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G.f.: Sum_{n>=0} a(n)*x^n/(1+x)^(2*n^2) = 1+x.
3

%I #2 Mar 30 2012 18:37:21

%S 1,1,2,13,166,3324,92718,3354712,150206430,8050991676,504049958320,

%T 36172232930282,2931474921768206,265078092222575572,

%U 26480336590135734816,2898139377307388441520,345055687960080723910286

%N G.f.: Sum_{n>=0} a(n)*x^n/(1+x)^(2*n^2) = 1+x.

%F a(n) = number of subpartitions of the partition [0,1,6,15,28,...,2(n-1)^2-(n-1)] for n>0 with a(0)=1. See A115728 for the definition of subpartitions.

%e 1+x = 1 + 1*x/(1+x)^2 + 2*x^2/(1+x)^8 + 13*x^3/(1+x)^18 + 166*x^4/(1+x)^32 + 3324*x^5/(1+x)^50 + 92718*x^6/(1+x)^72 +...

%o (PARI) {a(n)=local(F=1/(1+x+x*O(x^n)));polcoeff(1+x-sum(k=0,n-1,a(k)*x^k*F^(2*k^2)),n)}

%Y Cf. A177447, A177449, A177450.

%K nonn

%O 0,3

%A _Paul D. Hanna_, May 09 2010