%I #27 Mar 19 2015 20:43:38
%S 1,0,1,3,7,10,20,24,45,61,103,140,246,325,517,728,1086,1472,2184,2918,
%T 4197,5638,7875,10497,14625,19272,26354,34804,46992,61490,82471,
%U 107163,142128,184141,241701,311282,406164,519755,672726,858110,1102872
%N Number of partitions of n in which two summands (of each size) are designated.
%H Alois P. Heinz, <a href="/A255180/b255180.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: Product_{n>=1} 1 + x^(2*n)/(1 - x^n)^3.
%e a(4)=7. In order to designate two summands of each size, the multiplicity of each summand must be at least two. For n=4 we consider only the partitions 2+2 and 1+1+1+1. In the first case there is 1 way and in the second case there are 6 ways. 1 + 6 = 7.
%p b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
%p b(n, i-1)+add(b(n-i*j, i-1)*binomial(j, 2), j=2..n/i)))
%p end:
%p a:= n-> b(n$2):
%p seq(a(n), n=0..50); # _Alois P. Heinz_, Mar 19 2015
%t nn = 40; CoefficientList[Series[Product[1 + x^(2 n)/(1 - x^n)^3, {n, 1, nn}], {x, 0, nn}], x]
%Y Cf. A077285, A070933 (where any number of summands of any size are designated).
%K nonn
%O 0,4
%A _Geoffrey Critzer_, Mar 19 2015