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 1-4 of 4 results.

A306594 a(n) = Product_{i=1..n, j=1..n, k=1..n} (i + j + k).

Original entry on oeis.org

1, 3, 144000, 455282248974336000000, 9608917807566747651759509633033255126040576000000000000
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 27 2019

Keywords

Comments

Next term is too long to be included.

Crossrefs

Programs

  • Maple
    a:= n-> mul(mul(mul(i+j+k, i=1..n), j=1..n), k=1..n):
    seq(a(n), n=0..5);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Table[Product[i+j+k, {i, 1, n}, {j, 1, n}, {k, 1, n}], {n, 1, 6}]
    Table[Product[k^(3*(n - k + 1) (n - k + 2)/2), {k, 1, n}] * Product[k^((3*n - k + 1) (3*n - k + 2)/2), {k, 1, 3*n}] / Product[k^(3*(2*n - k + 1) (2*n - k + 2)/2), {k, 1, 2*n}], {n, 1, 6}]
    Clear[a]; a[n_] := a[n] = If[n == 1, 3, 3*n*a[n-1] * BarnesG[2+n]^3 * BarnesG[2+3*n]^3 * Gamma[1+2*n]^3 / (BarnesG[2+2*n]^6 * Gamma[1+3*n]^3)]; Table[a[n], {n, 1, 6}] (* Vaclav Kotesovec, Mar 28 2019 *)

Formula

a(n) = Product_{k=1..n} (BarnesG(k+2) * BarnesG(2*n+k+2) / BarnesG(n+k+2)^2).
a(n) = Product_{k=1..n} (k^(3*(n - k + 1)*(n - k + 2)/2)) * Product_{k=1..3*n} (k^((3*n - k + 1)*(3*n - k + 2)/2)) / Product_{k=1..2*n} (k^(3*(2*n - k + 1)*(2*n - k + 2)/2)).
a(n) ~ sqrt(Pi) * 3^(9*n^3/2 + 27*n^2/4 + 3*n + 3/8) * n^(n^3 + 3/8) / (A^(3/2) * 2^(4*n^3 + 9*n^2 + 6*n + 5/8) * exp(11*n^3/6 - Zeta(3)/(8*Pi^2) - 1/8)), where A is the Glaisher-Kinkelin constant A074962.

Extensions

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

A324425 a(n) = Product_{i=1..n, j=1..n, k=1..n} (i^2 + j^2 + k^2).

Original entry on oeis.org

1, 3, 5668704, 550388591715704109656479285248, 152455602303300418998634460043817052571893573096619261814850281699755319515987050496
Offset: 0

Views

Author

Vaclav Kotesovec, Feb 27 2019

Keywords

Comments

(a(n)^(1/n^3))/n^2 tends to 0.828859579669279... = A306617.

Crossrefs

Programs

  • Maple
    a:= n-> mul(mul(mul(i^2+j^2+k^2, i=1..n), j=1..n), k=1..n):
    seq(a(n), n=0..5);  # Alois P. Heinz, Jun 24 2023
  • Mathematica
    Table[Product[i^2+j^2+k^2, {i, 1, n}, {j, 1, n}, {k, 1, n}], {n, 1, 6}]
    Clear[a]; a[n_] := a[n] = If[n == 1, 3, a[n-1] * Product[k^2 + j^2 + n^2, {j, 1, n}, {k, 1, n}]^3 * (3*n^2) / (Product[k^2 + 2*n^2, {k, 1, n}]^3)]; Table[a[n], {n, 1, 6}] (* Vaclav Kotesovec, Mar 27 2019 *)

Extensions

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

A368723 a(n) = Product_{i=1..n, j=1..n, k=1..n} (i^4 + j^4 + k^4).

Original entry on oeis.org

1, 3, 30180180096, 130911253854794147456410254996552949923277899497472
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 04 2024

Keywords

Comments

Next term is too long to be included.
In general, for m>0, limit_{n->oo} (Product_{i=1..n, j=1..n, k=1..n} (i^m + j^m + k^m))^(1/(n^3)) / n^m = exp(Integral_{x=0..1, y=0..1, z=0..1} log(x^m + y^m + z^m) dz dy dx) = exp(Integral_{x=0..1, y=0..1} (log(1 + x^k + y^k) - k + k*hypergeom2F1(1/k, 1, (k+1)/k, -1/(x^k + y^k))) dy dx).

Crossrefs

Cf. A306594 (m=1), A324425 (m=2), A368722 (m=3).

Programs

  • Mathematica
    Table[Product[i^4 + j^4 + k^4, {i, 1, n}, {j, 1, n}, {k, 1, n}], {n, 0, 5}]

Formula

Limit_{n->oo} a(n)^(1/(n^3)) / n^4 = exp(Integral_{x=0..1, y=0..1, z=0..1} log(x^4 + y^4 + z^4) dz dy dx) = 0.3570458697635761757481417...

A368720 a(n) = Product_{j=1..n, k=1..n} (j^3 + k^3 + n^3).

Original entry on oeis.org

1, 3, 69360, 1522177267723200, 391047538356893112890665992192000, 6785985430272886334677590546861463643140253138288640000000
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 04 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Product[j^3 + k^3 + n^3, {j, 1, n}, {k, 1, n}], {n, 0, 6}]

Formula

Limit_{n->oo} a(n)^(1/(n^2)) / n^3 = exp(Integral_{x=0..1, y=0..1} log(x^3 + y^3 + 1) dy dx) = 1.4501318426779115107161463152054835452511102493859148975306452...
Showing 1-4 of 4 results.