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 1-3 of 3 results.

A346975 Expansion of e.g.f. log( 1 + (exp(x) - 1)^3 / 3! ).

Original entry on oeis.org

1, 6, 25, 80, 91, -1694, -22875, -193740, -1119569, -768394, 101162425, 1930987240, 23583202371, 181575384906, -306743537075, -45405986594980, -1070132302146089, -16439720013909794, -145808623945689375, 1048196472097011600, 84226169502099763051
Offset: 3

Views

Author

Ilya Gutkovskiy, Aug 09 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 23; CoefficientList[Series[Log[1 + (Exp[x] - 1)^3/3!], {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 3] &
    a[n_] := a[n] = StirlingS2[n, 3] - (1/n) Sum[Binomial[n, k] StirlingS2[n - k, 3] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 3, 23}]

Formula

a(n) = Stirling2(n,3) - (1/n) * Sum_{k=1..n-1} binomial(n,k) * Stirling2(n-k,3) * k * a(k).
a(n) ~ -(n-1)! * 2^(n+1) * cos(n*arctan(2*arctan(3^(5/6)/(2^(2/3) + 3^(1/3))) / log(1 + 6^(1/3) + 6^(2/3)))) / (4*arctan(3^(5/6)/(2^(2/3) + 3^(1/3)))^2 + log(1 + 6^(1/3) + 6^(2/3))^2)^(n/2). - Vaclav Kotesovec, Aug 09 2021
a(n) = Sum_{k=1..floor(n/3)} (-1)^(k-1) * (3*k)! * Stirling2(n,3*k)/(k * 6^k). - Seiichi Manyama, Jan 23 2025

A346954 Expansion of e.g.f. -log( 1 - (exp(x) - 1)^4 / 4! ).

Original entry on oeis.org

1, 10, 65, 350, 1736, 9030, 60355, 561550, 6188996, 69919850, 781211795, 8854058850, 106994019406, 1433756147470, 21287253921635, 339206526695750, 5630710652048216, 96341917117951890, 1708973354556320875, 31787279786739738250, 623964823224788294426
Offset: 4

Views

Author

Ilya Gutkovskiy, Aug 08 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 24; CoefficientList[Series[-Log[1 - (Exp[x] - 1)^4/4!], {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 4] &
    a[n_] := a[n] = StirlingS2[n, 4] + (1/n) Sum[Binomial[n, k] StirlingS2[n - k, 4] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 4, 24}]

Formula

a(n) = Stirling2(n,4) + (1/n) * Sum_{k=1..n-1} binomial(n,k) * Stirling2(n-k,4) * k * a(k).
a(n) ~ (n-1)! / (log(2^(3/4)*3^(1/4) + 1))^n. - Vaclav Kotesovec, Aug 09 2021
a(n) = Sum_{k=1..floor(n/4)} (4*k)! * Stirling2(n,4*k)/(k * 24^k). - Seiichi Manyama, Jan 23 2025

A346955 Expansion of e.g.f. -log( 1 - (exp(x) - 1)^5 / 5! ).

Original entry on oeis.org

1, 15, 140, 1050, 6951, 42651, 253660, 1594230, 12463451, 134921787, 1806513072, 25539589530, 355175465191, 4797717669123, 63797550625676, 860468790181686, 12275324511112971, 192498455326842819, 3353266112959628272, 63379650000684213834
Offset: 5

Views

Author

Ilya Gutkovskiy, Aug 08 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 24; CoefficientList[Series[-Log[1 - (Exp[x] - 1)^5/5!], {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 5] &
    a[n_] := a[n] = StirlingS2[n, 5] + (1/n) Sum[Binomial[n, k] StirlingS2[n - k, 5] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 5, 24}]

Formula

a(n) = Stirling2(n,5) + (1/n) * Sum_{k=1..n-1} binomial(n,k) * Stirling2(n-k,5) * k * a(k).
a(n) ~ (n-1)! / (log(120^(1/5) + 1))^n. - Vaclav Kotesovec, Aug 09 2021
a(n) = Sum_{k=1..floor(n/5)} (5*k)! * Stirling2(n,5*k)/(k * 120^k). - Seiichi Manyama, Jan 23 2025
Showing 1-3 of 3 results.