cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A324437 a(n) = Product_{i=1..n, j=1..n} (i^4 + j^4).

Original entry on oeis.org

1, 2, 18496, 189567553208832, 53863903330477722171391434817536, 4194051697335929481600368256016484482740174637152337920000, 530545265060440849231458462212366841894726534233233018777709463062563850450708386692464640000
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 28 2019

Keywords

Crossrefs

Programs

  • Maple
    a:= n-> mul(mul(i^4+j^4, i=1..n), j=1..n):
    seq(a(n), n=0..7);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Table[Product[i^4 + j^4, {i, 1, n}, {j, 1, n}], {n, 1, 6}]
  • Python
    from math import prod, factorial
    def A324437(n): return (prod(i**4+j**4 for i in range(1,n) for j in range(i+1,n+1))*factorial(n)**2)**2<Chai Wah Wu, Nov 26 2023

Formula

a(n) ~ c * 2^(n*(n+1)) * exp(Pi*n*(n+1)/sqrt(2) - 6*n^2) * (1 + sqrt(2))^(sqrt(2)*n*(n+1)) * n^(4*n^2 - 1), where c = A306620 = 0.23451584451404279281807143317500518660696293944961...
For n>0, a(n)/a(n-1) = A272247(n)^2 / (2*n^12). - Vaclav Kotesovec, Dec 01 2023

Extensions

a(0)=1 prepended by Alois P. Heinz, Jun 24 2023

A307209 Decimal expansion of Product_{i>=1, j>=1} (1 + 1/(i^3 + j^3)).

Original entry on oeis.org

3, 5, 0, 4, 7, 8, 2, 9, 9, 9, 3, 3, 9, 7, 2, 8, 3, 7, 5, 8, 9, 1, 1, 2, 0, 5, 7, 0, 4, 3, 8, 0, 6, 1, 2, 5, 5, 8, 3, 8, 9, 3, 2, 4, 7, 8, 6, 2, 7, 1, 2, 7, 5, 3, 5, 4, 1, 9, 9, 4, 6, 2, 6, 6, 1, 4, 0, 5, 8, 3, 8, 5, 0, 3, 5, 0, 3, 4, 7, 5, 6, 3, 5, 2, 7, 4, 7, 5, 0, 9, 5, 0, 5, 1, 3, 7, 8, 9, 1, 7, 8, 4, 5, 9, 7
Offset: 1

Views

Author

Vaclav Kotesovec, Mar 28 2019

Keywords

Comments

Product_{i>=1, j>=1} (1 + 1/(i^2 + j^2)) is divergent.
A324443(n) / A324403(n) ~ c * n^(Pi/2), where c = A306398 * 2^(3/4) * exp(-Pi/12) * Pi^(1/4) * Gamma(3/4) = 0.36753062884677326134620846786416595535234038999313...
Product_{i=1..n, j=1..n} (1 + 1/(i + j)) = A324444(n) / A079478(n) ~ 2^(2*n + 1) / (sqrt(Pi)*n^(3/2)).

Examples

			3.50478299933972837589112057043806125583893247862712753541994626614058385...
		

Crossrefs

Programs

  • Mathematica
    (* The iteration cycle: *) $MaxExtraPrecision = 1000; funs[n_] := Product[1 + 1/(i^3 + j^3), {i, 1, n}, {j, 1, n}]; Do[Print[N[Sum[(-1)^(m + j)*funs[j*Floor[200/m]] * j^(m - 1)/(j - 1)!/(m - j)!, {j, 1, m}], 100]], {m, 10, 100, 10}]
  • PARI
    default(realprecision, 50); exp(sumalt(k=1, -(-1)^k/k*sumnum(i=1, sumnum(j=1, 1/(i^3+j^3)^k)))) \\ 15 decimals correct

Formula

Equals limit_{n->infinity} A307210(n) / A324426(n).
Showing 1-2 of 2 results.