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.

A307210 a(n) = Product_{i=1..n, j=1..n} (i^3 + j^3 + 1).

Original entry on oeis.org

1, 3, 5100, 305727048000, 7748770873210669158912000, 476007332700693200670745550306381336371200000, 272661655519533773844144991586798737775635133552905539740860416000000000
Offset: 0

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Author

Vaclav Kotesovec, Mar 28 2019

Keywords

Comments

Product_{i=1..n, j=1..n} (1 + 1/(i + j)) = A324444(n) / A079478(n) ~ 2^(2*n + 1) / (sqrt(Pi)*n^(3/2)).
Product_{i=1..n, j=1..n} (1 + 1/(i^2 + j^2)) = A324443(n) / A324403(n) ~ c * n^(Pi/2), where c = A306398 * 2^(3/4) * exp(-Pi/12) * Pi^(1/4) * Gamma(3/4) = 0.36753062884677326134620846786416595535234038999313315993144237973600...

Crossrefs

Programs

  • Maple
    a:= n-> mul(mul(i^3+j^3+1, i=1..n), j=1..n):
    seq(a(n), n=0..7);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Table[Product[i^3 + j^3 + 1, {i, 1, n}, {j, 1, n}], {n, 1, 8}]

Formula

a(n) ~ A307209 * A324426(n).
a(n) ~ c * A * 2^(2*n*(n+1) + 1/4) * exp(Pi*(n*(n+1) + 1/6)/sqrt(3) - 9*n^2/2 - 1/12) * n^(3*n^2 - 3/4) / (3^(5/6) * Pi^(1/6) * Gamma(2/3)^2), where c = A307209 = Product_{i>=1, j>=1} (1 + 1/(i^3 + j^3)) = 3.504782999339728375891120570... and A is the Glaisher-Kinkelin constant A074962.

Extensions

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

A307215 Decimal expansion of Product_{i>=1, j>=1} (1 + 1/(i^4 + j^4)).

Original entry on oeis.org

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

Views

Author

Vaclav Kotesovec, Mar 29 2019

Keywords

Comments

Product_{i=1..n, j=1..n} (1 + 1/(i + j)) = A324444(n) / A079478(n) ~ 2^(2*n + 1) / (sqrt(Pi)*n^(3/2)).
Product_{i=1..n, j=1..n} (1 + 1/(i^2 + j^2)) = A324443(n) / A324403(n) ~ c * n^(Pi/2), where c = A306398 * 2^(3/4) * exp(-Pi/12) * Pi^(1/4) * Gamma(3/4) = 0.36753062884677326134620846786416595535234038999313315993144237973600...
Product_{i>=1, j>=1} (1 + 1/(i^3 + j^3)) = A307209 = 3.50478299933972837589112...

Examples

			1.94073028537236152995386077599647772038707968293217092813061397472522642172...
		

Crossrefs

Programs

  • Mathematica
    (* The iteration cycle: *) $MaxExtraPrecision = 1000; funs[n_] := Product[1 + 1/(i^4 + j^4), {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}]

Formula

Equals limit_{n->infinity} (Product_{i=1..n, j=1..n} (1 + i^4 + j^4)) / A324437(n).
Showing 1-2 of 2 results.