A367834 a(n) = Product_{i=1..n, j=1..n} (i^8 + j^8).

1, 2, 67634176, 1775927682136440882473213952, 22495149450984565292579847926810488282934424886723006835982336

Offset: 0

Author

Vaclav Kotesovec, Dec 02 2023

Keywords

• nonn

Comments

Next term is too long to be included.

For m > 0, Product_{j=1..n, k=1..n} (j^m + k^m) ~ c(m) * exp(n*(n+1)*s(m) - m*n*(n-2)/2) * n^(m*(n^2 - 1/4 - v)), where v = 0 if m > 1 and v = 1/6 if m = 1, s(m) = Sum_{j>=1} (-1)^(j+1)/(j*(1 + m*j)) and c(m) is a constant (dependent only on m). Equivalently, s(m) = log(2) - HurwitzLerchPhi(-1, 1, 1 + 1/m).

c(1) = A / (2^(1/12) * exp(1/12) * sqrt(Pi)).

c(2) = exp(Pi/12) * Gamma(1/4) / (2^(5/4) * Pi^(5/4)).

c(3) = A * 3^(1/6) * exp(Pi/(6*sqrt(3)) - 1/12) * Gamma(1/3)^2 / (2^(7/4) * Pi^(13/6)), where A = A074962 is the Glaisher-Kinkelin constant.

c(4) = A306620.

Crossrefs

Cf. A079478 (m=1), A324403 (m=2), A324426 (m=3), A324437 (m=4), A324438 (m=5), A324439 (m=6), A324440 (m=7).

Cf. A306620, A367670, A367833.

Programs

Mathematica

Table[Product[i^8 + j^8, {i, 1, n}, {j, 1, n}], {n, 0, 6}]

Python

from math import prod, factorial
def A367834(n): return (prod(i**8+j**8 for i in range(1,n) for j in range(i+1,n+1))*factorial(n)**4)**2<Chai Wah Wu, Dec 02 2023

Formula

For n>0, a(n)/a(n-1) = A367833(n)^2 / (2*n^24).

a(n) ~ c * 2^(n*(n+1)) * (1 + 1/(sqrt(1 - 1/sqrt(2)) - 1/2))^(sqrt(2 + sqrt(2))*n*((n+1)/2)) * (1 + 1/(sqrt(1 + 1/sqrt(2)) - 1/2))^(sqrt(2 - sqrt(2))*n*((n+1)/2)) * (n^(8*n^2 - 2) / exp(12*n^2 - Pi*sqrt(1 + 1/sqrt(2))*n*(n+1))), where c = 0.043985703178712025347328240881106818917398444790454628282522057393529338998...