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Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).
(Formerly M1606 N0628)
+20
126
0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719, 427859097160, 1120149658760
COMMENTS
a(n)/ A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005 Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005 A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009 Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012 In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013 For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014 [Note on how to calculate: take the two points (a,b) and (c,d) with a<b, c<d and a<d then subtract a from each: a-a=0, b-a=B, c-a=C, and d-a=D. The area is (D-(C-B)^2)/2.]
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017
REFERENCES
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
Ralph P. Grimaldi, Fibonacci and Catalan Numbers: An Introduction, (2012). See Property 4.6 at p. 10.
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).
FORMULA
a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2). For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002. G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). ( Simon Plouffe in his 1992 dissertation; see Comments to A055870), a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004 a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009 a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010 a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010 a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010 Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011 a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012 0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014 E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016 a(n) = A061646(n) - Fibonacci(n-1)^2.
EXAMPLE
G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...
MAPLE
with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
MATHEMATICA
(* Alternative: *)
Times@@@Partition[Fibonacci[Range[0, 30]], 2, 1] (* Harvey P. Dale, Aug 18 2011 *) (* Alternative: *)
Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)
PROG
(PARI) A001654(n)=fibonacci(n)*fibonacci(n+1); (PARI) b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
(Haskell)
a001654 n = a001654_list !! n
a001654_list = zipWith (*) (tail a000045_list) a000045_list
(Python)
from sympy import fibonacci as F
def a(n): return F(n)*F(n + 1)
(Python)
from math import prod
from gmpy2 import fib2
(Magma) I:=[0, 1, 2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018
Squared Fibonacci numbers: a(n) = F(n)^2 where F = A000045.
(Formerly M3364)
+20
121
0, 1, 1, 4, 9, 25, 64, 169, 441, 1156, 3025, 7921, 20736, 54289, 142129, 372100, 974169, 2550409, 6677056, 17480761, 45765225, 119814916, 313679521, 821223649, 2149991424, 5628750625, 14736260449, 38580030724, 101003831721, 264431464441, 692290561600
COMMENTS
a(n)*(-1)^(n+1) = (2*(1-T(n,-3/2))/5), n>=0, with Chebyshev's polynomials T(n,x) of the first kind, is the r=-1 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found. - Wolfdieter Lang, Oct 18 2004 Determinant of power series with alternate signs of gamma matrix with determinant 1!.
a(n) = Determinant(A - A^2 + A^3 - A^4 + A^5 - ... - (-1)^n*A^n) where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant.
A = [[1,1,1,1,1,1,...], [1,2,1,2,1,2,...], [1,2,3,1,2,3,...], [1,2,3,4,1,2,...], [1,2,3,4,5,1,...], [1,2,3,4,5,6,...], ...]; note: Determinant A(1..n,1..n) = (n-1)!.
a(n) is even with respect to signs of power of A.
Equals the INVERT transform of (1, 3, 2, 2, 2, ...). Example: a(7) = 169 = (1, 1, 4, 9, 25, 64) dot (2, 2, 2, 2, 3, 1) = (2 + 2 + 8 + 18 + 75 + 64). - Gary W. Adamson, Apr 27 2009 This is a divisibility sequence.
a(n+1)*(-1)^n, n>=0, is the sequence of the alternating row sums of the Riordan triangle A158454. - Wolfdieter Lang, Dec 18 2010 a(n+1) is the number of tilings of a 2 X 2n rectangle with n tetrominoes of any shape, cf. A230031. - Alois P. Heinz, Nov 29 2013 This is the case P1 = 1, P2 = -6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014 a(n+1) is the number of 2 X n matrices that can be obtained from a 2 X n matrix by moving each element to an adjacent position, horizontally or vertically. This is because F(n+1) is the number of domino tilings of that matrix, therefore with a checkerboard coloring and two domino tilings we can move the black element of each domino of the first tiling to the white element of the same domino and similarly move the white element of each domino of the second tiling to the black element of the same domino. - Fabio Visonà, May 04 2022 In general, squaring the terms of a second-order linear recurrence with signature (c,d) will result in a third-order linear recurrence with signature (c^2+d,(c^2+d)*d,-d^3). - Gary Detlefs, Jan 05 2023
REFERENCES
Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 8.
Ross Honsberger, Mathematical Gems III, M.A.A., 1985, p. 130.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Richard P. Stanley, Enumerative Combinatorics I, Example 4.7.14, p. 251.
LINKS
Dominique Foata and Guo-Niu Han, Nombres de Fibonacci et polynômes orthogonaux, in: Marcello Morelli and Marco Tangheroni (eds.), Leonardo Fibonacci : il tempo, le opere, l'ereditá scientifica, Pisa, 23-25 Marzo 1994, Pisa, Pacini Editore, 1994, pp. 179-200.
FORMULA
G.f.: x*(1-x)/((1+x)*(1-3*x+x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), n > 2. a(0)=0, a(1)=1, a(2)=1.
a(-n) = a(n) for all n in Z.
L.g.f.: 1/5*log((1+3*x+x^2)/(1-6*x+x^2)) = Sum_{n>=0} a(n)/n*x^n; special case of l.g.f. given in A079291. - Joerg Arndt, Apr 13 2011 a(0) = 0, a(1) = 1; a(n) = a(n-1) + Sum(a(n-i)) + k, 0 <= i < n where k = 1 when n is odd, or k = -1 when n is even. E.g., a(2) = 1 = 1 + (1 + 1 + 0) - 1, a(3) = 4 = 1 + (1 + 1 + 0) + 1, a(4) = 9 = 4 + (4 + 1 + 1 + 0) - 1, a(5) = 25 = 9 + (9 + 4 + 1 + 1 + 0) + 1. - Sadrul Habib Chowdhury (adil040(AT)yahoo.com), Mar 02 2004
a(n) = (2*Fibonacci(2*n+1) - Fibonacci(2*n) - 2*(-1)^n)/5. - Ralf Stephan, May 14 2004 Sum_{j=0..2*n} binomial(2*n,j)*a(j) = 5^(n-1)* A005248(n+1) for n >= 1 [P. Stanica]. Sum_{j=0..2*n+1} binomial(2*n+1,j)*a(j) = 5^n* A001519(n+1) [P. Stanica]. - R. J. Mathar, Oct 16 2006 a(n) = (-1)^k*(Fibonacci(n+k)^2-Fibonacci(k)*Fibonacci(2*n+k)), for any k. - Gary Detlefs, Dec 13 2010 a(n) = 3*a(n-1) - a(n-2) + 2*(-1)^(n+1), n > 1. - Gary Detlefs, Dec 20 2010 a(n) = Fibonacci(2*n-2) + a(n-2). - Gary Detlefs, Dec 20 2010 a(n) = (Fibonacci(3*n) - 3*(-1)^n*Fibonacci(n))/(5*Fibonacci(n)), n > 0. - Gary Detlefs, Dec 20 2010 a(n) = (Fibonacci(n)*Fibonacci(n+4) - 3*Fibonacci(n)*Fibonacci(n+1))/2. - Gary Detlefs, Jan 17 2011 a(n) = (((3+sqrt(5))/2)^n + ((3-sqrt(5))/2)^n - 2*(-1)^n)/5; without leading zero we would have a(n) = ((3+sqrt(5))*((3+sqrt(5))/2)^n + (3-sqrt(5))*((3-sqrt(5))/2)^n + 4*(-1)^n)/10. - Tim Monahan, Jul 17 2011 E.g.f.: (exp((phi+1)*x) + exp((2-phi)*x) - 2*exp(-x))/5, with the golden section phi:=(1+sqrt(5))/2. From the Binet-de Moivre formula for F(n). - Wolfdieter Lang, Jan 13 2012 a(0) = 0, a(1) = 1; a(n+1) = (a(n)^(1/2) + a(n-1)^(1/2))^2. - Thomas Ordowski, Jan 06 2013 a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), where alpha = 3/2 and beta = -1 and T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 3/2; 1, 1/2].
a(n) = U(n-1,i/2)*U(n-1,-i/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End) 0 = a(n)*(+a(n) - 2*a(n+1) - 2*a(n+2)) + a(n+1)*(+a(n+1) - 2*a(n+2)) + a(n+2)*(+a(n+2)) for all n in Z. - Michael Somos, Jun 03 2014 (F(n)*b(n+2))^2 + (F(n+1)*b(n-1))^2 = F(2*n+1)^3 = A001519(n+1)^3, with b(n) = a(n) + 2*(-1)^n and F(n) = A000045(n) (see Bruckman link). - Michel Marcus, Jan 24 2015 a(n) = 1/4*( a(n-2) - a(n-1) - a(n+1) + a(n+2) ). The same recurrence holds for A001254. - Peter Bala, Aug 18 2015 For n>2, a(n) = F(n-2)*(3*F(n-1) + F(n-3)) + F(2*n-5). Also, for n>2 a(n)=2*F(n-3)*F(n) + F(2*n-3) -(2)*(-1)^n. - J. M. Bergot, Nov 05 2015 a(n) = (F(n+2)^2 + L(n+1)^2) - 2*F(n+2)*L(n+1). - J. M. Bergot, Nov 08 2015 a(n) = F(n+3)^2 - 4*F(n+1)*F(n+2). - J. M. Bergot, Mar 17 2016 a(n) = (F(n-2)*F(n+2) + F(n-1)*F(n+1))/2. - J. M. Bergot, May 25 2017 a(n) = F(n+k)*F(n-k) + (-1)^(n+k)*a(k), for every integer k >= 0. - Federico Provvedi, Dec 10 2018 Sum_{n >= 3} 1/(a(n) - 1/a(n)) = 4/9.
Sum_{n >= 3} (-1)^n/(a(n) - 1/a(n)) = -(10 - 3*sqrt(5))/18.
Conjecture: Sum_{n >= 1, n != 2*k+1} 1/(a(n) + (-1)^n*a(2*k+1)) = 1/a(4*k+2) for k = 0,1,2,.... (End)
The conjecture is true (see Fried link). - Sela Fried, Nov 24 2025 Product_{n>=2} (1 + (-1)^n/a(n)) = phi ( A001622) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024
EXAMPLE
G.f. = x + x^2 + 4*x^3 + 9*x^4 + 25*x^5 + 64*x^6 + 169*x^7 + 441*x^8 + ...
MAPLE
with(combinat): seq(fibonacci(n)^2, n=0..27); # Zerinvary Lajos, Sep 21 2007
MATHEMATICA
LinearRecurrence[{2, 2, -1}, {0, 1, 1}, 41] (* Harvey P. Dale, May 18 2011 *)
PROG
(PARI) {a(n) = fibonacci(n)^2};
(PARI) concat(0, Vec(x*(1-x)/((1+x)*(1-3*x+x^2)) + O(x^30))) \\ Altug Alkan, Nov 06 2015 (SageMath) [(fibonacci(n))^2 for n in range(0, 28)]# Zerinvary Lajos, May 15 2009 (Haskell)
(SageMath) [fibonacci(n)^2 for n in range(30)] # G. C. Greubel, Dec 10 2018 (GAP) List([0..30], n -> Fibonacci(n)^2); # G. C. Greubel, Dec 10 2018 (Python)
from sympy import fibonacci
CROSSREFS
Cf. A000032, A000045, A001254, A001622, A001654, A047946, A056570, A059260, A061646, A065885, A079291, A080097, A105393.
Fibonacci entry points: a(n) = least k >= 1 such that n divides Fibonacci number F_k (= A000045(k)).
(Formerly M2314 N0914)
+20
81
1, 3, 4, 6, 5, 12, 8, 6, 12, 15, 10, 12, 7, 24, 20, 12, 9, 12, 18, 30, 8, 30, 24, 12, 25, 21, 36, 24, 14, 60, 30, 24, 20, 9, 40, 12, 19, 18, 28, 30, 20, 24, 44, 30, 60, 24, 16, 12, 56, 75, 36, 42, 27, 36, 10, 24, 36, 42, 58, 60, 15, 30, 24, 48, 35, 60, 68, 18, 24, 120
COMMENTS
In the formula, the relation a(p^e) = p^(e-1)*a(p) is called Wall's conjecture, which has been verified for primes up to 10^14. See A060305. Primes for which this relation fails are called Wall-Sun-Sun primes. - T. D. Noe, Mar 03 2009 All solutions to F_m == 0 (mod n) are given by m == 0 (mod a(n)). For a proof see, e.g., Vajda, p. 73. [Old comment changed by Wolfdieter Lang, Jan 19 2015] If p is a prime of the form 10n +- 1 then a(p) is a divisor of p-1. If q is a prime of the form 10n +- 3 then a(q) is a divisor of q+1. - Robert G. Wilson v, Jul 07 2007 Definition 1 in Riasat (2011) calls this k(n), or sometimes just k. Corollary 1 in the same paper, "every positive integer divides infinitely many Fibonacci numbers," demonstrates that this sequence is infinite. - Alonso del Arte, Jul 27 2013 If p is a prime then a(p) <= p+1. This is because if p is a prime then exactly one of the following Fibonacci numbers is a multiple of p: F(p-1), F(p) or F(p+1). - Dmitry Kamenetsky, Jul 23 2015 From Renault 1996:
1. a(lcm(n,m)) = lcm(a(n), a(m)).
2. if n|m then a(n)|a(m).
3. if m has prime factorization m=p1^e1 * p2^e2 * ... * pn^en then a(m) = lcm(a(p1^e1), a(p2^e2), ..., a(pn^en)). - Dmitry Kamenetsky, Jul 23 2015 a(n)=n if and only if n=5^k or n=12*5^k for some k >= 0 (see Marques 2012). - Dmitry Kamenetsky, Aug 08 2015 Every positive integer (except 2) eventually appears in this sequence. This is because every Fibonacci number bigger than 1 (except Fibonacci(6)=8 and Fibonacci(12)=144) has at least one prime factor that is not a factor of any earlier Fibonacci number (see Knott reference). Let f(n) be such a prime factor for Fibonacci(n); then a(f(n))=n. - Dmitry Kamenetsky, Aug 08 2015 We can reconstruct the Fibonacci numbers from this sequence using the formula Fibonacci(n+2) = 1 + Sum_{i: a(i) <= n} phi(i)*floor(n/a(i)), where phi(n) is Euler's totient function A000010 (see the Stroinski link). For example F(6) = 1 + phi(1)*floor(4/a(1)) + phi(2)*floor(4/a(2)) + phi(3)*floor(4/a(4)) = 1 + 1*4 + 1*1 + 2*1 = 8. - Peter Bala, Sep 10 2015 a(F_m) = m for all m > 1. Indeed, let (b(j)) be defined by b(1)=b(2)=1, and b(j+2) = (b(j) + b(j+1)) mod n. Then a(n) equals the index of the first occurrence of 0 in (b(j)). Example: if n=4 then b = A079343 = 1,1,2,3,1,0,1,1,..., so a(4)=6. If n is a Fibonacci number n=F_m, then obviously a(n)=m. Note that this gives a simple proof of the fact that all integers larger than 2 occur in (a(n)). - Michel Dekking, Nov 10 2017
REFERENCES
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 25.
B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers. Report ORNL-4261, Oak Ridge National Laboratory, Oak Ridge, Tennessee, June 1968.
Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Afterword by Herbert A. Hauptman, Nobel Laureate, 2. 'The Minor Modulus m(n)', Prometheus Books, NY, 2007, page 329-342.
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).
S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
N. N. Vorob'ev, Fibonacci numbers, Blaisdell, NY, 1961.
LINKS
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014; see also, J. Int. Seq. 17 (2014) # 14.8.5.
FORMULA
a(n) = n if and only if n is of form 5^k or 12*5^k (proved in Marques paper), a(n) = n - 1 if and only if n is in A106535, a(n) = n + 1 if and only if n is in A000057, a(n) = n + 5 if and only if n is in 5* A000057, ... - Benoit Cloitre, Feb 10 2007 a(1) = 1, a(2) = 3, a(4) = 6 and for e > 2, a(2^e) = 3*2^(e-2); a(5^e) = 5^e; and if p is an odd prime not 5, then a(p^e) = p^max(0, e-s)*a(p) where s = valuation( A000045(a(p)), p) (Wall's conjecture states that s = 1 for all p). If (m, n) = 1 then a(m*n) = lcm(a(m), a(n)). See Posamentier & Lahmann. - Robert G. Wilson v, Jul 07 2007; corrected by Max Alekseyev, Oct 19 2007, Jun 24 2011
EXAMPLE
a(4) = 6 because the smallest Fibonacci number that 4 divides is F(6) = 8.
a(5) = 5 because the smallest Fibonacci number that 5 divides is F(5) = 5.
a(6) = 12 because the smallest Fibonacci number that 6 divides is F(12) = 144.
a(2) = 3, hence 2 | F(m) iff m = 2*k, for k >= 0;
a(3) = 4, hence 3 | F(m) iff m = 4*k, for k >= 0;
etc. See a comment above with the Vajda reference. (End)
MAPLE
for k from 1 do
if combinat[fibonacci](k) mod n = 0 then
return k;
end if;
end do:
N:= 1000: # to get a(1) to a(N)
L:= ilcm($1..N):
count:= 0:
for n from 1 while count < N do
fn:= igcd(L, combinat:-fibonacci(n));
divs:= select(`<=`, numtheory:-divisors(fn), N);
for d in divs do if not assigned(A[d]) then count:= count+1; A[d]:= n fi od:
od:
MATHEMATICA
fibEntry[n_] := Block[{k = 1}, While[ Mod[ Fibonacci@k, n] != 0, k++ ]; k]; Array[fibEntry, 74] (* Robert G. Wilson v, Jul 04 2007 *)
PROG
(PARI) a(n)=if(n<0, 0, s=1; while(fibonacci(s)%n>0, s++); s) \\ Benoit Cloitre, Feb 10 2007 (PARI) ap(p)=my(k=p+[0, -1, 1, 1, -1][p%5+1], f=factor(k)); for(i=1, #f[, 1], for(j=1, f[i, 2], if((Mod([1, 1; 1, 0], p)^(k/f[i, 1]))[1, 2], break); k/=f[i, 1])); k
a(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e14, ap(f[i, 1]^f[i, 2]), ap(f[i, 1])*f[i, 1]^(f[i, 2]-1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<<max(f[1, 2]-2, 1)); lcm(v) \\ Charles R Greathouse IV, May 08 2017 (Haskell)
a001177 n = head [k | k <- [1..], a000045 k `mod` n == 0]
(Python)
CROSSREFS
Various derived sequences:
Analogous sequence for the tribonacci numbers: A046737, for Lucas numbers: A223486, for Pell numbers: A214028. Cf. also A000057, A106535, A120255, A120256, A175026, A213648, A214031, A214781, A214783, A230359, A233283, A233285, A233287. (End)
n occurs Fibonacci(n) times (cf. A000045).
+20
76
1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10
COMMENTS
Number of digits in Zeckendorf-binary representation of n. E.g., the Zeckendorf representation of 12 is 8+3+1, which in binary notation is 10101, which consists of 5 digits. - Clark Kimberling, Jun 05 2004 First position where value n occurs is A000045(n+1), i.e., a( A000045(n)) = n-1, for n >= 2 and a( A000045(n)-1) = n-2, for n >= 3. This is the number of distinct Fibonacci numbers greater than 0 which are less than or equal to n. - Robert G. Wilson v, Dec 10 2006 The smallest nondecreasing sequence a(n) such that a(Fibonacci(n-1)) = n. - Tanya Khovanova, Jun 20 2007
FORMULA
G.f.: (Sum_{n>1} x^Fibonacci(n))/(1-x). - Michael Somos, Apr 25 2003 a(n) = floor(log_phi((sqrt(5)*n + sqrt(5*n^2+4))/2)) - 1, where phi is A001622. a(n) = floor(arcsinh(sqrt(5)*n/2)/log(phi)) - 1.
a(n) = -1 + floor( log_phi( (n+0.2)*sqrt(5) ) ), where log_phi(x) is the logarithm to the base (1+sqrt(5))/2. - Ralf Stephan, May 14 2007
EXAMPLE
1, 1, then F(2) 2's, then F(3) 3's, then F(4) 4's, ..., then F(k) k's, ...
MAPLE
local j;
for j from ilog[(1+sqrt(5))/2](n) while combinat[fibonacci](j+1)<=n do
end do;
j-1
end proc:
MATHEMATICA
a[n_] := Module[{j}, For[j = Floor@Log[GoldenRatio, n], Fibonacci[j+1] <= n, j++]; j-1];
PROG
(PARI) a(n) = -1+floor(log(((n+0.2)*sqrt(5)))/log((1+sqrt(5))/2))
(PARI) a(n)=local(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2)
(Haskell)
a072649 n = a072649_list !! (n-1)
a072649_list = f 1 where
f n = (replicate (fromInteger $ a000045 n) n) ++ f (n+1)
(Python)
from sympy import fibonacci
def a(n):
if n<1: return 0
m=0
while fibonacci(m)<=n: m+=1
return m-2
(Python)
a, b, c = 0, 1, -2
while a <= n:
a, b = b, a+b
c += 1
(Scheme) (define ( A072649 n) (let ((b ( A072648 n))) (+ -1 b (floor->exact (/ n ( A000045 (1+ b))))))) ;; (The implementation below is better) (Scheme) (define ( A072649 n) (if (<= n 3) n (let loop ((k 5)) (if (> ( A000045 k) n) (- k 2) (loop (+ 1 k)))))) ;; (Use this with the memoized implementation of A000045 given under that entry. No floating point arithmetic is involved). - Antti Karttunen, Oct 06 2017
a(n+2) = -a(n+1) + a(n) (signed Fibonacci numbers) with a(-2) = a(-1) = 1; or Fibonacci numbers ( A000045) extended to negative indices.
+20
63
1, 1, 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
COMMENTS
Knuth defines the negaFibonacci numbers as follows: F(-1) = 1, F(-2) = -1, F(-3) = 2, F(-4) = -3, F(-5) = 5, ..., F(-n) = (-1)^(n-1) F(n). See A215022, A215023 for the negaFibonacci representation of n. - N. J. A. Sloane, Aug 03 2012 The sequence a(n), n >= 0 := 0, 1, -1, 2, -3, 5, -8, 13, ... is the inverse binomial transform of A000045. - Philippe Deléham, Oct 28 2008 Equals the INVERTi transform of A038754, assuming that an additional A038754(0) = 1 is added in front of A038754, and that the a(n) are prefixed with another 1 and then get offset 0. If we remove a(-2) and then set the offset to 0, we have the INVERT transform of a signed A011782: (1, -1, 2, -4, 8, -16, 32, ...). (End) The sequence 0, 1, -1, 2, -3, 5, -8, 13, -21, 34, -55, 89, -144, ... (starting at offset 0) is the Lucas U(-1,-1) sequence. - R. J. Mathar, Jan 08 2013 This sequence appears in the formula for 1/rho(5)^n, with rho(5) = (1 + sqrt(5))/2 = phi (golden section), when written in the power basis <1, rho(5)> of the quadratic number field Q(rho(5)): 1/rho(5)^n = a(n+1) * 1 + a(n) * rho(5), n >= -2. - Wolfdieter Lang, Nov 04 2013 The sequence 1, -1, 2, -3, 5, -8, 13, -21, 34, -55, 89, -144, ... (starting at offset 1) is the reversion of the g.f. for the "shadows" of Motzkin numbers with offset 1 (see A343773). - Gennady Eremin, Jul 16 2021 A Pisano sequence (modulo m) terminates with (..., 13, -8, 5, -3, 2, -1, 1, 0). - Roman Stawski, Nov 11 2024
REFERENCES
Donald E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7.1.3, p. 168, Eq. (145).
FORMULA
G.f.: (1+2*x)/(x^2*(1+x-x^2)).
a(n) = ((phi - 1)^n + 1/phi*(-(1/phi) - 1)^(n+1))/sqrt(5), where phi = (1 + sqrt(5))/2. - Arkadiusz Wesolowski, Oct 28 2012 a(n) = Sum_{k = 1..n} binomial(n-1, k-1)*Fibonacci(k)*(-1)^(n-k), n > 0, a(0) = 1. - Perminova Maria, Jan 22 2013 G.f.: 1 + x/(Q(0) - x) where Q(k) = 1 - x/(x*k - 1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Feb 23 2013 G.f.: 2 - 2/(Q(0) + 1) where Q(k) = 1 + 2*x/(1 - x/(x + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Apr 05 2013 G.f.: 1 + x^2 + x^3 + 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.: 1/(G(0)*x^3) + (2*x^2+x-1)/x^3, where G(k) = 1 + 2*x*(k+1)/(k + 2 - x*(k+2)*(k+3)/(x*(k+3) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 27 2013 G.f.: Q(0)/x - 1/x + 1+ x, where Q(k) = 1 + x^2 + x^3 + k*x*(1+x^2) - x^2*(1 + x*(k+2))*(1+k*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 13 2014 a(n) = -(-1)^n* A000045(n), at least for all n >= 0 (and also for n < 0 if A000045 is extended to negative indices). - M. F. Hasler, May 10 2017 a(n) = Sum_{k=0..n-1} (-1)^k * 2^(n-1-k) * Lucas(k+1) for n >= 0, where Lucas(k) = A000032(k) (Martinjak, 2015). a(n) = Sum_{k=0..n-1} (-1)^k * (Lucas(k+1) - Fibonacci(k)) for n >= 0 (Bhatnagar , 2016). (End)
a(n) = (phi^(-n) - phi^(n)*exp(-I*Pi*n))/sqrt(5) where phi = (1 + sqrt(5))/2.
Written in this form the function a(x) satisfies the Cassini analog:
a(x + 1) * a(x - 1) - a(x)^2 = exp(-I*Pi*x). (End)
EXAMPLE
With the golden section phi = rho(5) = (1 + sqrt(5))/2:
n = -2: phi^2 = a(-1)*1 + a(-2)*phi = 1 + phi,
n = -1: phi = a(0)*1 + a(-1)*phi = phi, (trivial)
n = 0: 1/phi^0 = a(1)*1 + a(0)*phi = 1, (trivial)
n = 1: 1/phi = a(2)*1 + a(1)*phi = -1 + phi,
n = 2: 1/phi^2 = a(3)*1 + a(2)*phi = 2 - phi, ... (End)
G.f. = x^-2 + x^-1 + x - x^2 + 2*x^3 - 3*x^4 + 5*x^5 - 8*x^6 + 13*x^7 - ...
MAPLE
a:= n-> (Matrix([[0, 1], [1, -1]])^n) [1, 2]: seq(a(n), n=-2..50); # Alois P. Heinz, Nov 01 2008
PROG
(PARI) a(n) = fibonacci(-n);
(Haskell)
a039834 n = a039834_list !! (n+2)
a039834_list = 1 : 1 : zipWith (-) a039834_list (tail a039834_list)
(SageMath)
x, y = 1, 1
while True:
yield x
x, y = y, x - y
(SageMath)
R.<t> = LaurentSeriesRing(ZZ, 't', default_prec = len)
f = (-2*t-1)/(t^4-t^3-t^2)
return f.list()
(Python)
from sympy import fibonacci
CROSSREFS
Cf. A000032, A000045, A001622, A038754, A011782, A055830, A130595, A215022, A215023, A227431, A343773.
AUTHOR
Alexander Grasser (pyropunk(AT)usa.net)
0, 1, 2, 16, 162, 3600, 147456, 12320100, 2058386904, 701841817600, 488286500625000, 696425232679321600, 2038348954317776486400, 12259459134020160144810000, 151596002479762016373851690400, 3855806813438155578522841251840000
COMMENTS
The obverse convolution of sequences
s = (s(0), s(1), ...) and t = (t(0), t(1), ...)
is introduced here as the sequence s**t given by
s**t(n) = (s(0)+t(n)) * (s(1)+t(n-1)) * ... * (s(n)+t(0)).
Swapping * and + in the representation s(0)*t(n) + s(1)*t(n-1) + ... + s(n)*t(0)
of ordinary convolution yields s**t.
If x is an indeterminate or real (or complex) variable, then for every sequence t of real (or complex) numbers, s**t is a sequence of polynomials p(n) in x, and the zeros of p(n) are the numbers -t(0), -t(1), ..., -t(n).
Following are abbreviations in the guide below for triples (s, t, s**t):
F = (0,1,1,2,3,5,...) = A000045, Fibonacci numbers L = (2,1,3,4,7,11,...) = A000032, Lucas numbers P = (2,3,5,7,11,...) = A000040, primes T = (1,3,6,10,15,...) = A000217, triangular numbers C = (1,2,6,20,70, ...) = A000984, central binomial coefficients LW = (1,3,4,6,8,9,...) = A000201, lower Wythoff sequence UW = (2,5,7,10,13,...) = A001950, upper Wythoff sequence [ ] = floor
In the guide below, sequences s**t are identified with index numbers Axxxxxx; in some cases, s**t and Axxxxxx differ in one or two initial terms.
Table 1. s = A000012 = (1,1,1,1...) = (1); The obverse convolutions listed in Table 1 are, trivially, divisibility sequences. Likewise, if s = (-1,-1,-1,...) instead of s = (1,1,1,...), then s**t is a divisibility sequence for every choice of t; e.g. if s = (-1,-1,-1,...) and t = A279019, then s**t = A130031. Table 2. s = A000027 = (0,1,2,3,4,5,...) = (n); Table 3. s = A000290 = (0,1,4,9,16,...) = (n^2); Table 4. s = t;
s = t = A000045, F s**s = this sequence If n = 2*k+1, then s**s(n) is a square; specifically,
s**s(n) = ((s(0)+s(n))*(s(1)+s(n-1))*...*(s(k)+s(k+1)))^2.
If n = 2*k, then s**s(n) has the form 2*s(k)*m^2, where m is an integer.
Table 5. Others
Table 6. Arrays of coefficients of s(x)**t(x), where s(x) and t(x) are polynomials
s(x) t(x) s(x)**t(x)
FORMULA
a(n) ~ c * phi^(3*n^2/4 + n) / 5^((n+1)/2), where c = QPochhammer(-1, 1/phi^2)^2/2 if n is even and c = phi^(1/4) * QPochhammer(-phi, 1/phi^2)^2 / (phi + 1)^2 if n is odd, and phi = A001622 is the golden ratio. - Vaclav Kotesovec, Aug 01 2024
EXAMPLE
a(0) = 0 + 0 = 0
a(1) = (0+1) * (1+0) = 1
a(2) = (0+1) * (1+1) * (1+0) = 2
a(3) = (0+2) * (1+1) * (1+1) * (2+0) = 16
As noted above, a(2*k+1) is a square for k>=0. The first 5 squares are 1, 16, 3600, 12320100, 701841817600, with corresponding square roots 1, 4, 60, 3510, 837760.
If n = 2*k, then s**s(n) has the form 2*F(k)*m^2, where m is an integer and F(k) is the k-th Fibonacci number; e.g., a(6) = 2*F(3)*(192)^2.
MAPLE
a:= n-> (F-> mul(F(n-j)+F(j), j=0..n))(combinat[fibonacci]):
MATHEMATICA
s[n_] := Fibonacci[n]; t[n_] := Fibonacci[n];
u[n_] := Product[s[k] + t[n - k], {k, 0, n}];
Table[u[n], {n, 0, 20}]
PROG
(PARI) a(n)=prod(k=0, n, fibonacci(k) + fibonacci(n-k)) \\ Andrew Howroyd, Jul 31 2024
Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).
+20
49
3, 3, 5, 9, 8, 8, 5, 6, 6, 6, 2, 4, 3, 1, 7, 7, 5, 5, 3, 1, 7, 2, 0, 1, 1, 3, 0, 2, 9, 1, 8, 9, 2, 7, 1, 7, 9, 6, 8, 8, 9, 0, 5, 1, 3, 3, 7, 3, 1, 9, 6, 8, 4, 8, 6, 4, 9, 5, 5, 5, 3, 8, 1, 5, 3, 2, 5, 1, 3, 0, 3, 1, 8, 9, 9, 6, 6, 8, 3, 3, 8, 3, 6, 1, 5, 4, 1, 6, 2, 1, 6, 4, 5, 6, 7, 9, 0, 0, 8, 7, 2, 9, 7, 0, 4
COMMENTS
André-Jeannin proved that this constant is irrational.
This constant does not belong to the quadratic number field Q(sqrt(5)) (Bundschuh and Väänänen, 1994). - Amiram Eldar, Oct 30 2020
REFERENCES
Daniel Duverney, Number Theory, World Scientific, 2010, 5.22, pp.75-76.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 358.
LINKS
Richard André-Jeannin, Problem H-450, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 29, No. 1 (1991), p. 89; Comparable, Solution to Problem H-450 by Paul S. Bruckman, ibid., Vol. 30, No. 2 (1992), p. 191-192. Paul S. Bruckman, Problem B-602, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 25, No. 3 (1987), p. 279; Fibonacci Infinite Series, Solution to Problem B-602 by C. Georghiou, ibid., Vol. 26, No. 3 (1988), pp. 281-282.
FORMULA
Alternating series representation: 3 + Sum_{k >= 1} (-1)^(k+1)/(F(k)*F(k+1)*F(k+2)). - Peter Bala, Nov 30 2013 Equals sqrt(5) * Sum_{k>=0} (1/(phi^(2*k+1) - 1) - 2*phi^(2*k+1)/(phi^(4*(2*k+1)) - 1)), where phi is the golden ratio ( A001622) (Greig, 1977). Equals sqrt(5) * Sum_{k>=0} (-1)^k/(phi^(2*k+1) - (-1)^k) (Griffin, 1992).
Equals 1 + c1*(c2 + 32*Integral_{x=0..infinity} f(x) dx),
f(x) = sin(x)*(4+cos(2*x))/((exp(Pi*x/log(phi))-1)*(2*cos(2*x)+3)*(7-2*cos(2*x))) (End)
Equals 3 + 2 * Sum_{k>=1} 1/(F(2*k-1)*F(2*k+1)*F(2*k+2)) (Bruckman, 1987).
Equals 2 + Sum_{k>=1} 1/ A350901(k) (André-Jeannin, Problem H-450, 1991). Equals sqrt(5/4)*Sum_{j>=1} i^(1-j)/sin(j*c) where c = Pi/2 + i*arccsch(2). - Peter Luschny, Nov 15 2023
EXAMPLE
3.35988566624317755317201130291892717968890513373...
MAPLE
Digits := 120: c := Pi/2 + I*arccsch(2):
Jeannin := n -> sqrt(5/4)*add(I^(1-j)/sin(j*c), j = 1..n):
MATHEMATICA
digits = 105; Sqrt[5]*NSum[(-1)^n/(GoldenRatio^(2*n + 1) - (-1)^n), {n, 0, Infinity}, WorkingPrecision -> digits, NSumTerms -> digits] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Apr 09 2013 *) (* Alternative: *)
First@RealDigits[Sqrt[5]/4 ((Log[5] + 2 QPolyGamma[1, 1/GoldenRatio^4] - 4 QPolyGamma[1, 1/GoldenRatio^2])/(2 Log[GoldenRatio]) + EllipticTheta[2, 0, 1/GoldenRatio^2]^2), 10, 105] (* Vladimir Reshetnikov, Nov 18 2015 *)
PROG
(PARI) /* Fast computation without splitting into even and odd indices, see the Arndt reference */
lambert2(x, a, S)=
{
/* Return G(x, a) = Sum_{n>=1} a*x^n/(1-a*x^n) (generalized Lambert series)
computed as Sum_{n=1..S} x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) )
As series in x correct up to order S^2.
We also have G(x, a) = Sum_{n>=1} a^n*x^n/(1-x^n) */
return( sum(n=1, S, x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) ) ) );
}
inv_fib_sum(p=1, q=1, S)=
{
/* Return Sum_{n>=1} 1/f(n) where f(0)=0, f(1)=1, f(n) = p*f(n-1) + q*f(n-1)
computed using generalized Lambert series.
Must have p^2+4*q > 0 */
my(al, be);
\\ Note: the q here is -q in the Horadam paper.
\\ The following numerical examples are for p=q=1:
al=1/2*(p+sqrt(p^2+4*q)); \\ == +1.6180339887498...
be=1/2*(p-sqrt(p^2+4*q)); \\ == -0.6180339887498...
return( (al-be)*( 1/(al-1) + lambert2(be/al, 1/al, S) ) ); \\ == 3.3598856...
}
default(realprecision, 100);
S = 1000; /* (be/al)^S == -0.381966^S == -1.05856*10^418 << 10^-100 */
inv_fib_sum(1, 1, S) /* 3.3598856... */ /* Joerg Arndt, Jan 30 2011 */ (PARI) suminf(k=1, 1/(fibonacci(k))) \\ Michel Marcus, Feb 19 2019 (SageMath) m=120; numerical_approx(sum(1/fibonacci(k) for k in (1..10*m)), digits=m) # G. C. Greubel, Feb 20 2019
CROSSREFS
Cf. A000045, A000796, A001622, A002163, A002390, A084119, A093540, A016628, A105199, A153386, A153387.
Third power of Fibonacci numbers ( A000045).
+20
35
0, 1, 1, 8, 27, 125, 512, 2197, 9261, 39304, 166375, 704969, 2985984, 12649337, 53582633, 226981000, 961504803, 4073003173, 17253512704, 73087061741, 309601747125, 1311494070536, 5555577996431, 23533806109393, 99690802348032, 422297015640625, 1788878864685457, 7577812474746632
COMMENTS
Divisibility sequence; that is, if n divides m, then a(n) divides a(m).
In general, cubing the terms of a Horadam sequence with signature (c,d) will result in a fourth-order recurrence with signature (c^3+2*c*d, c^4*d+3*(c*d)^2+2*d^3, -(c*d)^3-2*c*d^4, -d^6). - Gary Detlefs, Nov 12 2021 a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/3,2/3)-fences and third-squares (1/3 X 1 pieces, always placed so that the shorter sides are horizontal). A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/6,1/3)-fences and (1/6,5/6)-fences. - Michael A. Allen, Jan 11 2022
REFERENCES
Răzvan Gelca and Titu Andreescu, Putnam and Beyond, Springer 2007, pp. 99, 461-462.
Ralph P. Grimaldi, Fibonacci and Catalan Numbers: An Introduction, (2012). See Exercise 6 at p. 11.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 85, (exercise 1.2.8. Nr. 30) and p. 492 (solution).
FORMULA
G.f.: x*p(3, x)/q(3, x) with p(3, x) = Sum_{m=0..2} A056588(2, m)*x^m = 1 -2*x -x^2 and q(3, x) = Sum_{m=0..4} A055870(4, m)*x^m = 1 -3*x -6*x^2 +3*x^3 +x^4 = (1+x-x^2)*(1-4*x-x^2) (factorization deduced from Riordan result). Recursion (cf. Knuth's exercise): 1*a(n) -3*a(n-1) -6*a(n-2) +3*a(n-3) +1*a(n-4) = 0, n >= 4, a(0)=0, a(1)=a(2)=1, a(3)=2^3. See 5th row of signed Fibonomial triangle for coefficients: A055870(4, m), m=0..4 a(n) = (Fibonacci(3n) - 3(-1)^n*Fibonacci(n))/5. - Ralf Stephan, May 14 2004 a(n) and a(n+1) are found as rightmost and leftmost terms (respectively) in M^n * [1 0 0 0] where M = the 4 X 4 upper triangular Pascal's triangle matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., Ma(4) = 27, a(5) = 125. M^4 * [1 0 0 0] = [125 75 45 27]; where 75 = A066259(4) and 45 = A066258(3). The characteristic polynomial of M = x^4 - 3x^3 - 6x^2 + 3x + 1. a(n)/a(n-1) of the sequence and companions tend to 2+sqrt(5) = 4.2360679..., an eigenvalue of M and a root of the polynomial. - Gary W. Adamson, Oct 31 2004 Sum_{j=0..n} (-1)^j*binomial(n,j)*a(j) = ((-2)^n* A000045(n) - 3* A001906(n))/5. (End) G.f.: x*(1-2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)). - Colin Barker, Feb 28 2012 a(n) = F(n-2)*F(n+1)^2 + F(n-1)*(-1)^n. - J. M. Bergot, Mar 17 2016 a(n) = ((-3*(1/2*(-1-sqrt(5)))^n-(2-sqrt(5))^n+3*(1/2*(-1+sqrt(5)))^n+(2+sqrt(5))^n)) / (5*sqrt(5)). - Colin Barker, Jun 04 2016 a(n) = F(n-1)*F(n)*F(n+1) + F(n)*(-1)^(n-1). - Tony Foster III, Apr 11 2018 5*a(n) = L(2*n-1)*F(n+2) - L(2*n+1)*F(n-2) - 7*(-1)^n*F(n), where L(n) = A000032(n). - Peter Bala, Nov 12 2019 F(n+1)*F(n)*F(n-1) = 2*Sum_{j=1..n-1} P(j)*a(n-j) for n>0, where Pell number P(n) = A000129(n). - Michael A. Allen, Jan 11 2022 E.g.f.: (6*exp(-x/2)*sinh(sqrt(5)*x/2) + 2*exp(2*x)*sinh(sqrt(5)*x)) / (5*sqrt(5)).
a(n) = a(n-1) + a(n-2) + 3*F(n-2)*F(n-1)*F(n) for n > 1. (End)
a(n) = A014445(n-1) - a(n-1) + a(n+2), with a(0)=0 and a(1)=1 (see Gelca and Andreescu). - Stefano Spezia, Aug 21 2026
EXAMPLE
a(4) = 27 because the fourth Fibonacci number is 3 and 3^3 = 27.
a(5) = 125 because the fifth Fibonacci number is 5 and 5^3 = 125.
MAPLE
A056570 := proc(n) combinat[fibonacci](n)^3 ; end proc:
PROG
(PARI) concat(0, Vec(x*(1-2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)) + O(x^30))) \\ Colin Barker, Jun 04 2016 (SageMath) [fibonacci(n)^3 for n in (0..30)] # G. C. Greubel, Feb 20 2019
The first Fibonacci based variant of arithmetic derivative: a(p) = A000045(p) for prime p, a(u*v) = a(u)*v + u*a(v), with a(0) = a(1) = 0.
+20
34
0, 0, 1, 2, 4, 5, 7, 13, 12, 12, 15, 89, 20, 233, 33, 25, 32, 1597, 33, 4181, 40, 53, 189, 28657, 52, 50, 479, 54, 80, 514229, 65, 1346269, 80, 289, 3211, 100, 84, 24157817, 8381, 725, 100, 165580141, 127, 433494437, 400, 105, 57337, 2971215073, 128, 182, 125, 4825, 984, 53316291173, 135, 500, 188, 12581, 1028487, 956722026041, 160
FORMULA
a(n) = n * Sum e_j * A000045(p_j)/p_j for n = Product p_j^e_j.
MATHEMATICA
A328845[n_] := If[n <= 1, 0, n*Total[MapApply[#2*Fibonacci[#]/# &, FactorInteger[n]]]];
PROG
(PARI) A328845(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]*fibonacci(f[i, 1])/f[i, 1]));
a(n) is the index of F(n+1) at the unique occurrence of the ordered pair of reversed consecutive terms (F(n+1),F(n)) in Stern's diatomic sequence A002487, where F(k) denotes the k-th term of the Fibonacci sequence A000045.
+20
33
1, 3, 5, 13, 21, 53, 85, 213, 341, 853, 1365, 3413, 5461, 13653, 21845, 54613, 87381, 218453, 349525, 873813, 1398101, 3495253, 5592405, 13981013, 22369621, 55924053, 89478485, 223696213, 357913941, 894784853, 1431655765, 3579139413
COMMENTS
If the Fibonacci pairs are kept in the natural order (F(n),F(n+1)), it appears that the first term of the pair occurs in A002487 at the index given by A061547(n). Starting at n=3, begin subtracting from (2^(n-1)-1)/2^(n-1): 3/4 - 1/2 = 1/4 with 1+4=5=a(3); 7/8 - 1/4 = 5/8 with 5+8=13=a(4); 15/16 - 5/8 = 5/16 with 5+16=21= a(5); 31/32 - 5/16 = 21/32 with 21+32=53=a(6); 63/64 - 21/32 = 21/64 with 21+64=85=a(7) and so on. For n odd in the first fraction (2^(n-1)-1)/2^(n-1), the result approaches 1/3, and for n even in the first fraction, the result approaches 2/3. - J. M. Bergot, May 08 2015 Also, the decimal representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017
FORMULA
It appears that a(n)=(4^((n+1)/2)-1)/3 if n is odd and a(n)=(a(n-1)+a(n+1))/2 if n is even.
G.f.: (1+2*x-2*x^2)/((1-x)*(1-4*x^2)); a(n) = 2^(n-1)(3-(-1)^n/3)-1/3 (offset 0); a(n) = Sum{k=0..n+1, 4^floor(k/2)/2} (offset 0); a(2n) = A002450(n+1) (offset 0); a(2n+1) = A072197(n) (offset 0). - Paul Barry, May 21 2004 a(n+2) = 4*a(n) + 1, a(1) = 1, a(2) = 3, n > 0. - Yosu Yurramendi, Mar 07 2017
EXAMPLE
A002487 begins 0,1,1,2,1,3,2,... with offset 0. Thus a(1)=1 since (F(2),F(1)) = (1,1) occurs at term 1 of A002487. Similarly, a(2)=3 and a(3)=5, since (F(3),F(2))=(2,1) occurs at term 3 and (F(4),F(3))=(3,2) at term 5 of A002487.
MATHEMATICA
f[n_] := Module[{a = 1, b = 0, m = n}, While[m > 0, If[OddQ@ m, b = a + b, a = a + b]; m = Floor[m/2]]; b]; a = Table[f[n], {n, 0, 10^6}]; b = Reverse /@ Partition[Map[Fibonacci, Range[Ceiling@ Log[GoldenRatio, Max@ a] + 1]], 2, 1]; Map[If[Length@ # > 0, #[[1, 1]] - 1, 0] &@ SequencePosition[a, #] &, b] (* Michael De Vlieger, Mar 15 2017, Version 10.1, after Jean-François Alcover at A002487 *)
CROSSREFS
Positive terms of A096773 in ascending order.
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