A117524 Total number of parts of multiplicity 3 in all partitions of n.

0, 0, 1, 0, 1, 2, 3, 3, 7, 8, 13, 17, 25, 32, 48, 59, 83, 108, 145, 183, 247, 310, 406, 512, 659, 824, 1055, 1307, 1651, 2047, 2558, 3146, 3913, 4788, 5904, 7202, 8821, 10707, 13054, 15770, 19118, 23027, 27775, 33312, 40029, 47835, 57231, 68182, 81261

Offset: 1

Author

Vladeta Jovovic, Apr 26 2006

Keywords

• easy
• nonn

Examples

			a(9) = 7 because among the 30 (=A000041(9)) partitions of 9 only [6,(1,1,1)],[4,2,(1,1,1)],[(3,3,3)],[3,3,(1,1,1)],[3,(2,2,2)],[(2,2,2),(1,1,1)] contain parts of multiplicity 3 and their total number is 7 (shown between parentheses)
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A024786, A116646. Column k=3 of A197126.

Programs

Maple

g:=(x^3/(1-x^3)-x^4/(1-x^4))/product(1-x^i,i=1..65): gser:=series(g,x=0,62): seq(coeff(gser,x,n),n=1..58); # Emeric Deutsch, Apr 29 2006

Formula

G.f. for total number of parts of multiplicity m in all partitions of n is (x^m/(1-x^m)-x^(m+1)/(1-x^(m+1)))/Product(1-x^i,i=1..infinity).

a(n) = Sum(k*A118806(n,k), k>=0). - Emeric Deutsch, Apr 29 2006

a(n) ~ exp(Pi*sqrt(2*n/3)) / (24*Pi*sqrt(2*n)). - Vaclav Kotesovec, May 24 2018