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Number of partitions of n into distinct terms of (1,2)-Ulam sequence, cf. A002858.
+20
7
1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 14, 14, 16, 16, 17, 19, 20, 22, 23, 25, 26, 27, 29, 30, 31, 34, 35, 38, 40, 41, 45, 45, 48, 51, 52, 57, 60, 62, 66, 68, 71, 75, 78, 83, 86, 93, 97, 100, 107, 109, 115, 120, 124, 132, 138
EXAMPLE
The first terms of A002858 are 1, 2, 3, 4, 6, 8, 11, 13, 16, 18, ... a(10) = #{8+2, 6+4, 6+3+1, 4+3+2+1} = 4;
a(11) = #{11, 8+3, 8+2+1, 6+4+1, 6+3+2} = 5;
a(12) = #{11+1, 8+4, 8+3+1, 6+4+2, 6+3+2+1} = 5.
PROG
(Haskell)
a199017 = p a002858_list where
p _ 0 = 1
p (u:us) m | m < u = 0
| otherwise = p us (m - u) + p us m
Number of partitions of n into terms of (1,2)-Ulam sequence, cf. A002858.
+20
5
1, 1, 2, 3, 5, 6, 10, 12, 18, 22, 30, 37, 50, 60, 78, 94, 120, 143, 179, 213, 262, 309, 376, 440, 531, 618, 737, 855, 1012, 1167, 1372, 1575, 1840, 2104, 2442, 2783, 3214, 3649, 4193, 4746, 5430, 6126, 6980, 7853, 8914, 10002, 11311, 12660, 14274, 15934
EXAMPLE
The first terms of A002858 are 1, 2, 3, 4, 6, 8, 11, 13, 16, 18, ... a(6) = #{6, 4+2, 4+1+1, 3+3, 3+2+1, 3+1+1+1, 2+2+2, 2+2+1+1, 2+1+1+1+1, 1+1+1+1+1+1} = 10;
a(7) = #{6+1, 4+3, 4+2+1, 4+1+1+1, 3+3+1, 3+2+2, 3+2+1+1, 3+1+1+1+1, 2+2+2+1, 2+2+1+1+1, 2+1+1+1+1+1, 1+1+1+1+1+1+1} = 12.
PROG
(Haskell)
199016 = p a002858_list where
p _ 0 = 1
p us'@(u:us) m | m < u = 0
| otherwise = p us' (m - u) + p us m
First differences of Ulam's sequence A002858.
+20
4
1, 1, 1, 2, 2, 3, 2, 3, 2, 8, 2, 8, 2, 9, 1, 5, 4, 5, 7, 3, 5, 5, 5, 10, 2, 3, 4, 8, 12, 5, 7, 7, 3, 7, 20, 2, 3, 2, 7, 8, 9, 3, 10, 2, 15, 2, 3, 2, 10, 5, 2, 13, 9, 27, 7, 3, 5, 15, 2, 15, 2, 5, 7, 12, 8, 10, 2, 7, 3, 2, 15, 2, 3, 7, 10, 5, 27, 2, 12, 5, 20, 2, 20, 2, 20, 2, 17, 17, 3, 2, 5, 12, 3
COMMENTS
No pattern in this sequence has ever been observed, according to Finch.
In the first 49,999,999 terms, 1 occurs four times, 36.9% of the terms are 2, and some terms have not occurred. - Jud McCranie, Mar 02 2012 Even though some values seem to get larger as the sequence grows, it appears that the combined frequency of the values 2 and 3 approaches 50%. This shows up even in the first 10000 terms (combined frequency: 49.9%). - Enrique Navarrete, May 08 2017
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 145-151.
Index in A002858 of the smaller of the two Ulam numbers that sum to the n-th Ulam number.
+20
4
0, 0, 1, 1, 2, 2, 3, 2, 3, 2, 6, 2, 6, 2, 7, 1, 5, 4, 11, 9, 3, 6, 8, 10, 12, 2, 3, 4, 6, 18, 19, 13, 16, 3, 17, 20, 2, 3, 2, 24, 6, 11, 3, 8, 2, 15, 2, 3, 2, 15, 20, 2, 8, 27, 13, 25, 3, 6, 18, 2, 15, 2, 15, 25, 11, 6, 10, 2, 17, 3, 2, 15, 2, 3, 27, 20, 15, 20, 2, 25, 27, 42, 2, 15
COMMENTS
The first two are zero because the Ulam sequence ( A002858) is initialized for those terms. The terms of the sum are in A080328 and A080329 and the indices are in A080330 and A080331.
EXAMPLE
The 11th Ulam number (26) is the sum of the 6th Ulam number (8) and the 10th Ulam number (18), so a(11)=6.
PROG
(PARI) See Links section.
Index in A002858 of the larger of the two Ulam numbers that sum to the n-th Ulam number.
+20
4
0, 0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 13, 15, 15, 17, 13, 17, 20, 20, 20, 20, 20, 25, 26, 27, 28, 20, 20, 27, 25, 33, 27, 28, 36, 37, 38, 27, 40, 38, 42, 42, 44, 40, 46, 47, 48, 42, 40, 51, 52, 38, 53, 44, 56, 56, 54, 59, 55, 61, 56, 53, 61, 65, 65, 67, 61, 69, 70
COMMENTS
The first two terms in this sequence are zero because the Ulam sequence ( A002858) is initialized for those terms. The terms of the sum are in A080328 and A080329 and the indices are in A080330 and A080331.
EXAMPLE
The 11th Ulam number (26) is the sum of the 6th Ulam number (8) and the 10th Ulam number (18), so a(11)=10.
PROG
(PARI) See Links section.
Indices of prime Ulam numbers: numbers k such that A002858(k) is prime.
+20
3
2, 3, 7, 8, 15, 17, 25, 31, 41, 48, 69, 73, 91, 97, 106, 107, 123, 125, 138, 167, 172, 177, 181, 193, 194, 241, 242, 246, 267, 280, 286, 287, 297, 306, 312, 322, 323, 338, 340, 343, 353, 354, 382, 388, 393, 398, 403, 411, 412, 415, 416, 433, 444, 448, 460
MATHEMATICA
ulams = {1, 2}; Do[AppendTo[ulams, n = Last[ulams]; While[n++; Length[ DeleteCases[Intersection[ulams, n - ulams], n/2, 1, 1]] != 2]; n], {100}]; Flatten[Position[ulams, _?(PrimeQ[#] &)]] (* after Jean-François Alcover at A002858 *)
Numbers that are not differences between successive Ulam numbers ( A002858).
+20
2
COMMENTS
Unfortunately, it appears that the entries are all conjectural. I am not aware of any proofs that the listed numbers can never appear. The terms shown are those listed by Pickover. - N. J. A. Sloane, Jun 19 2008 However, a gap of 35 doesn't occur until after the 35755308th term (483379914), so these missing gaps could eventually occur. - Jud McCranie, Jun 13 2008 The sequence is conjectured to continue: 23, 26, 28, 31, 33. These gaps do not appear in the first 158000000 terms, so are candidates for the sequence. [ Jud McCranie, Sep 12 2013] The conjectured list of gaps that don't appear holds through the first 28 billion Ulam numbers. - Jud McCranie, Jan 07 2016
REFERENCES
Clifford A. Pickover, "Wonders of Numbers, ...", Oxford University Press, 2000
LINKS
Shyam Sunder Gupta, Ulam Numbers. In: Exploring the Beauty of Fascinating Numbers. Springer Praxis Books(). Springer, Singapore, (2025). C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review
For n => 1, the number of distinct summands u and v that can be used in the representation of n as u+v, where u and v are two (possibly equal) Ulam numbers A002858.
+20
2
0, 1, 2, 3, 4, 3, 4, 3, 4, 4, 2, 5, 2, 6, 4, 3, 6, 2, 8, 4, 4, 5, 0, 6, 0, 3, 4, 2, 8, 4, 4, 5, 0, 6, 0, 3, 4, 2, 8, 4, 4, 6, 0, 8, 0, 4, 2, 2, 8, 4, 6, 5, 2, 10, 4, 7, 2, 4, 6, 4, 6, 2, 6, 10, 6, 8, 0, 4, 2, 6, 4, 3, 10, 6, 10, 5, 2, 6, 4, 8, 4, 2, 10, 4, 12
COMMENTS
An odd number in the sequence means that there exists the "pseudo-representation" u + u, where u is an Ulam number. For example, a(22)=5 since 22 = 18 + 4 = 16 + 6 = 11 + 11, and the 5 distinct summands 18,4,16,6,11 are Ulam numbers.
Note that both 2 and 3 are values for Ulam numbers since, by the previous comment, a value of 3 means that the Ulam number has the additional "pseudo-representation" u + u (see the Examples).
It seems that all nonnegative integers occur as values of this sequence.
EXAMPLE
a(23) = 0 since 23 can't be written as the sum of two distinct Ulam numbers. This type of numbers are in A033629. a(94) = 1 since 94 = 47 + 47, where 47 is an Ulam number. This type of numbers are in A287611. a(11) = 2 since 11 has the unique representation 11 = 8 + 3, where 8,3 are Ulam numbers. If such n is also an Ulam number (such as 11), then it is in A002858. a(8) = 3 since it has the representation 8 = 6 + 2 and also the additional "pseudo-representation" 8 = 4 + 4, where 6, 2, and 4 are Ulam numbers. If n has such a "pseudo-representation" and is an Ulam number, then it is in A068799.
Indices of twin Ulam primes: numbers k such that both u(k) and u(k+1) are primes, where u(k) = A002858(k) are the Ulam numbers.
+20
2
2, 7, 106, 193, 241, 286, 322, 353, 411, 415, 753, 858, 859, 1086, 1164, 1305, 1547, 1548, 1625, 1631, 1648, 1678, 1896, 1972, 2007, 2103, 2406, 2503, 2515, 2516, 2530, 2553, 2638, 2714, 3003, 3059, 3060, 3337, 3903, 4012, 4072, 4299, 4386, 4404, 4625, 4698
MATHEMATICA
ulams = {1, 2}; Do[AppendTo[ulams, n = Last[ulams]; While[n++; Length[ DeleteCases[Intersection[ulams, n - ulams], n/2, 1, 1]] != 2]; n], {1000}]; p=PrimeQ[ulams]; len=Length[p]; s={}; Do[If[p[[n]]&&p[[n+1]], AppendTo[s, n]], {n, 1, len-1}]; s (* after Jean-François Alcover at A002858 *)
Number of Ulam numbers u ( A002858) between powers of 2, 2^n < u <= 2^(n+1).
+20
2
1, 2, 2, 3, 3, 7, 11, 20, 31, 47, 92, 162, 312, 632, 1235, 2460, 4844, 9665, 19335, 38727, 77569, 155729, 310405, 620596, 1240580, 2481645, 4966229, 9926596, 19855760, 39717367, 79428417
COMMENTS
Conjecture 1: For all m > 1 there is always at least one Ulam number u(j) such that m < u(j) < 2m.
Conjecture 2: For all m > 4 there is always at least two Ulam numbers u(j), u(j+1) such that m < u(j) < u(j+1) < 2m.
This sequence illustrates how far these conjectures are oversatisfied.
Conjecture 1 implies that Ulam numbers form a complete sequence because u(1) = 1 and 2u(j) >= u(j+1).
Conjecture 2 implies that three consecutive Ulam numbers satisfies the triangle inequality because 2u(j) > u(j+2) > u(j+1) > u(j) and u(j) + u(j+1) > 2u(j) > u(j+2). It further implies that n consecutive Ulam numbers can always form an n-gon.
EXAMPLE
a(6) = 11 because the Ulam numbers between 64 and 128 are (69, 72, 77, 82, 87, 97, 99, 102, 106, 114, 126).
MATHEMATICA
ulams = {1, 2}; Do[AppendTo[ulams, n = Last[ulams]; While[n++; Length[DeleteCases[Intersection[ulams, n-ulams], n/2, 1, 1]]!=2]; n], {10^4}]; ulst = ulams; (* Jean-François Alcover, Sep 08 2011 *) upi[n_] := Module[{p = 1}, While[ulst[[p]] <= n, p++]; p - 1]; Table[upi[2^(n + 1)] - upi[2^n], {n, 0, 16}]
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