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A359914
a(n) = coefficient of x^n in A(x) such that 2 = Sum_{n=-oo..+oo} (-1)^n * x^(n*(3*n-1)/2) * A(x)^(3*n) * (1 + x^n*A(x)).
6
1, 2, 4, 30, 154, 1078, 7046, 50766, 364268, 2713444, 20384884, 155954760, 1204192106, 9400024042, 73945396990, 586088682472, 4673927031694, 37484566094970, 302098932029282, 2445538771089012, 19875632898821430, 162118004651048048, 1326658157736876148
OFFSET
0,2
COMMENTS
Conjecture: a(n)/2 == A000041(n) (mod 2) for n >= 1; that is, a(n) is even, and a(n)/2 has the same parity as the number of partitions of n, for n >= 1.
Conjecture: a(n) == A361552(n+1) (mod 4) for n >= 0. - Paul D. Hanna, Mar 19 2023
LINKS
Eric Weisstein's World of Mathematics, Quintuple Product Identity.
FORMULA
G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following.
(1) 2 = Sum_{n=-oo..+oo} (-1)^n * x^(n*(3*n-1)/2) * A(x)^(3*n) * (1 + x^n*A(x)).
(2) 2 = Product_{n>=1} (1 - x^n) * (1 + x^(n-1)*A(x)) * (1 + x^n/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) ~ c * d^n / n^(3/2), where d = 8.762505108391427770669887... and c = 0.25785454119524349137288... - Vaclav Kotesovec, Jan 24 2023
Formula (2) can be rewritten as the functional equation QPochhammer(x) * QPochhammer(-y/x, x)/(1 + y/x) * QPochhammer(-1/y, x)/(1 + 1/y) * QPochhammer(y^2/x, x^2)/(1 - y^2/x) * QPochhammer(1/(x*y^2), x^2)/(1 - 1/(x*y^2)) = 2. - Vaclav Kotesovec, Jan 19 2024
EXAMPLE
G.f.: A(x) = 1 + 2*x + 4*x^2 + 30*x^3 + 154*x^4 + 1078*x^5 + 7046*x^6 + 50766*x^7 + 364268*x^8 + 2713444*x^9 + 20384884*x^10 + ...
where A = A(x) satisfies the doubly infinite sum
2 = ... + x^26/A^12*(1 + 1/x^4*A) - x^15/A^9*(1 + 1/x^3*A) + x^7/A^6*(1 + 1/x^2*A) - x^2/A^3*(1 + 1/x^1*A) + x^0*A^0*(1 + x^0*A) - x^1*A^3*(1 + x^1*A) + x^5*A^6*(1 + x^2*A) - x^12*A^9*(1 + x^3*A) + x^22*A^12*(1 + x^4*A) + ... + (-1)^n*x^(n*(3*n+1)/2)*A^(3*n)*(1 + x^n*A) + ...
also, by the Watson quintuple product identity,
2 = (1-x)*(1+1*A)*(1+x/A)*(1-x*A^2)*(1-x/A^2) * (1-x^2)*(1+x*A)*(1+x^2/A)*(1-x^3*A^2)*(1-x^3/A^2) * (1-x^3)*(1+x^2*A)*(1+x^3/A)*(1-x^5*A^2)*(1-x^5/A^2) * (1-x^4)*(1+x^3*A)*(1+x^4/A)*(1-x^7*A^2)*(1-x^7/A^2) * ...
SPECIFIC VALUES.
A(x) at x = 100/876 diverges.
A(100/877) = 1.557056751214068970380867667963285879403994350720494...
A(1/9) = 1.450191456209956107571253359997937360795442585014595870...
A(1/10) = 1.324252801492679846747365280925526932201317768972870665...
MATHEMATICA
(* Calculation of constant d: *) 1/r /. FindRoot[{r^3 * s^3 * QPochhammer[r] * QPochhammer[1/(r*s^2), r^2] * QPochhammer[-1/s, r] * QPochhammer[-s/r, r] * QPochhammer[s^2/r, r^2] / ((1 + s)*(r + s)*(-r + s^2)*(-1 + r*s^2)) == -2, (-3*r^2 - 2*r*(1 + r)*s + r^3*s^2 - s^4 + 2*r*(1 + r)*s^5 + 3*r*s^6) * Log[r] + (1 + s)*(r + s)*(r - s^2)*(-1 + r*s^2) * (QPolyGamma[0, Log[-1/s]/Log[r], r] + QPolyGamma[0, -1/2 - Log[s]/Log[r], r^2] - QPolyGamma[0, -1/2 + Log[s]/Log[r], r^2] - QPolyGamma[0, Log[-s/r]/Log[r], r]) == 0}, {r, 1/8}, {s, 2}, WorkingPrecision -> 70] (* Vaclav Kotesovec, Jan 19 2024 *)
PROG
(PARI) /* Using the doubly infinite series */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(2 - sum(m=-#A, #A, (-1)^m * x^(m*(3*m-1)/2) * Ser(A)^(3*m) * (1 + x^m*Ser(A)) ), #A-1) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))
(PARI) /* Using the quintuple product */
{a(n) = my(A=[1]); for(i=1, n, A = concat(A, 0);
A[#A] = polcoeff(2 - prod(m=1, #A, (1 - x^m) * (1 + x^(m-1)*Ser(A)) * (1 + x^m/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), ", "))
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jan 23 2023
STATUS
approved