A116646 Number of doubletons in all partitions of n. 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)] has two doubletons, shown between parentheses).

0, 0, 1, 0, 2, 2, 4, 5, 10, 11, 20, 25, 38, 49, 73, 91, 131, 167, 228, 291, 392, 493, 653, 822, 1065, 1336, 1714, 2131, 2706, 3354, 4209, 5193, 6471, 7934, 9817, 11990, 14725, 17909, 21875, 26477, 32172, 38797, 46893, 56339, 67804, 81147, 97260, 116017

Offset: 0

Author

Emeric Deutsch, Feb 20 2006

Keywords

• nonn

Comments

a(n) = (the number of 2's in all partitions of n) - (the number of 3's in all partitions of n). - Gregory L. Simay, Jul 28 2020

Examples

			a(6) = 4 because in the partitions of 6, namely [6],[5,1],[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] and [1,1,1,1,1,1], we have a total of 4 doubletons (shown between parentheses).
		

Links

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

Crossrefs

Cf. A116644. Column k=2 of A197126.

Programs

Maple

f:= x^2/(1+x)/(1-x^3)/product(1-x^j, j=1..70): fser:= series(f,x=0,70): seq(coeff(fser,x,n), n=0..55);

Mathematica

nmax = 50; CoefficientList[Series[x^2/((1+x)*(1-x^3)) * Product[1/(1-x^k), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Mar 07 2016 *)
Table[Sum[PartitionsP[n-6*m-2] - PartitionsP[n-6*m-3] + PartitionsP[n-6*m-4], {m, 0, Floor[n/6]}], {n, 0, 50}] (* Vaclav Kotesovec, Mar 07 2016 *)

Formula

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

a(n) = Sum_{k>=0} k * A116644(n,k).

a(n) ~ exp(Pi*sqrt(2*n/3)) / (3*2^(5/2)*Pi*sqrt(n)). - Vaclav Kotesovec, Mar 07 2016