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

1, 1, 2, 5, 14, 50, 200, 919, 4633, 25361, 148606, 923394, 6043996, 41447150, 296571213, 2206965193, 17034374165, 136066491764, 1122656493744, 9552206133005, 83695193972045, 754199756930791, 6981787930209535, 66327351641879318, 646031757787129761, 6445726513363688990, 65825739028009602120, 687540665329016479660

Offset: 0

Author

Paul D. Hanna, Oct 25 2018

Keywords

• nonn

Examples

			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 + ...
such that
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 + ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A303058, A301929.

Programs

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]}
for(n=0, 30, print1(a(n), ", "))

Formula

Given g.f. A(x) then

(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.

(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.