A359919 a(n) = coefficient of x^n in A(x) such that x^2 = Sum_{n=-oo..+oo} x^(n*(3*n+1)/2) * (A(x)^(3*n) - 1/A(x)^(3*n+1)).

1, 0, 1, 5, 19, 65, 211, 681, 2255, 7830, 28786, 111230, 443789, 1795972, 7284981, 29466755, 118834438, 479034654, 1936617163, 7872885832, 32226147305, 132808096158, 550444192577, 2291095125465, 9564074472264, 40005894288101, 167610376198140, 703308153554903

Offset: 0

Author

Paul D. Hanna, Jan 22 2023

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x^2 + 5*x^3 + 19*x^4 + 65*x^5 + 211*x^6 + 681*x^7 + 2255*x^8 + 7830*x^9 + 28786*x^10 + 111230*x^11 + 443789*x^12 + ...
where A = A(x) satisfies the doubly infinite sum
x^2 = ... + 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,
x^2 = (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 = 0..200
• Eric Weisstein's World of Mathematics, Quintuple Product Identity.

Crossrefs

Cf. A359920, A359921, A359924, A359915, A359916.

Cf. A359914.

Programs

PARI

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

PARI

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

Formula

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

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

(2) x^2 = 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.