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A320954 G.f. A(x) satisfies: 1/(1-x) = Sum_{n>=0} (1+x)^(n^2) * x^n / A(x)^n. 1

%I #4 Oct 25 2018 16:10:14

%S 1,1,2,5,14,50,200,919,4633,25361,148606,923394,6043996,41447150,

%T 296571213,2206965193,17034374165,136066491764,1122656493744,

%U 9552206133005,83695193972045,754199756930791,6981787930209535,66327351641879318,646031757787129761,6445726513363688990,65825739028009602120,687540665329016479660

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

%H Paul D. Hanna, <a href="/A320954/b320954.txt">Table of n, a(n) for n = 0..200</a>

%F Given g.f. A(x) then

%F (1) 1/(1-x) = 1/(1 - q*x/(A(x) - q*(q^2-1)*x/(1 - q^5*x/(A(x) - q^3*(q^4-1)*x/(1 - q^9*x/(A(x) - q^5*(q^6-1)*x/(1 - q^13*x/(A(x) - q^7*(q^8-1)*x/(1 - ...))))))))), where q = (1+x), a continued fraction due to a partial elliptic theta function identity.

%F (2) 1/(1-x) = Sum_{n>=0} x^n/A(x)^n * (1+x)^n * Product_{k=1..n} (A(x) - x*(1+x)^(4*k-3)) / (A(x) - x*(1+x)^(4*k-1)), due to a q-series identity.

%e G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 50*x^5 + 200*x^6 + 919*x^7 + 4633*x^8 + 25361*x^9 + 148606*x^10 + 923394*x^11 + 6043996*x^12 + ...

%e such that

%e 1/(1-x) = 1 + (1+x)*x/A(x) + (1+x)^4*x^2/A(x)^2 + (1+x)^9*x^3/A(x)^3 + (1+x)^16*x^4/A(x)^4 + (1+x)^25*x^5/A(x)^5 + (1+x)^36*x^6/A(x)^6 + ...

%o (PARI) {a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0); A[#A] = -1 + Vec(sum(n=0, #A, ((1+x)^n +x*O(x^#A))^n * x^n/Ser(A)^n ) )[#A+1] ); A[n+1]}

%o for(n=0, 30, print1(a(n), ", "))

%Y Cf. A303058, A301929.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 25 2018

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