5)*x/2). - Len Smiley, Nov 30 2001 [0 1; 1 1]^n [0 1] = [F(n); F(n+1)]
x | F(n) ==> x | F(kn).
A sufficient condition for F(m) to be divisible by a prime p is (p - 1) divides m, if p == 1 or 4 (mod 5); (p + 1) divides m, if p == 2 or 3 (mod 5); or 5 divides m, if p = 5. (This is essentially Theorem 180 in Hardy and Wright.) - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 29 2001
a(n)=F(n) has the property: F(n)*F(m) + F(n+1)*F(m+1) = F(n+m+1). - Miklos Kristof, Nov 13 2003 Conjecture 1: for n >= 2, sqrt(F(2n+1) + F(2n+2) + F(2n+3) + F(2n+4) + 2*(-1)^n) = (F(2n+1) + 2*(-1)^n)/F(n-1). [For a proof see Comments section.]
Conjecture 2: for n >= 0, (F(n+2)*F(n+3)) - (F(n+1)*F(n+4)) + (-1)^n = 0.
Theorem 1: for n >= 0, (F(n+3)^ 2 - F(n+1)^ 2)/F(n+2) = (F(n+3)+ F(n+1)).
Theorem 2: for n >= 0, F(n+10) = 11*F(n+5) + F(n).
Theorem 3: for n >= 6, F(n) = 4*F(n-3) + F(n-6). (End)
Conjecture 2 of Rashid is actually a special case of the general law F(n)*F(m) + F(n+1)*F(m+1) = F(n+m+1) (take n <- n+1 and m <- -(n+4) in this law). - Harmel Nestra (harmel.nestra(AT)ut.ee), Apr 22 2005
Conjecture 2 of Rashid Kurmang simplified: F(n)*F(n+3) = F(n+1)*F(n+2)-(-1)^n. Follows from d'Ocagne's identity: m=n+2. - Alex Ratushnyak, May 06 2012 Conjecture: for all c such that 2-phi <= c < 2*(2-phi) we have F(n) = floor(phi*a(n-1)+c) for n > 2. - Gerald McGarvey, Jul 21 2004 For x > phi, Sum_{n>=0} F(n)/x^n = x/(x^2 - x - 1). - Gerald McGarvey, Oct 27 2004 F(n+1) = exponent of the n-th term in the series f(x, 1) determined by the equation f(x, y) = xy + f(xy, x). - Jonathan Sondow, Dec 19 2004 a(n-1) = Sum_{k=0..n} (-1)^k*binomial(n-ceiling(k/2), floor(k/2)). - Benoit Cloitre, May 05 2005 F(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)(1+(-1)^(n-k))/2. - Paul Barry, Aug 28 2005 Fibonacci(n) = Product_{j=1..ceiling(n/2)-1} (1 + 4(cos(j*Pi/n))^2). [Bicknell and Hoggatt, pp. 47-48.] - Emeric Deutsch, Oct 15 2006 F(n) = 2^(-(n-1))*Sum_{k=0..floor((n-1)/2)} binomial(n,2*k+1)*5^k. - Hieronymus Fischer, Feb 07 2006 F(n*m) = Sum_{k = 0..m} binomial(m,k)*F(n-1)^k*F(n)^(m-k)*F(m-k). The generating function of F(n*m) (n fixed, m = 0,1,2,...) is G(x) = F(n)*x / ((1 - F(n-1)*x)^2 - F(n)*x*(1 - F(n-1)*x) - (F(n)*x)^2). E.g., F(15) = 610 = F(5*3) = binomial(3,0)* F(4)^0*F(5)^3*F(3) + binomial(3,1)* F(4)^1*F(5)^2*F(2) + binomial(3,2)* F(4)^2*F(5)^1*F(1) + binomial(3,3)* F(4)^3*F(5)^0*F(0) = 1*1*125*2 + 3*3*25*1 + 3*9*5*1 + 1*27*1*0 = 250 + 225 + 135 + 0 = 610. - Miklos Kristof, Feb 12 2007 Let L(n) = A000032(n) = Lucas numbers. Then: For a >= b and odd b, F(a+b) + F(a-b) = L(a)*F(b).
For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).
For a >= b and odd b, F(a+b) - F(a-b) = F(a)*L(b).
For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).
F(n+m) + (-1)^m*F(n-m) = F(n)*L(m);
F(n+m) - (-1)^m*F(n-m) = L(n)*F(m);
F(n+m+k) + (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = F(n)*L(m)*L(k);
F(n+m+k) - (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = L(n)*L(m)*F(k);
F(n+m+k) + (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = L(n)*F(m)*L(k);
F(n+m+k) - (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = 5*F(n)*F(m)*F(k). (End)
A corollary to Kristof 2007 is 2*F(a+b) = F(a)*L(b) + L(a)*F(b). - Graeme McRae, Apr 24 2014 For n > m, the sum of the 2m consecutive Fibonacci numbers F(n-m-1) thru F(n+m-2) is F(n)*L(m) if m is odd, and L(n)*F(m) if m is even (see the McRae link). - Graeme McRae, Apr 24 2014. F(n) = b(n) + (p-1)*Sum_{k=2..n-1} floor(b(k)/p)*F(n-k+1) where b(k) is the digital sum analog of the Fibonacci recurrence, defined by b(k) = ds_p(b(k-1)) + ds_p(b(k-2)), b(0)=0, b(1)=1, ds_p=digital sum base p. Example for base p=10: F(n) = A010077(n) + 9*Sum_{k=2..n-1} A059995( A010077(k))*F(n-k+1). - Hieronymus Fischer, Jul 01 2007 F(n) = b(n)+p*Sum_{k=2..n-1} floor(b(k)/p)*F(n-k+1) where b(k) is the digital product analog of the Fonacci recurrence, defined by b(k) = dp_p(b(k-1)) + dp_p(b(k-2)), b(0)=0, b(1)=1, dp_p=digital product base p. Example for base p=10: F(n) = A074867(n) + 10*Sum_{k=2..n-1} A059995( A074867(k))*F(n-k+1). - Hieronymus Fischer, Jul 01 2007 a(n) = denominator of continued fraction [1,1,1,...] (with n ones); e.g., 2/3 = continued fraction [1,1,1]; where barover[1] = [1,1,1,...] = 0.6180339.... - Gary W. Adamson, Nov 29 2007 F(n + 3) = 2F(n + 2) - F(n), F(n + 4) = 3F(n + 2) - F(n), F(n + 8) = 7F(n + 4) - F(n), F(n + 12) = 18F(n + 6) - F(n). - Paul Curtz, Feb 01 2008 a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}... Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n), where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any l_i + l_(i+1) >= 2 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009 a(n+1) = 2^n sqrt(Product_{k=1..n} cos(k Pi/(n+1))^2+1/4) (Kasteleyn's formula specialized). - Sarah-Marie Belcastro, Jul 04 2009 If p[i] = modp(i,2) and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n >= 1, a(n)=det A. - Milan Janjic, May 02 2010 Limit_{k->oo} F(k+n)/F(k) = (L(n) + F(n)*sqrt(5))/2 with the Lucas numbers L(n) = A000032(n). - Johannes W. Meijer, May 27 2010 For n >= 1, F(n) = round(log_2(2^(phi*F(n-1)) + 2^(phi*F(n-2)))), where phi is the golden ratio. - Vladimir Shevelev, Jun 24 2010, Jun 27 2010 For n >= 1, a(n+1) = ceiling(phi*a(n)), if n is even and a(n+1) = floor(phi*a(n)), if n is odd (phi = golden ratio). - Vladimir Shevelev, Jul 01 2010 a(n) = 2*a(n-2) + a(n-3), n > 2. - Gary Detlefs, Sep 08 2010 a(n)^2 - a(n-1)^2 = a(n+1)*a(n-2), see A121646. a(n) = sqrt((-1)^k*(a(n+k)^2 - a(k)*a(2n+k))), for any k. - Gary Detlefs, Dec 03 2010 (-1)^(n+1) = F(n)^2 + F(n)*F(1+n) - F(1+n)^2.
F(n) = F(n+2) - 1 + (F(n+1))^4 + 2*(F(n+1)^3*F(n+2)) - (F(n+1)*F(n+2))^2 - 2*F(n+1)(F(n+2))^3 + (F(n+2))^4 - F(n+1). (End)
F(n) = 4*F(n-2) - 2*F(n-3) - F(n-6). - Gary Detlefs, Apr 01 2012 G.f.: A(x) = x/(1-x-x^2) = G(0)/sqrt(5) where G(k) = 1 - ((-1)^k)*2^k/(a^k - b*x*a^k*2^k/(b*x*2^k - 2*((-1)^k)*c^k/G(k+1))) and a=3+sqrt(5), b=1+sqrt(5), c=3-sqrt(5); (continued fraction, 3rd kind, 3-step).
Let E(x) be the e.g.f., i.e.,
E(x) = 1*x + (1/2)*x^2 + (1/3)*x^3 + (1/8)*x^4 + (1/24)*x^5 + (1/90)*x^6 + (13/5040)*x^7 + ...; then
E(x) = G(0)/sqrt(5); G(k) = 1 - ((-1)^k)*2^k/(a^k - b*x*a^k*2^k/(b*x*2^k - 2*((-1)^k)*(k+1)*c^k/G(k+1))), where a=3+sqrt(5), b=1+sqrt(5), c=3-sqrt(5); (continued fraction, 3rd kind, 3-step).
(End)
F(n) = 1 + Sum_{j_1=1..n-2} 1 + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} 1 + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} Sum_{j_3=1..j_2-2} 1 + ... + Sum_{j_1=1..n-2} Sum_{j_2=1..j_1-2} Sum_{j_3=1..j_2-2} ... Sum_{j_k=1..j_(k-1)-2} 1, where k = floor((n-1)/2).
Example: F(6) = 1 + Sum_{j=1..4} 1 + Sum_{j=1..4} Sum_{k=1..(j-2)} 1 + 0 = 1 + (1 + 1 + 1 + 1) + (1 + (1 + 1)) = 8.
F(n) = Sum_{j=0..k} S(j+1,n-2j), where k = floor((n-1)/2) and the S(j,n) are the n-th j-simplex sums: S(1,n) = 1 is the 1-simplex sum, S(2,n) = Sum_{k=1..n} S(1,k) = 1+1+...+1 = n is the 2-simplex sum, S(3,n) = Sum_{k=1..n} S(2,k) = 1+2+3+...+n is the 3-simplex sum (= triangular numbers = A000217), S(4,n) = Sum_{k=1..n} S(3,k) = 1+3+6+...+n(n+1)/2 is the 4-simplex sum (= tetrahedral numbers = A000292) and so on. Since S(j,n) = binomial(n-2+j,j-1), the formula above equals the well-known binomial formula, essentially. (End)
G.f.: A(x) = x / (1 - x / (1 - x / (1 + x))). - Michael Somos, Jan 04 2013 Sum_{n >= 1} (-1)^(n-1)/(a(n)*a(n+1)) = 1/phi (phi=golden ratio). - Vladimir Shevelev, Feb 22 2013 For odd k, Sum_{n >= 1} a(k)^2*(-1)^(n-1)/(a(k*n)*a(k*n+k)) = phi^(-k).
For even k, Sum_{n >= 1} a(k)^2/(a(k*n)*a(k*n+k)) = phi^(-k). (End)
(1) Expression a(n+1) via a(n): a(n+1) = (a(n) + sqrt(5*(a(n))^2 + 4*(-1)^n))/2;
(2) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(3) a(n)/a(n+1) = 1/phi + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
F(n+1) = F(n)/2 + sqrt((-1)^n + 5*F(n)^2/4), n >= 0. F(n+1) = U_n(i/2)/i^n, (U:= Chebyshev polynomial of the 2nd kind, i=sqrt(-1)). - Bill Gosper, Mar 04 2013 G.f.: -Q(0) where Q(k) = 1 - (1+x)/(1 - x/(x - 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Mar 06 2013 G.f.: x - 1 - 1/x + (1/x)/Q(0), where Q(k) = 1 - (k+1)*x/(1 - x/(x - (k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 23 2013 G.f.: x*G(0), where G(k) = 1 + x*(1+x)/(1 - x*(1+x)/(x*(1+x) + 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 08 2013 G.f.: x^2 - 1 + 2*x^2/(W(0)-2), where W(k) = 1 + 1/(1 - x*(k + x)/( x*(k+1 + x) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013 G.f.: Q(0) - 1, where Q(k) = 1 + x^2 + (k+2)*x - x*(k+1 + x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013 Let b(n) = b(n-1) + b(n-2), with b(0) = 0, b(1) = phi. Then, for n >= 2, F(n) = floor(b(n-1)) if n is even, F(n) = ceiling(b(n-1)), if n is odd, with convergence. - Richard R. Forberg, Jan 19 2014 a(n) = Sum_{t1*g(1)+t2*g(2)+...+tn*g(n)=n} multinomial(t1+t2+...+tn,t1,t2,...,tn), where g(k)=2*k-1. - Mircea Merca, Feb 27 2014 F(n) = round(sqrt(F(n-1)^2 + F(n)^2 + F(n+1)^2)/2), for n > 0. This rule appears to apply to any sequence of the form a(n) = a(n-1) + a(n-2), for any two values of a(0) and a(1), if n is sufficiently large. - Richard R. Forberg, Jul 27 2014 F(n) = round(2/(1/F(n) + 1/F(n+1) + 1/F(n+2))), for n > 0. This rule also appears to apply to any sequence of the form a(n) = a(n-1) + a(n-2), for any two values of a(0) and a(1), if n is sufficiently large. - Richard R. Forberg, Aug 03 2014 a(n) = hypergeometric([-n/2+1/2, -n/2+1], [-n+1], -4) for n >= 2. - Peter Luschny, Sep 19 2014 Limit_{n -> oo} (log F(n+1)/log F(n))^n = e. - Thomas Ordowski, Oct 06 2014 F(n) = (L(n+1)^2 - L(n-1)^2)/(5*L(n)), where L(n) is A000032(n), with a similar inverse relationship. - Richard R. Forberg, Nov 17 2014 Consider the graph G[1-vertex;1-loop,2-loop] in comment above. Construct the power matrix array T(n,j) = [A^*j]*[S^*(j-1)] where A=(1,1,0,...) and S=(0,1,0,...)( A063524). [* is convolution operation] Define S^*0=I with I=(1,0,...). Then T(n,j) counts n-walks containing (j) loops and a(n-1) = Sum_{j=1..n} T(n,j). - David Neil McGrath, Nov 21 2014 Define F(-n) to be F(n) for n odd and -F(n) for n even. Then for all n and k, F(n) = F(k)*F(n-k+3) - F(k-1)*F(n-k+2) - F(k-2)*F(n-k) + (-1)^k*F(n-2k+2). - Charlie Marion, Dec 04 2014 F(2*n) = F(n+1)^2 - F(n-1)^2, similar to Koshy (D) and Forberg 2011, but different. - Hermann Stamm-Wilbrandt, Aug 12 2015 F(n+1) = ceiling( (1/phi)*Sum_{k=0..n} F(k) ). - Tom Edgar, Sep 10 2015 F(n) = (F(2n+k+1) - F(n+1)*F(n+k+1))/F(n+k), k >= 0.
Thus when k=0: F(n) = sqrt(F(2n+1) - F(n+1)^2).
F(n) = (F(3n) - F(n+1)^3 + F(n-1)^3)^(1/3).
F(n+2k) = binomial transform of any subsequence starting with F(n). Example F(6)=8: 1*8 = F(6)=8; 1*8 + 1*13 = F(8)=21; 1*8 + 2*13 + 1*21 = F(10)=55; 1*8 + 3*13 + 3*21 + 1*34 = F(12)=144, etc. This formula applies to Fibonacci-type sequences with any two seed values for a(0) and a(1) (e.g., Lucas sequence A000032: a(0)=2, a(1)=1). (End)
F(n) = L(k)*F(n-k) + (-1)^(k+1)*F(n-2k) for all k >= 0, where L(k) = A000032(k). - Anton Zakharov, Aug 02 2016 a(n) = F_n(1), where F_n(x) are the Fibonacci polynomials.
Inverse binomial transform of A001906. Number of zeros in substitution system {0 -> 11, 1 -> 1010} at step n from initial string "1" (1 -> 1010 -> 101011101011 -> ...) multiplied by 1/ A000079(n). (End) For n >= 2, a(n) = 2^(n^2+n) - (4^n-2^n-1)*floor(2^(n^2+n)/(4^n-2^n-1)) - 2^n*floor(2^(n^2) - (2^n-1-1/2^n)*floor(2^(n^2+n)/(4^n-2^n-1))). - Benoit Cloitre, Apr 17 2017 f(n+1) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2j,k)*binomial(j,k). - Tony Foster III, Sep 04 2017 F(n) = Sum_{k=0..floor((n-1)/2)} ( (n-k-1)! / ((n-2k-1)! * k!) ). - Zhandos Mambetaliyev, Nov 08 2017 For x even, F(n) = (F(n+x) + F(n-x))/L(x). For x odd, F(n) = (F(n+x) - F(n-x))/L(x) where n >= x in both cases. Therefore F(n) = F(2*n)/L(n) for n >= 0. - David James Sycamore, May 04 2018 Let [a/p] denote the Legendre symbol. Then, for an odd prime p:
F(p+n) == [5/p]*F([5/p]+n) (mod p), if [5/p] = 1 or -1.
F(p+n) == 3*F(n) (mod p), if [5/p] = 0 (i.e., p = 5).
This is true for negative-indexed terms as well, if this sequence is extended by the negafibonacci numbers (i.e., F(-n) = A039834(n)). (End) a(n) = (L(n-3) + L(n-2) + L(n-1) + L(n))/5 with L(n)= A000032(n). - Art Baker, Jan 04 2019 F(n) = F(k-1)*F(abs(n-k-2)) + F(k-1)*F(n-k-1) + F(k)*F(abs(n-k-2)) + 2*F(k)*F(n-k-1), for n > k > 0. - Joseph M. Shunia, Aug 12 2019 F(n) = F(n-k+2)*F(k-1) + F(n-k+1)*F(k-2) for all k such that 2 <= k <= n. - Michael Tulskikh, Oct 09 2019 F(n)^2 - F(n+k)*F(n-k) = (-1)^(n+k) * F(k)^2 for 2 <= k <= n [Catalan's identity]. - Hermann Stamm-Wilbrandt, May 07 2021 a(n) = Product_{d|n} b(d) = Product_{k=1..n} b(gcd(n,k))^(1/phi(n/gcd(n,k))) = Product_{k=1..n} b(n/gcd(n,k))^(1/phi(n/gcd(n,k))) where b(n) = A061446(n) = primitive part of a(n), phi(n) = A000010(n). - Richard L. Ollerton, Nov 08 2021 a(n) = 2*i^(1-n)*sin(n*arccos(i/2))/sqrt(5), i=sqrt(-1). - Bill Gosper, May 05 2022 a(n) = i^(n-1)*sin(n*c)/sin(c) = i^(n-1)*sin(c*n)*csc(c), where c = Pi/2 + i*arccsch(2). - Peter Luschny, May 23 2022 F(2n) = Sum_{k=1..n} (k/5)*binomial(2n, n+k), where (k/5) is the Legendre or Jacobi Symbol; F(2n+1)= Sum_{k=1..n} (-(k+2)/5)*binomial(2n+1, n+k), where (-(k+2)/5) is the Legendre or Jacobi Symbol. For example, F(10) = 1*binomial(10,6) - 1*binomial(10,7) - 1*binomial(10,8) + 1*binomial(10,9) + 0*binomial(10,10), F(11) = 1*binomial(11,6) - 1*binomial(11,7) + 0*binomial(11,8) - 1*binomial(11,9) + 1*binomial(11,10) + 1*binomial(11,11). - Yike Li, Aug 21 2022 For n > 0, 1/F(n) = Sum_{k>=1} F(n*k)/(F(n+2)^(k+1)). - Diego Rattaggi, Oct 26 2022 For n == 0 (mod 4): F(n) = F((n+2)/2)*( F(n/2) + F((n/2)-2) ) + 1;
For n == 1 (mod 4): F(n) = F((n-1)/2)*( F((n-1)/2) + F(2+(n-1)/2) ) + 1;
For n == 2 (mod 4): F(n) = F((n-2)/2)*( F(n/2) + F((n/2)+2) ) + 1;
For n == 3 (mod 4): F(n) = F((n-1)/2)*( F((n-1)/2) + F(2+(n-1)/2) ) - 1. (End)
F(n) = Sum_{i=0..n-1} L(n-i)*(-2)^i. - Greg Dresden and Xiaoya Gao, Oct 11 2025 For n>=2, a(n) = 2^(n-2)*hypergeometric([(1-n)/3,(2-n)/3,1-n/3,2-n],[(2-n)/2,(3-n)/2,1-n],27/32). - Victor Petrescu, Oct 25 2025 Sum_{n>=1} 1/a(n)^s = 5^(s/2) * Sum_{k>=0} C(-s, k) * 1/(phi^(s+2k) - (-1)^k), see Eq. 5 in Navas.
Sum_{n>=1} (-1)^(n+1)/a(n)^s = 5^(s/2) * Sum_{k>=0} C(-s, k) * 1/(phi^(s+2k) + (-1)^k). (End)
EXAMPLE
For x = 0,1,2,3,4, x=1/(x+1) = 1, 1/2, 2/3, 3/5, 5/8. These fractions have numerators 1,1,2,3,5, which are the 2nd to 6th terms of the sequence. - Cino Hilliard, Sep 15 2008 There are a(7)=13 compositions of 7 where there is a drop between every second pair of parts, starting with the first and second part:
01: [ 2 1 2 1 1 ]
02: [ 2 1 3 1 ]
03: [ 2 1 4 ]
04: [ 3 1 2 1 ]
05: [ 3 1 3 ]
06: [ 3 2 2 ]
07: [ 4 1 2 ]
08: [ 4 2 1 ]
09: [ 4 3 ]
10: [ 5 1 1 ]
11: [ 5 2 ]
12: [ 6 1 ]
13: [ 7 ]
There are abs(a(6+1))=13 compositions of 6 where there is no rise between every second pair of parts, starting with the second and third part:
01: [ 1 2 1 2 ]
02: [ 1 3 1 1 ]
03: [ 1 3 2 ]
04: [ 1 4 1 ]
05: [ 1 5 ]
06: [ 2 2 1 1 ]
07: [ 2 3 1 ]
08: [ 2 4 ]
09: [ 3 2 1 ]
10: [ 3 3 ]
11: [ 4 2 ]
12: [ 5 1 ]
13: [ 6 ]
(End)
Partially ordered partitions of (n-1) into parts 1,2,3 where only the order of the adjacent 1's and 2's are unimportant. E.g., a(8)=21. These are (331),(313),(133),(322),(232),(223),(3211),(2311),(1321),(2131),(1132),(2113),(31111 ),(13111),(11311),(11131),(11113),(2221),(22111),(211111),(1111111). - David Neil McGrath, Jul 25 2015Consider the partitions of 7 with summands initially listed in nonincreasing order. Keep the 1's frozen in position (indicated by "[]") and then allow the other summands to otherwise vary their order: 7; 6,[1]; 5,2; 2,5; 4,3; 3,4; 5,[1,1], 4,2,[1]; 2,4,[1]; 3,3,[1]; 3,3,2; 3,2,3; 2,3,3; 4,[1,1,1]; 3,2,[1,1]; 2,3,[1,1]; 2,2,2,[1]; 3,[1,1,1,1]; 2,2,[1,1,1]; 2,[1,1,1,1,1]; [1,1,1,1,1,1,1]. There are 21 = a(7+1) arrangements in all. - Gregory L. Simay, Jun 14 2016
MAPLE
A000045 := proc(n) combinat[fibonacci](n); end;
# Alternative:
ZL:=[S, {a = Atom, b = Atom, S = Prod(X, Sequence(Prod(X, b))), X = Sequence(b, card >= 1)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=0..38); # Zerinvary Lajos, Apr 04 2008 # Alternative:
spec := [B, {B=Sequence(Set(Z, card>1))}, unlabeled ]: seq(combstruct[count](spec, size=n), n=1..39); # Zerinvary Lajos, Apr 04 2008 # Alternative: The following Maple command isFib(n) yields true or false depending on whether n is a Fibonacci number or not.
isFib:= proc(n) (t-> issqr(t+4) or issqr(t-4))(5*n^2) end proc:
MATHEMATICA
(* Alternative: *)
Table[2^n Sqrt @ Product[(Cos[Pi k/(n + 1)]^2 + 1/4), {k, n}] // FullSimplify, {n, 15}]; (* Kasteleyn's formula specialized, Sarah-Marie Belcastro, Jul 04 2009 *) (* Alternative: *)
LinearRecurrence[{1, 1}, {0, 1}, 40] (* Harvey P. Dale, Aug 03 2014 *) (* Alternative: *)
(* Alternative: *)
CoefficientList[Series[-(x/(-1 + x + x^2)), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 22 2017 *)
PROG
(Axiom) [fibonacci(n) for n in 0..50]
(Magma) [Fibonacci(n): n in [0..38]];
(Maxima) makelist(fib(n), n, 0, 100); /* Martin Ettl, Oct 21 2012 */ (PARI) a(n) = fibonacci(n)
(PARI) a(n) = imag(quadgen(5)^n)
from itertools import islice
def fib_gen():
x, y = 0, 1
while True:
yield x
x, y = y, x + y
fib_list = lambda n: list(islice(fib_gen(), n))
(Python)
A000045 = lambda n: (1<<((n+1)*(n+1)))//((4<<(2*n))-(2<<n)-1)&((2<<n)-1) # Adapted from Sage code below by M. F. Hasler, Feb 20 2025. Corrected by Tyler Busby, Nov 24 2025.
(Python) [(i:=-1)+(j:=1)] + [(j:=i+j)+(i:=j-i) for _ in range(100)] # Jwalin Bhatt, Apr 03 2025 (SageMath) # Demonstration program from Jaap Spies: a = sloane. A000045; # choose sequence print(a) # This returns the name of the sequence.
print(a(38)) # This returns the 38th term of the sequence.
print(a.list(39)) # This returns a list of the first 39 terms.
(SageMath)
a = BinaryRecurrenceSequence(1, 1); print([a(n) for n in range(20)])
# Closed form integer formula with F(1) = 0 from Paul Hankin (see link).
F = lambda n: (4<<(n-1)*(n+2))//((4<<2*(n-1))-(2<<(n-1))-1)&((2<<(n-1))-1)
(SageMath) print(list(fibonacci_sequence(0, 40))) # Bruno Berselli, Jun 26 2014 (Haskell)
-- Based on code from http://www.haskell.org/haskellwiki/The_Fibonacci_sequence
-- which also has other versions.
fib :: Int -> Integer
fib n = fibs !! n
where
fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
(Julia)
function fib(n)
F = BigInt[1 1; 1 0]
Fn = F^n
Fn[2, 1]
end
(Julia) # faster
function fibrec(n::Int)
n == 0 && return (BigInt(0), BigInt(1))
a, b = fibrec(div(n, 2))
c = a * (b * 2 - a)
d = a * a + b * b
iseven(n) ? (c, d) : (d, c + d)
end
fibonacci(n::Int) = fibrec(n)[1]
println([fibonacci(n) for n in 0:40]) # Peter Luschny, Apr 03 2022 (GAP)
Fib:=[0, 1];; for n in [3..10^3] do Fib[n]:=Fib[n-1]+Fib[n-2]; od; Fib; # Muniru A Asiru, Sep 03 2017 (Scala) def fibonacci(n: BigInt): BigInt = {
val zero = BigInt(0)
def fibTail(n: BigInt, a: BigInt, b: BigInt): BigInt = n match {
case `zero` => a
case _ => fibTail(n - 1, b, a + b)
}
fibTail(n, 0, 1)
} // Based on "Case 3: Tail Recursion" from Carrasquel (2016) link
CROSSREFS
Cf. A001622 (phi), A039834 (signed Fibonacci numbers), A001519 (F(2n-1)), A001906 (F(2n)), A001690 (complement), A000213, A000288, A000322, A000383, A060455, A030186, A020695, A020701, A060280 (inv. Eul. Trans), A071679, A099731, A100492, A094216, A094638, A000108, A101399, A101400, A001611, A000071, A157725, A001911, A157726, A006327, A157727, A157728, A157729, A167616, A059929, A144152, A152063, A114690, A003893, A000032, A060441, A000930, A003269, A000957, A057078, A007317, A091867, A104597, A249548, A262342, A001060, A022095, A072649, A163733, A073133, A166861 (Euler Transform), A337009 (Multiset Transf.). Cf. also A001175 (Pisano periods), A001177 (Entry points), A001176 (number of zeros in a fundamental period). Fibonacci-Cayley triangle: A327992. Cf. A061446 (primitive part of Fibonacci numbers), A000010 (comments on product formulas).
KEYWORD
nonn, core, nice, easy, hear, changed
The cubes: a(n) = n^3.
(Formerly M4499 N1905)
+10
1121
0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507
COMMENTS
a(n) is the sum of the next n odd numbers; i.e., group the odd numbers so that the n-th group contains n elements like this: (1), (3, 5), (7, 9, 11), (13, 15, 17, 19), (21, 23, 25, 27, 29), ...; then each group sum = n^3 = a(n). Also the median of each group = n^2 = mean. As the sum of first n odd numbers is n^2 this gives another proof of the fact that the n-th partial sum = (n(n + 1)/2)^2. - Amarnath Murthy, Sep 14 2002 Total number of triangles resulting from criss-crossing cevians within a triangle so that two of its sides are each n-partitioned. - Lekraj Beedassy, Jun 02 2004. See Propp and Propp-Gubin for a proof. Also structured triakis tetrahedral numbers (vertex structure 7) (cf. A100175 = alternate vertex); structured tetragonal prism numbers (vertex structure 7) (cf. A100177 = structured prisms); structured hexagonal diamond numbers (vertex structure 7) (cf. A100178 = alternate vertex; A000447 = structured diamonds); and structured trigonal anti-diamond numbers (vertex structure 7) (cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004 Schlaefli symbol for this polyhedron: {4, 3}.
Least multiple of n such that every partial sum is a square. - Amarnath Murthy, Sep 09 2005 Draw a regular hexagon. Construct points on each side of the hexagon such that these points divide each side into equally sized segments (i.e., a midpoint on each side or two points on each side placed to divide each side into three equally sized segments or so on), do the same construction for every side of the hexagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least one side of the polygon. The result is a triangular tiling of the hexagon and the creation of a number of smaller regular hexagons. The equation gives the total number of regular hexagons found where n = the number of points drawn + 1. For example, if 1 point is drawn on each side then n = 1 + 1 = 2 and a(n) = 2^3 = 8 so there are 8 regular hexagons in total. If 2 points are drawn on each side then n = 2 + 1 = 3 and a(n) = 3^3 = 27 so there are 27 regular hexagons in total. - Noah Priluck (npriluck(AT)gmail.com), May 02 2007
The solutions of the Diophantine equation: (X/Y)^2 - X*Y = 0 are of the form: (n^3, n) with n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^2k - XY = 0 are of the form: (m*n^(2k + 1), m*n^(2k - 1)) with m >= 1, k >= 1 and n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^(2k + 1) - XY = 0 are of the form: (m*n^(k + 1), m*n^k) with m >= 1, k >= 1 and n >= 1. - Mohamed Bouhamida, Oct 04 2007 Except for the first two terms, the sequence corresponds to the Wiener indices of C_{2n} i.e., the cycle on 2n vertices (n > 1). - K.V.Iyer, Mar 16 2009 Totally multiplicative sequence with a(p) = p^3 for prime p. - Jaroslav Krizek, Nov 01 2009 One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010 Numbers n for which order of torsion subgroup t of the elliptic curve y^2 = x^3 - n is t = 2. - Artur Jasinski, Jun 30 2010 The sequence with the lengths of the Pisano periods mod k is 1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, ... for k >= 1, apparently multiplicative and derived from A000027 by dividing every ninth term through 3. Cubic variant of A186646. - R. J. Mathar, Mar 10 2011 The number of atoms in a bcc (body-centered cubic) rhombic hexahedron with n atoms along one edge is n^3 (T. P. Martin, Shells of atoms, eq. (8)). - Brigitte Stepanov, Jul 02 2011 Twice the area of a triangle with vertices at (0, 0), (t(n - 1), t(n)), and (t(n), t(n - 1)), where t = A000217 are triangular numbers. - J. M. Bergot, Jun 25 2013 For n > 2, a(n) = twice the area of a triangle with vertices at points (binomial(n,3),binomial(n+2,3)), (binomial(n+1,3),binomial(n+1,3)), and (binomial(n+2,3),binomial(n,3)). - J. M. Bergot, Jun 14 2014 Determinants of the spiral knots S(4,k,(1,1,-1)). a(k) = det(S(4,k,(1,1,-1))). - Ryan Stees, Dec 14 2014 One of the oldest-known examples of this sequence is shown in the Senkereh tablet, BM 92698, which displays the first 32 terms in cuneiform. - Charles R Greathouse IV, Jan 21 2015 We construct a number triangle from the integers 1, 2, 3, ... 2*n-1 as follows. The first column contains all the integers 1, 2, 3, ... 2*n-1. Each succeeding column is the same as the previous column but without the first and last items. The last column contains only n. The sum of all the numbers in the triangle is n^3.
Here is the example for n = 4, where 1 + 2*2 + 3*3 + 4*4 + 3*5 + 2*6 + 7 = 64 = a(4):
1
2 2
3 3 3
4 4 4 4
5 5 5
6 6
7
(End)
For n > 0, a(n) is the number of compositions of n+11 into n parts avoiding parts 2 and 3. - Milan Janjic, Jan 07 2016 Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017 Number of inequivalent face colorings of the cube using at most n colors such that each color appears at least twice. - David Nacin, Feb 22 2017 Consider A = {a,b,c} a set with three distinct members. The number of subsets of A is 8, including {a,b,c} and the empty set. The number of subsets from each of those 8 subsets is 27. If the number of such iterations is n, then the total number of subsets is a(n-1). - Gregory L. Simay, Jul 27 2018 By Fermat's Last Theorem, these are the integers of the form x^k with the least possible value of k such that x^k = y^k + z^k never has a solution in positive integers x, y, z for that k. - Felix Fröhlich, Jul 27 2018
REFERENCES
Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 191.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 43, 64, 81.
Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), pages 99, 139-140, 306-307.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 292.
T. Aaron Gulliver, "Sequences from cubes of integers", International Mathematical Journal, 4 (2003), no. 5, 439 - 445. See http://www.m-hikari.com/z2003.html for information about this journal. [I expanded the reference to make this easier to find. - N. J. A. Sloane, Feb 18 2019] J. Propp and A. Propp-Gubin, "Counting Triangles in Triangles", Pi Mu Epsilon Journal (to appear).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Murray R. Spiegel, Calculus of Finite Differences and Difference Equations, "Schaum's Outline Series", McGraw-Hill, 1971, p. 33.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 6-7.
D. Wells, You Are A Mathematician, pp. 238-241, Penguin Books 1995.
LINKS
N. Brothers, S. Evans, L. Taalman, L. Van Wyk, D. Witczak, and C. Yarnall, Spiral knots, Missouri J. of Math. Sci., 22 (2010). M. DeLong, M. Russell, and J. Schrock, Colorability and determinants of T(m,n,r,s) twisted torus knots for n equiv. +/-1(mod m), Involve, Vol. 8 (2015), No. 3, 361-384. Charles Hutton, POWER. A mathematical and philosophical dictionary Vol. 2 (1795), 270-272. T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (8). Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)]
FORMULA
G.f.: x*(1+4*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation a(n) = lcm(n, (n - 1)^2) - (n - 1)^2. E.g.: lcm(1, (1 - 1)^2) - (1 - 1)^2 = 0, lcm(2, (2 - 1)^2) - (2 - 1)^2 = 1, lcm(3, (3 - 1)^2) - (3 - 1)^2 = 8, ... - Mats Granvik, Sep 24 2007 Starting (1, 8, 27, 64, 125, ...), = binomial transform of [1, 7, 12, 6, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007 a(n) = binomial(n+2,3) + 4*binomial(n+1,3) + binomial(n,3). [Worpitzky's identity for cubes. See. e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019] a(n) = n + 6*binomial(n+1,3) = binomial(n,1)+6*binomial(n+1,3). - Ron Knott, Jun 10 2019 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6. - Ant King Apr 29 2013 a(k) = det(S(4,k,(1,1,-1))) = k*b(k)^2, where b(1)=1, b(2)=2, b(k) = 2*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Jon Tavasanis, Feb 21 2016 a(n) = n + Sum_{j=0..n-1} Sum_{k=1..2} binomial(3,k)*j^(3-k). - Patrick J. McNab, Mar 28 2016 a(n) = n*binomial(n+1, 2) + 2*binomial(n+1, 3) + binomial(n,3). - Tony Foster III, Nov 14 2017 Sum_{n>=1} 1/a(n) = zeta(3) ( A002117). Sum_{n>=1} (-1)^(n+1)/a(n) = 3*zeta(3)/4 ( A197070). (End) Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi.
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)/(3*Pi). (End)
EXAMPLE
For k=3, b(3) = 2 b(2) - b(1) = 4-1 = 3, so det(S(4,3,(1,1,-1))) = 3*3^2 = 27.
For n=3, a(3) = 3 + (3*0^2 + 3*0 + 3*1^2 + 3*1 + 3*2^2 + 3*2) = 27. - Patrick J. McNab, Mar 28 2016
MAPLE
isA000578 := proc(r)
local p;
if r = 0 or r =1 then
true;
else
for p in ifactors(r)[2] do
if op(2, p) mod 3 <> 0 then
return false;
end if;
end do:
true ;
end if;
MATHEMATICA
(* Alternative: *)
CoefficientList[Series[x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jul 05 2014 *) (* Alternative: *)
Accumulate[Join[{0}, Table[3n^2+3n+1, {n, 0, 20}]]] (* Harvey P. Dale, Aug 18 2018 *) (* Alternative: *)
LinearRecurrence[{4, -6, 4, -1}, {0, 1, 8, 27, 64}, 20] (* Harvey P. Dale, Aug 18 2018 *)
PROG
(Haskell)
a000578 = (^ 3)
a000578_list = 0 : 1 : 8 : zipWith (+)
(map (+ 6) a000578_list)
(map (* 3) $ tail $ zipWith (-) (tail a000578_list) a000578_list)
(Magma) I:=[0, 1, 8, 27]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jul 05 2014 (Python)
A000578_list, m = [], [6, -6, 1, 0]
for _ in range(10**2):
for i in range(3):
CROSSREFS
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Mersenne exponents: primes p such that 2^p - 1 is prime. Then 2^p - 1 is called a Mersenne prime.
(Formerly M0672 N0248)
+10
716
2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161, 74207281, 77232917, 82589933
COMMENTS
Equivalently, integers k such that 2^k - 1 is prime.
It is believed (but unproved) that this sequence is infinite. The data suggest that the number of terms up to exponent N is roughly K log N for some constant K.
Length of prime repunits in base 2.
In one of his first publications Euler found the numbers up to 31 but erroneously included 41 and 47.
Equals number of bits in binary expansion of n-th Mersenne prime ( A117293). - Artur Jasinski, Feb 09 2007 Number of divisors of n-th even perfect number, divided by 2. Number of divisors of n-th even perfect number that are powers of 2. Number of divisors of n-th even perfect number that are multiples of n-th Mersenne prime A000668(n). - Omar E. Pol, Feb 24 2008 Number of divisors of n-th even superperfect number A061652(n). Numbers of divisors of n-th superperfect number A019279(n), assuming there are no odd superperfect numbers. - Omar E. Pol, Mar 01 2008 Differences between exponents when the even perfect numbers are represented as differences of powers of 2, for example: The 5th even perfect number is 33550336 = 2^25 - 2^12 then a(5)=25-12=13 (see A135655, A133033, A090748). - Omar E. Pol, Mar 01 2008 Number of 1's in binary expansion of n-th even perfect number (see A135650). Number of 1's in binary expansion of divisors of n-th even perfect number that are multiples of n-th Mersenne prime A000668(n) (see A135652, A135653, A135654, A135655). - Omar E. Pol, May 04 2008 Indices of the numbers A006516 that are also even perfect numbers. - Omar E. Pol, Aug 30 2008 The (prime) number p appears in this sequence if and only if there is no prime q<2^p-1 such that the order of 2 modulo q equals p; a special case is that if p=4k+3 is prime and also q=2p+1 is prime then the order of 2 modulo q is p so p is not a term of this sequence. - Joerg Arndt, Jan 16 2011 Primes p such that sigma(2^p) - sigma(2^p-1) = 2^p-1. - Jaroslav Krizek, Aug 02 2013 Integers k such that every degree k irreducible polynomial over GF(2) is also primitive, i.e., has order 2^k-1. Equivalently, the integers k such that A001037(k) = A011260(k). - Geoffrey Critzer, Dec 08 2019 Conjecture: for k > 1, 2^k-1 is (a Mersenne) prime or k = 2^(2^m)+1 (is a Fermat number) if and only if (k-1)^(2^k-2) == 1 (mod (2^k-1)k^2). - Thomas Ordowski, Oct 05 2023 Conjecture: for p prime, 2^p-1 is (a Mersenne) prime or p = 2^(2^m)+1 (is a Fermat number) if and only if (p-1)^(2^p-2) == 1 (mod 2^p-1). - David Barina, Nov 25 2024 Already as of Dec. 2020, all exponents up to 10^8 had been verified, implying that 74207281, 77232917 and 82589933 are indeed the next three terms. As of today, all exponents up to 130439863 have been tested at least once, see the GIMPS Milestones Report. - M. F. Hasler, Apr 11 2025 On June 23. 2025 all exponents up to 74340751 have been verified, confirming that 74207281 is the exponent of the 49th Mersenne Prime. - Rodolfo Ruiz-Huidobro, Jun 23 2025 According to the Milestones Report link, on Sep 08 2025, all tests below M(77232917) are verified, officially making it the 50th Mersenne prime. - Robert Israel, Sep 15 2025 Conjecture: a(1) = 2; and then, for n > 0, a(n+1) is the least k such that lpf(2^k - 1) > lpf(2^a(n) - 1), where lpf(m) is the least prime factor of m. See A049479 (look at the record value positions). - Thomas Ordowski, Sep 03 2026
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 4.
J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, p. 79.
R. K. Guy, Unsolved Problems in Number Theory, Section A3.
F. Lemmermeyer, Reciprocity Laws From Euler to Eisenstein, Springer-Verlag, 2000, p. 57.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 19.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 47.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 132-134.
B. Tuckerman, The 24th Mersenne prime, Notices Amer. Math. Soc., 18 (Jun, 1971), Abstract 684-A15, p. 608.
J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 82.
LINKS
P. T. Bateman, J. L. Selfridge, and S. S. Wagstaff, Jr., The new Mersenne conjecture, Amer. Math. Monthly 96 (1989), no. 2, 125--128. MR0992073 (90c:11009). J. Brillhart et al., Factorizations of b^n +- 1, Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 3rd edition, 2002. Albert A. Mullin, Letter to the editor, about "The new Mersenne conjecture" [Amer. Math. Monthly 96 (1989), no. 2, 125-128; MR0992073 (90c:11009)] by P. T. Bateman, J. L. Selfridge and S. S. Wagstaff, Jr., Amer. Math. Monthly 96 (1989), no. 6, 511. MR0999415 (90f:11008). Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)] Eric Weisstein's World of Mathematics, Repunit
FORMULA
a(n) = log((1/2)*(1+sqrt(1+8* A000396(n))))/log(2). - Artur Jasinski, Sep 23 2008 (under the assumption there are no odd perfect numbers, Joerg Arndt, Feb 23 2014)
EXAMPLE
Corresponding to the initial terms 2, 3, 5, 7, 13, 17, 19, 31 ... we get the Mersenne primes 2^2 - 1 = 3, 2^3 - 1 = 7, 2^5 - 1 = 31, 127, 8191, 131071, 524287, 2147483647, ... (see A000668).
MATHEMATICA
MersennePrimeExponent[Range[48]] (* Eric W. Weisstein, Jul 17 2017; updated Oct 21 2024 *)
PROG
(PARI) is(n)=my(h=Mod(2, 2^n-1)); for(i=1, n-2, h=2*h^2-1); h==0||n==2 \\ Lucas-Lehmer test for exponent e. - Joerg Arndt, Jan 16 2011, and Charles R Greathouse IV, Jun 05 2013 forprime(e=2, 5000, if(is(e), print1(e, ", "))); /* terms < 5000 */
(Python)
from sympy import isprime, prime
for n in range(1, 100):
if isprime(2**prime(n)-1):
CROSSREFS
Cf. A028335 (integer lengths of Mersenne primes). Cf. A001348 (Mersenne numbers with n prime). Cf. A016027, A046051, A057429, A057951- A057958, A066408, A117293, A127962, A127963, A127964, A127965, A127961, A000979, A000978, A124400, A124401, A127955, A127956, A127957, A127958, A127936, A134458, A000225, A000396, A090748, A133033, A135655, A006516, A019279, A061652, A133033, A135650, A135652, A135653, A135654, A260073, A050475, A379590.
KEYWORD
hard,nonn,nice,core,changed
EXTENSIONS
a(46) = 42643801 and a(47) = 43112609, whose ordinal positions in the sequence are now confirmed, communicated by Eric W. Weisstein, Apr 12 2018 a(48) = 57885161, whose ordinal position in the sequence is now confirmed, communicated by Benjamin Przybocki, Jan 05 2022 As of Jan 31 2025, 48 terms are known, and are shown in the DATA section. Four additional numbers are known to be in the sequence, namely 74207281, 77232917, 82589933, and 136279841, but they may not be the next terms. See the GIMPS website for the latest information. - N. J. A. Sloane, Jan 31 2025
Mersenne primes (primes of the form 2^n - 1).
(Formerly M2696 N1080)
+10
661
3, 7, 31, 127, 8191, 131071, 524287, 2147483647, 2305843009213693951, 618970019642690137449562111, 162259276829213363391578010288127, 170141183460469231731687303715884105727
COMMENTS
For a Mersenne number 2^n - 1 to be prime, the exponent n must itself be prime.
Primes that are repunits in base 2.
Define f(k) = 2k+1; begin with k = 2, a(n+1) = least prime of the form f(f(f(...(a(n))))). - Amarnath Murthy, Dec 26 2003 Mersenne primes other than the first are of the form 6n+1. - Lekraj Beedassy, Aug 27 2004. Mersenne primes other than the first are of the form 24n+7; see also A124477. - Artur Jasinski, Nov 25 2007 Mersenne primes are solutions to sigma(n+1)-sigma(n) = n as perfect numbers ( A000396(n)) are solutions to sigma(n) = 2n. In fact, appears to give all n such that sigma(n+1)-sigma(n) = n. - Benoit Cloitre, Aug 27 2002 If n is in the sequence then sigma(sigma(n)) = 2n+1. Is it true that this sequence gives all numbers n such that sigma(sigma(n)) = 2n+1? - Farideh Firoozbakht, Aug 19 2005 It is easily proved that if n is a Mersenne prime then sigma(sigma(n)) - sigma(n) = n. Is it true that Mersenne primes are all the solutions of the equation sigma(sigma(x)) - sigma(x) = x? - Farideh Firoozbakht, Feb 12 2008 Sum of divisors of n-th even superperfect number A061652(n). Sum of divisors of n-th superperfect number A019279(n), if there are no odd superperfect numbers. - Omar E. Pol, Mar 11 2008 Indices of both triangular numbers and generalized hexagonal numbers ( A000217) that are also even perfect numbers. - Omar E. Pol, May 10 2008, Sep 22 2013 Number of positive integers (1, 2, 3, ...) whose sum is the n-th perfect number A000396(n). - Omar E. Pol, May 10 2008 Vertex number where the n-th perfect number A000396(n) is located in the square spiral whose vertices are the positive triangular numbers A000217. - Omar E. Pol, May 10 2008 The digital roots are 1 if p == 1 (mod 6) and 4 if p == 5 (mod 6). [T. Koshy, Math Gaz. 89 (2005) p. 465]
Primes p such that for all primes q < p, p XOR q = p - q. - Brad Clardy, Oct 26 2011 All prime numbers p can be classified by k = (p mod 12) into four classes: k=1, 5, 7, 11. The Mersennne prime numbers 2^p-1, p > 2 are in the class k=7 with p=12*(n-1)+7, n=1,2,.... As all 2^p (p odd) are in class k=8 it follows that all 2^p-1, p > 2 are in class k=7. - Freimut Marschner, Jul 27 2013 From "The Guinness Book of Primes": "During the reign of Queen Elizabeth I, the largest known prime number was the number of grains of rice on the chessboard up to and including the nineteenth square: 524,287 [= 2^19 - 1]. By the time Lord Nelson was fighting the Battle of Trafalgar, the record for the largest prime had gone up to the thirty-first square of the chessboard: 2,147,483,647 [= 2^31 - 1]. This ten-digits number was proved to be prime in 1772 by the Swiss mathematician Leonard Euler, and it held the record until 1867." [du Sautoy] - Robert G. Wilson v, Nov 26 2013 If n is in the sequence then A024816(n) = antisigma(n) = antisigma(n+1) - 1. Is it true that this sequence gives all numbers n such that antisigma(n) = antisigma(n+1) - 1? Are there composite numbers with this property? - Jaroslav Krizek, Jan 24 2014 If n is in the sequence then phi(n) + sigma(sigma(n)) = 3n. Is it true that Mersenne primes are all the solutions of the equation phi(x) + sigma(sigma(x)) = 3x? - Farideh Firoozbakht, Sep 03 2014 Equivalently, prime powers of the form 2^n - 1, see Theorem 2 in Lemos & Cambraia Junior. - Charles R Greathouse IV, Jul 07 2016 Primes whose sum of divisors is a power of 2. Primes p such that p + 1 is a power of 2. Primes in A046528. - Omar E. Pol, Jul 09 2016 Primes p such that sigma(p+1) = 2p+1.
Primes p such that A051027(p) = sigma(sigma(p)) = 2^k-1 for some k > 1. Primes p of the form sigma(2^prime(n)-1)-1 for some n. Corresponding values of numbers n are in A016027. Primes p of the form sigma(2^(n-1)) for some n > 1. Corresponding values of numbers n are in A000043 (Mersenne exponents). Primes of the form sigma(2^(n+1)) for some n > 1. Corresponding values of numbers n are in A153798 (Mersenne exponents-2). Primes p of the form sigma(n) where n is even; subsequence of A023195. Primes p of the form sigma(n) for some n. Conjecture: 31 is the only prime p such that p = sigma(x) = sigma(y) for distinct numbers x and y; 31 = sigma(16) = sigma(25). Conjecture: numbers n such that n = sigma(sigma(n+1)-n-1)-1, i.e., A072868(n)-1. Conjecture: primes of the form sigma(4*(n-1)) for some n. Corresponding values of numbers n are in A281312. (End) [Conjecture] For n > 2, the Mersenne number M(n) = 2^n - 1 is a prime if and only if 3^M(n-1) == -1 (mod M(n)). - Thomas Ordowski, Aug 12 2018 [This needs proof! - Joerg Arndt, Mar 31 2019] Named "Mersenne's numbers" by W. W. Rouse Ball (1892, 1912) after Marin Mersenne (1588-1648). - Amiram Eldar, Feb 20 2021 Theorem. Let b = 2^p - 1 (where p is a prime). Then b is a Mersenne prime iff (c = 2^p - 2 is totient or a term of A002202). Otherwise, if c is (nontotient or a term of A005277) then b is composite. Proof. Trivial, since, while b = v^g - 1 where v is even, v > 2, g is an integer, g > 1, b is always composite, and c = v^g - 2 is nontotient (or a term of A005277), and so is for any composite b = 2^g - 1 (in the last case, c = v^g - 2 is also nontotient, or a term of A005277). - Sergey Pavlov, Aug 30 2021 [Disclaimer: This proof has not been checked. - N. J. A. Sloane, Oct 01 2021]
REFERENCES
Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 4.
John Brillhart, D. H. Lehmer, J. L. Selfridge, Bryant Tuckerman and S. S. Wagstaff, Jr., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 135-136.
William Dunham, Journey Through Genius, Wiley, 1990, Chapter 3, pp. 75, 81-82.
Graham Everest, Alf van der Poorten, Igor Shparlinski and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 76.
Marcus P. F. du Sautoy, The Number Mysteries, A Mathematical Odyssey Through Everyday Life, Palgrave Macmillan, First published in 2010 by the Fourth Estate, an imprint of Harper Collins UK, 2011, p. 46. - Robert G. Wilson v, Nov 26 2013 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Bryant Tuckerman, The 24th Mersenne prime, Notices Amer. Math. Soc., 18 (Jun, 1971), Abstract 684-A15, p. 608.
J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 82.
LINKS
W. W. Rouse Ball, Mersenne's numbers, Messenger of Mathematics (1892) Vol. 21, pp. 34-40, 121. John Brillhart, D. H. Lehmer, J. L. Selfridge, Bryant Tuckerman and S. S. Wagstaff, Jr., Factorizations of b^n +- 1, Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 3rd edition, 2002.
MAPLE
i := 2^(ithprime(n))-1:
if (isprime(i)) then
return i
fi: end:
# Alternative:
seq(numtheory:-mersenne([i]), i=1..26); # Robert Israel, Jul 13 2014
MATHEMATICA
2^Array[MersennePrimeExponent, 18] - 1 (* Jean-François Alcover, Feb 17 2018, Mersenne primes with less than 1000 digits *) (* Alternative: *)
PROG
(PARI) LL(e) = my(n, h); n = 2^e-1; h = Mod(2, n); for (k=1, e-2, h=2*h*h-1); return(0==h) \\ after Joerg Arndt in A000043forprime(p=1, , if(LL(p), print1(p, ", "))) \\ Felix Fröhlich, Feb 17 2018 (GAP)
(Python)
from sympy import isprime, primerange
print([2**n-1 for n in primerange(1, 1001) if isprime(2**n-1)]) # Karl V. Keller, Jr., Jul 16 2020
CROSSREFS
Cf. A001348 (Mersenne numbers with n prime). Cf. A000040, A000203, A000217, A000396, A003056, A007947, A016027, A019279, A023195, A023758, A028335 (lengths), A034876, A046051, A057951- A057958, A059305, A061652, A083420, A085104, A124477, A135659, A173898.
Fermat primes: primes of the form 2^(2^k) + 1, for some k >= 0.
+10
395
COMMENTS
It is conjectured that there are only 5 terms. Currently it has been shown that 2^(2^k) + 1 is composite for 5 <= k <= 32 (see Eric Weisstein's Fermat Primes link). - Dmitry Kamenetsky, Sep 28 2008 No Fermat prime is a Brazilian number. So Fermat primes belong to A220627. For proof see Proposition 3 page 36 in "Les nombres brésiliens" in Links. - Bernard Schott, Dec 29 2012 This sequence and A001220 are disjoint (see "Other theorems about Fermat numbers" in Wikipedia link). - Felix Fröhlich, Sep 07 2014 Numbers n > 1 such that n * 2^(n-2) divides (n-1)! + 2^(n-1). - Thomas Ordowski, Jan 15 2015 Primes p such that phi(p) = 2*phi(p-1); primes from A171271. Primes p such that sigma(p-1) = 2p - 3.
Primes p such that sigma(p-1) = 2*sigma(p) - 5.
For n > 1, a(n) = primes p such that p = 4 * phi((p-1) / 2) + 1.
Conjectures:
1) primes p such that phi(p) = 2*phi(p-2).
2) primes p such that phi(p) = 2*phi(p-1) = 2*phi(p-2).
3) primes p such that p = sigma(phi(p-2)) + 2.
4) primes p such that phi(p-1) + 1 divides p + 1.
5) numbers n such that sigma(n-1) = 2*sigma(n) - 5. (End)
Odd primes p such that ratio of the form (the number of nonnegative m < p such that m^q == m (mod p))/(the number of nonnegative m < p such that -m^q == m (mod p)) is a divisor of p for all nonnegative q. - Juri-Stepan Gerasimov, Oct 13 2020 Numbers n such that tau(n)*(number of distinct ratio (the number of nonnegative m < n such that m^q == m (mod n))/(the number of nonnegative m < n such that -m^q == m (mod n))) for nonnegative q is equal to 4. - Juri-Stepan Gerasimov, Oct 22 2020 The numbers of primitive roots for the five known terms are 1, 2, 8, 128, 32768. - Gary W. Adamson, Jan 13 2022 Prime numbers such that every residue is either a primitive root or a quadratic residue. - Keith Backman, Jul 11 2022 If there are only 5 Fermat primes, then there are only 31 odd order groups which have a 2-group automorphism group. See the Miles Englezou link for a proof. - Miles Englezou, Mar 10 2025
REFERENCES
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 137-141, 197.
John Derbyshire, Unknown Quantity: A Real and Imaginary History of Algebra. Washington, DC: Joseph Henry Press (2006): 112-113.
Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), page 268.
William Dunham, Journey Through Genius, Wiley, 1990, Chapter 10, pp. 229-236.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
C. F. Gauss, Disquisitiones Arithmeticae, Yale, 1965; see Table 1, p. 458.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, pp. 78-79.
Richard K. Guy, Unsolved Problems in Number Theory, A3.
Hardy and Wright, An Introduction to the Theory of Numbers, bottom of page 18 in the sixth edition, gives an heuristic argument that this sequence is finite.
Stephen Hawking, God Created the Integers: The Mathematical Breakthroughs That Changed History, Running Press, 2007. See p. 593.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 7, 70.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.
LINKS
Bernard Schott, Les nombres brésiliens, Quadrature (avril-juin 2010) No. 76, 30-38. [ Local copy, included here with permission from the editors of Quadrature.]
PROG
(PARI) for(i=0, 10, isprime(2^2^i+1) && print1(2^2^i+1, ", ")) \\ M. F. Hasler, Nov 21 2009 (SageMath) [2^(2^n)+1 for n in (0..4) if is_prime(2^(2^n)+1)] # G. C. Greubel, Mar 07 2019
Fermat pseudoprimes to base 2, also called Sarrus numbers or Poulet numbers.
(Formerly M5441 N2365)
+10
374
341, 561, 645, 1105, 1387, 1729, 1905, 2047, 2465, 2701, 2821, 3277, 4033, 4369, 4371, 4681, 5461, 6601, 7957, 8321, 8481, 8911, 10261, 10585, 11305, 12801, 13741, 13747, 13981, 14491, 15709, 15841, 16705, 18705, 18721, 19951, 23001, 23377, 25761, 29341
COMMENTS
A composite number n is a Fermat pseudoprime to base b if and only if b^(n-1) == 1 (mod n). Fermat pseudoprimes to base 2 are often simply called pseudoprimes.
Theorem: If both numbers q and 2q - 1 are primes (q is in the sequence A005382) and n = q*(2q-1) then 2^(n-1) == 1 (mod n) (n is in the sequence) if and only if q is of the form 12k + 1. The sequence 2701, 18721, 49141, 104653, 226801, 665281, 721801, ... is related. This subsequence is also a subsequence of the sequences A005937 and A020137. - Farideh Firoozbakht, Sep 15 2006 Also, composite odd numbers n such that n divides 2^n - 2 (cf. A006935). It is known that all primes p divide 2^(p-1) - 1. There are only two known numbers n such that n^2 divides 2^(n-1) - 1, A001220(n) = {1093, 3511} Wieferich primes p: p^2 divides 2^(p-1) - 1. 1093^2 and 3511^2 are the terms of a(n). - Alexander Adamchuk, Nov 06 2006 An odd composite number 2n + 1 is in the sequence if and only if multiplicative order of 2 (mod 2n+1) divides 2n. - Ray Chandler, May 26 2008 The Carmichael numbers A002997 are a subset of this sequence. For the Sarrus numbers which are not Carmichael numbers, see A153508. - Artur Jasinski, Dec 28 2008 An odd number n greater than 1 is a Fermat pseudoprime to base b if and only if ((n-1)! - 1)*b^(n-1) == -1 (mod n). - Arkadiusz Wesolowski, Aug 20 2012 The name "Sarrus numbers" is after Frédéric Sarrus, who, in 1819, discovered that 341 is a counterexample to the "Chinese hypothesis" that n is prime if and only if 2^n is congruent to 2 (mod n). - Alonso del Arte, Apr 28 2013 The name "Poulet numbers" appears in Monografie Matematyczne 42 from 1932, apparently because Poulet in 1928 produced a list of these numbers (cf. Miller, 1975). - Felix Fröhlich, Aug 18 2014 Numbers n > 2 such that (n-1)! + 2^(n-1) == 1 (mod n). Composite numbers n such that (n-2)^(n-1) == 1 (mod n). - Thomas Ordowski, Feb 20 2016 The only twin pseudoprimes up to 10^13 are 4369, 4371. - Thomas Ordowski, Feb 12 2016 [There are no other pairs below 2^64. - Amiram Eldar, Jan 14 2026] Theorem (A. Rotkiewicz, 1965): (2^p-1)*(2^q-1) is a pseudoprime if and only if p*q is a pseudoprime, where p,q are different primes. - Thomas Ordowski, Mar 31 2016 Theorem (W. Sierpiński, 1947): if n is a pseudoprime (odd or even), then 2^n-1 is a pseudoprime. - Thomas Ordowski, Apr 01 2016 If 2^n-1 is a pseudoprime, then n is a prime or a pseudoprime (odd or even). - Thomas Ordowski, Sep 05 2016 Erdős (1950) called these numbers "almost primes".
According to Erdős (1949) and Piza (1954), the term "pseudoprime" was coined by D. H. Lehmer.
Named after the French mathematician Pierre de Fermat (1607-1665), or, alternatively, after the Belgian mathematician Paul Poulet (1887-1946), or, the French mathematician Pierre Frédéric Sarrus (1798-1861). (End)
If m is a term of this sequence, then (m-1)/ord(2,m) >= 5, where ord(a,m) is the multiplicative order of a modulo m; see my link below. Actually, it seems that we always have (m-1)/ord(2,m) >= 9. - Jianing Song, Nov 04 2024
REFERENCES
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, p. 80.
Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section A12, pp. 44-50.
George P. Loweke, The Lore of Prime Numbers. New York: Vantage Press (1982), p. 22.
Øystein Ore, Number Theory and Its History, McGraw-Hill, 1948.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 88-92.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 145.
FORMULA
Sum_{n>=1} 1/a(n) is in the interval (0.015260, 33) (Bayless and Kinlaw, 2017). The upper bound was reduced to 0.0911 by Kinlaw (2023). - Amiram Eldar, Oct 15 2020, Feb 24 2024
MAPLE
select(t -> not isprime(t) and 2 &^(t-1) mod t = 1, [seq(i, i=3..10^5, 2)]); # Robert Israel, Feb 18 2016
MATHEMATICA
Select[Range[3, 30000, 2], ! PrimeQ[ # ] && PowerMod[2, (# - 1), # ] == 1 &] (* Farideh Firoozbakht, Sep 15 2006 *)
PROG
(PARI) q=1; vector(50, i, until( !isprime(q) & (1<<(q-1)-1)%q == 0, q+=2); q) \\ M. F. Hasler, May 04 2007 (PARI) is_ A001567(n)={Mod(2, n)^(n-1)==1 && !isprime(n) && n>1} \\ M. F. Hasler, Oct 07 2012, updated to current PARI syntax and to exclude even pseudoprimes on Mar 01 2019 (Magma) [n: n in [3..3*10^4 by 2] | IsOne(Modexp(2, n-1, n)) and not IsPrime(n)]; // Bruno Berselli, Jan 17 2013 (Python)
from itertools import islice
from sympy import nextprime
def A001567_gen(): # generator of terms p, q = 7, 11
while True:
for n in range(p+2, q, 2):
if pow(2, n, n) == 2:
yield n
p, q = q, nextprime(q)
CROSSREFS
Cf. A001220 = Wieferich primes p: p^2 divides 2^(p-1) - 1.
EXTENSIONS
"Poulet numbers" added to name by Joerg Arndt, Aug 18 2014
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