A116645 Number of partitions of n having no doubletons. By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] of 18 has two doubletons, shown between parentheses).

1, 1, 1, 3, 3, 5, 8, 10, 13, 20, 26, 33, 46, 58, 75, 101, 125, 157, 206, 253, 317, 403, 494, 608, 760, 926, 1131, 1393, 1685, 2038, 2487, 2985, 3585, 4331, 5168, 6172, 7392, 8771, 10410, 12382, 14622, 17258, 20400, 23975, 28159, 33115, 38739, 45298, 53000

Offset

0,4

Comments

Number of partitions of n having no part that appears exactly twice.

Infinite convolution product of [1,1,0,1,1,1,1,1,1,1] aerated n-1 times. I.e., [1,1,0,1,1,1,1,1,1,1] * [1,0,1,0,0,0,1,0,1,0] * [1,0,0,1,0,0,0,0,0,1] * ... - Mats Granvik, Gary W. Adamson, Aug 07 2009

Links

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Formula

G.f.: Product_{j>=1} (1-x^(2j)+x^(3j))/(1-x^j).

G.f. for the number of partitions of n having no part that appears exactly m times is Product_{k>0} (1/(1-x^k)-x^(m*k)).

a(n) = A000041(n) - A183559(n) = A183568(n,0) - A183568(n,2). - Alois P. Heinz, Oct 09 2011

Example

a(4) = 3 because we have [4],[3,1] and [1,1,1,1] (the partitions [2,2] and [2,1,1] do not qualify since each of them has a doubleton).

Maple

h:=product((1-x^(2*j)+x^(3*j))/(1-x^j), j=1..60): hser:=series(h, x=0, 60): seq(coeff(hser, x, n), n=0..56);

Mathematica

nn=48; CoefficientList[Series[Product[1/(1-x^i)-x^(2i), {i, 1, nn}], {x, 0, nn}], x] (* Geoffrey Critzer, Sep 30 2013 *)

Crossrefs

Column 0 of A116644.

Cf. A000041, A007690, A116595, A183559, A183568.

Sequence in context: A015723 A333150 A342343 * A177739 A323581 A327731

Adjacent sequences: A116642 A116643 A116644 * A116646 A116647 A116648

Keyword

nonn

Author

Emeric Deutsch and Vladeta Jovovic, Feb 20 2006

Status

approved