A118083 Number of partitions of n such that largest part k occurs at least floor(k/2) times.

1, 1, 2, 3, 4, 5, 7, 8, 11, 13, 17, 20, 26, 30, 38, 45, 55, 64, 79, 91, 110, 128, 152, 176, 209, 240, 282, 325, 379, 434, 505, 576, 666, 760, 873, 993, 1139, 1290, 1473, 1668, 1897, 2141, 2430, 2736, 3095, 3481, 3925, 4404, 4958, 5550, 6232, 6968, 7805, 8710

Offset: 0

Author

Emeric Deutsch, Apr 12 2006

Keywords

• nonn

Comments

Also number of partitions of n such that if the number of parts is k, then the smallest part is at least floor(k/2). Example: a(8)=11 because we have [8],[7,1],[6,2],[5,3],[4,4],[6,1,1],[5,2,1],[4,3,1],[4,2,2],[3,3,2] and [2,2,2,2].

Also number of partitions of 2*n into distinct parts with either all parts odd or all parts even. - Vladeta Jovovic, Jul 03 2007

Examples

			a(8)=11 because we have [4,4],[3,3,2],[3,3,1,1],[3,2,2,1],[3,2,1,1,1],[3,1,1,1,1,1],[2,2,2,2],[2,2,2,1,1],[2,2,1,1,1,1],[2,1,1,1,1,1,1] and [1,1,1,1,1,1,1,1].
		

Crossrefs

Cf. A118082, A118084.

Programs

Maple

g:=sum(x^(k*floor(k/2))/product(1-x^j,j=1..k),k=1..15): gser:=series(g,x=0,65): seq(coeff(gser,x,n),n=0..60);

Formula

G.f.=sum(x^(k*floor(k/2))/product(1-x^j, j=1..k), k=1..infinity).

a(n) = A000700(2*n) + A000009(n), n>0. - Vladeta Jovovic, Jul 03 2007

a(n) ~ (2 + sqrt(2)) * exp(sqrt(n/3)*Pi) / (8*3^(1/4)*n^(3/4)). - Vaclav Kotesovec, Mar 06 2020