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.

A346946 Expansion of e.g.f. log( 1 + log(1 + x)^4 / 4! ).

Original entry on oeis.org

1, -10, 85, -735, 6734, -66024, 693230, -7774250, 92759821, -1172483598, 15630569591, -218793782025, 3201481037819, -48746860400024, 768683653934928, -12487871805640344, 207761719406853466, -3513910668343842900, 59833161662103132050, -1011244718827893629750
Offset: 4

Views

Author

Ilya Gutkovskiy, Aug 08 2021

Keywords

Crossrefs

Cf. A000454, A003713, A346944, A346945, A346947.
Cf. A346976.

Programs

  • Mathematica
    nmax = 23; CoefficientList[Series[Log[1 + Log[1 + x]^4/4!], {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 4] &
a[n_] := a[n] = StirlingS1[n, 4] - (1/n) Sum[Binomial[n, k] StirlingS1[n - k, 4] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 4, 23}]

Formula

a(n) = Stirling1(n,4) - (1/n) * Sum_{k=1..n-1} binomial(n,k) * Stirling1(n-k,4) * k * a(k).
a(n) = Sum_{k=1..floor(n/4)} (-1)^(k-1) * (4*k)! * Stirling1(n,4*k)/(k * 24^k). - Seiichi Manyama, Jan 23 2025

A346945 Expansion of e.g.f. log( 1 + log(1 + x)^3 / 3! ).

Original entry on oeis.org

1, -6, 35, -235, 1834, -16352, 164044, -1830630, 22513326, -302700926, 4419167532, -69637654996, 1178377833424, -21315571470320, 410529985172400, -8388475139138320, 181270810764205440, -4130796696683135280, 99008773205008777760, -2490134250475836315120
Offset: 3

Views

Author

Ilya Gutkovskiy, Aug 08 2021

Keywords

Crossrefs

Cf. A000399, A003713, A081051, A346944, A346946, A346947.
Cf. A346975.

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[Log[1 + Log[1 + x]^3/3!], {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 3] &
a[n_] := a[n] = StirlingS1[n, 3] - (1/n) Sum[Binomial[n, k] StirlingS1[n - k, 3] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 3, 22}]

Formula

a(n) = Stirling1(n,3) - (1/n) * Sum_{k=1..n-1} binomial(n,k) * Stirling1(n-k,3) * k * a(k).
a(n) = Sum_{k=1..floor(n/3)} (-1)^(k-1) * (3*k)! * Stirling1(n,3*k)/(k * 6^k). - Seiichi Manyama, Jan 23 2025

A346947 Expansion of e.g.f. log( 1 + log(1 + x)^5 / 5! ).

Original entry on oeis.org

1, -15, 175, -1960, 22449, -269451, 3423860, -46238280, 664233856, -10143487354, 164423204582, -2823783679080, 51273355515264, -982236541934430, 19809898439192946, -419752648063849626, 9325875631405818996, -216846992855331506052, 5267598064689049209252
Offset: 5

Views

Author

Ilya Gutkovskiy, Aug 08 2021

Keywords

Crossrefs

Cf. A000482, A003713, A346944, A346945, A346946.

Programs

  • Mathematica
    nmax = 23; CoefficientList[Series[Log[1 + Log[1 + x]^5/5!], {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 5] &
a[n_] := a[n] = StirlingS1[n, 5] - (1/n) Sum[Binomial[n, k] StirlingS1[n - k, 5] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 5, 23}]

Formula

a(n) = Stirling1(n,5) - (1/n) * Sum_{k=1..n-1} binomial(n,k) * Stirling1(n-k,5) * k * a(k).
a(n) = Sum_{k=1..floor(n/5)} (-1)^(k-1) * (5*k)! * Stirling1(n,5*k)/(k * 120^k). - Seiichi Manyama, Jan 23 2025

A346966 Expansion of e.g.f. -log( 1 - log(1 - x)^2 / 2 ).

Original entry on oeis.org

1, 3, 14, 80, 559, 4599, 43665, 470196, 5666586, 75600690, 1106587008, 17636532264, 304092954138, 5640892517610, 112029356591862, 2371963759970352, 53338181764577304, 1269586152655203672, 31891196481381667008, 843109673024218773600, 23400930987874505081160
Offset: 2

Views

Author

Ilya Gutkovskiy, Aug 09 2021

Keywords

Crossrefs

Cf. A000254, A003713, A346921, A346944.

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[-Log[1 - Log[1 - x]^2/2], {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 2] &
a[n_] := a[n] = Abs[StirlingS1[n, 2]] + (1/n) Sum[Binomial[n, k] Abs[StirlingS1[n - k, 2]] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 2, 22}]

Formula

a(n) = |Stirling1(n,2)| + (1/n) * Sum_{k=1..n-1} binomial(n,k) * |Stirling1(n-k,2)| * k * a(k).
a(n) ~ (n-1)! / (1 - exp(-sqrt(2)))^n. - Vaclav Kotesovec, Jun 04 2022
a(n) = Sum_{k=1..floor(n/2)} (2*k)! * |Stirling1(n,2*k)|/(k * 2^k). - Seiichi Manyama, Jan 23 2025
Showing 1-4 of 4 results.

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