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.

Previous Showing 11-14 of 14 results.

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...

A368721 a(n) = Product_{j=1..n, k=1..n} (j^4 + k^4 + n^4).

Original entry on oeis.org

1, 3, 940896, 18425962131085183248, 652934720004728520613911984092239003385856, 433324200327440062759688153700055880769227264159137063987248492437306880000
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 04 2024

Keywords

Comments

In general, for m>0, limit_{n->oo} (Product_{j=1..n, k=1..n} (j^m + k^m + n^m))^(1/(n^2)) / n^m = exp(Integral_{x=0..1, y=0..1} log(x^m + y^m + 1) dy dx) = 3 / exp(HurwitzLerchPhi(-1/2, 1, 1 + 1/m)/2 + Integral_{x=0..1} HurwitzLerchPhi(-1/(1 + x^m), 1, 1 + 1/m) / (1 + x^m) dx).

Crossrefs

Cf. A368685 (m=1), A368622 (m=2), A368720 (m=3).

Programs

  • Mathematica
    Table[Product[j^4 + k^4 + n^4, {j, 1, n}, {k, 1, n}], {n, 0, 6}]

Formula

Limit_{n->oo} a(n)^(1/(n^2)) / n^4 = exp(Integral_{x=0..1, y=0..1} log(x^4 + y^4 + 1) dy dx) = 1.35451345305131009729671041498902524074679186355643287514556358...

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).

A367679 a(n) = Product_{i=1..n, j=1..n} (i^4 - i^3*j + i^2*j^2 - i*j^3 + j^4).

Original entry on oeis.org

1, 1936, 1765124816400, 19271059559619728900751360000, 25048411180596698786915756280274804766474649600000000, 23045227505577134384745253646275782332295626096040088365089618773238077194240000000000
Offset: 1

Views

Author

Vaclav Kotesovec, Nov 26 2023

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Product[i^4 - i^3*j + i^2*j^2 - i*j^3 + j^4, {i, 1, n}, {j, 1, n}], {n, 1, 6}]
  • Python
    from math import prod, factorial
    def A367679(n): return (prod(i*(i*(i*(i-j)+j**2)-j**3)+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) = A324438(n) / A079478(n).
a(n) ~ c * n^(4*n^2 - 5/6) * phi^(sqrt(5)*n*(n+1)) / exp(6*n^2 - sqrt(phi)*Pi*n*(n+1)/5^(1/4)), where phi = A001622 is the golden ratio and c = 0.2505211390193028244009922677012518708897316924498037078191143761182342931773594...
Previous Showing 11-14 of 14 results.