A359921 a(n) = coefficient of x^n in A(x) such that 1/x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).

1, 1, 9, 80, 774, 8077, 89059, 1021106, 12048985, 145347965, 1784282449, 22217589408, 279934808090, 3562376922346, 45721210139842, 591139659619262, 7692224199601436, 100663182977093130, 1323944771879772911, 17491108974090887920, 232015023433972687373, 3088855705228007528177

Offset: 1

Author

Paul D. Hanna, Jan 22 2023

Keywords

• nonn

Examples

			G.f.: A(x) = x + x^2 + 9*x^3 + 80*x^4 + 774*x^5 + 8077*x^6 + 89059*x^7 + 1021106*x^8 + 12048985*x^9 + 145347965*x^10 + ...
where A = A(x) satisfies the doubly infinite sum
1/x = ... + x^12*(1/A^9 - A^8) + x^5*(1/A^6 - A^5) + x*(1/A^3 - A^2) + (1 - 1/A) + x^2*(A^3 - 1/A^4) + x^7*(A^6 - 1/A^7) + x^15*(A^9 - 1/A^10) + ... + x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)) + ...
also, by the Watson quintuple product identity,
1/x = (1-x)*(1-x*A)*(1-1/A)*(1-x*A^2)*(1-x/A^2) * (1-x^2)*(1-x^2*A)*(1-x/A)*(1-x^3*A^2)*(1-x^3/A^2) * (1-x^3)*(1-x^3*A)*(1-x^2/A)*(1-x^5*A^2)*(1-x^5/A^2) * (1-x^4)*(1-x^4*A)*(1-x^3/A)*(1-x^7*A^2)*(1-x^7/A^2) * ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..200
• Eric Weisstein's World of Mathematics, Quintuple Product Identity.

Crossrefs

Cf. A359920, A359924, A361050, A361052, A361538.

Programs

PARI

/* Using the doubly infinite series */
{a(n) = my(A=[0,1]); for(i=1,n, A = concat(A,0);
A[#A] = polcoeff(1/x - sum(m=-#A,#A, (Ser(A)^(3*m) - 1/Ser(A)^(3*m+1)) * x^(m*(3*m+1)/2) ),#A-4) ); A[n+1]}
for(n=1,30, print1(a(n),", "))

PARI

/* Using the quintuple product */
{a(n) = my(A=[0,1]); for(i=1,n, A = concat(A,0);
A[#A] = polcoeff(1/x - prod(m=1,#A, (1 - x^m) * (1 - x^m*Ser(A)) * (1 - x^(m-1)/Ser(A)) * (1 - x^(2*m-1)*Ser(A)^2) * (1 - x^(2*m-1)/Ser(A)^2) ),#A-4) ); A[n+1]}
for(n=1,30, print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following.

(1) 1/x = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).

(2) 1/x = Product_{n>=1} (1 - x^n) * (1 - x^n*A(x)) * (1 - x^(n-1)/A(x)) * (1 - x^(2*n-1)*A(x)^2) * (1 - x^(2*n-1)/A(x)^2), by the Watson quintuple product identity.

a(n) = Sum_{k=0..n-1} A361050(n,k) for n >= 1. - Paul D. Hanna, Mar 19 2023

a(n) ~ c * d^n / n^(3/2), where d = 14.308864552026948863076624... and c = 0.01145810893741095458355... - Vaclav Kotesovec, Mar 19 2023