A015737 Number of 3's in partitions of n into distinct parts.

0, 0, 1, 1, 1, 1, 1, 2, 3, 4, 4, 5, 6, 8, 10, 12, 14, 17, 20, 24, 29, 34, 40, 47, 55, 64, 75, 87, 101, 117, 135, 155, 179, 205, 235, 269, 307, 350, 399, 453, 514, 583, 660, 746, 843, 950, 1070, 1205, 1354, 1520, 1705, 1910, 2138, 2392

Offset

1,8

Links

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (first 70 terms from Vincenzo Librandi)

Formula

G.f.: (x^3/(1 + x^3)) * Product_{j >= 1} (1 + x^j). - Emeric Deutsch, Apr 17 2006

Corresponding g.f. for "number of k's" is (x^k/(1 + x^k)) * Product_{j >= 1} (1 + x^j). - Joerg Arndt, Feb 20 2014

a(n) ~ exp(Pi*sqrt(n/3)) / (8*3^(1/4)*n^(3/4)). - Vaclav Kotesovec, Oct 30 2015

Example

a(9) = 3 because in the eight partitions of 9 into distinct parts, namely [9], [8, 1], [7, 2], [6, 3], [6, 2, 1], [5, 4], [5, 3, 1] and [4, 3, 2], only three contain 3.

Maple

g:=x^3*product(1+x^j, j=1..60)/(1+x^3): gser:=series(g, x=0, 57): seq(coeff(gser, x, n), n=1..54); # Emeric Deutsch, Apr 17 2006

Mathematica

Table[Count[Select[IntegerPartitions[n], Length[Union[#]] == Length[#] &], _?(MemberQ[#, 3] &)], {n, 60}] (* Harvey P. Dale, Aug 19 2011 *)
nmax = 100; Rest[CoefficientList[Series[x^3/(1 + x^3) * Product[1 + x^k, {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Oct 30 2015 *)

Crossrefs

Cf. A000009.

Sequence in context: A065328 A049877 A029063 * A015745 A017854 A261171

Adjacent sequences: A015734 A015735 A015736 * A015738 A015739 A015740

Keyword

nonn

Author

Clark Kimberling

Extensions

Example clarified by Harvey P. Dale, Aug 19 2011

Status

approved