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.

A347022 Expansion of e.g.f. 1 / (1 - 5 * log(1 + x))^(1/5).

Original entry on oeis.org

1, 1, 5, 50, 720, 13650, 320370, 8967720, 291538080, 10795026840, 448484788680, 20658543923280, 1044915105622800, 57572197848878400, 3432143603792520000, 220109018869587398400, 15110184224165199667200, 1105545474191480800492800, 85881534014930659599571200
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 11 2021

Keywords

Crossrefs

Cf. A006252, A008548, A320343, A346984, A346987, A347020, A347021, A347023.

Programs

  • Mathematica
    nmax = 18; CoefficientList[Series[1/(1 - 5 Log[1 + x])^(1/5), {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[StirlingS1[n, k] 5^k Pochhammer[1/5, k], {k, 0, n}], {n, 0, 18}]

Formula

a(n) = Sum_{k=0..n} Stirling1(n,k) * A008548(k).
a(n) ~ n! * exp(1/25) / (Gamma(1/5) * 5^(1/5) * n^(4/5) * (exp(1/5) - 1)^(n + 1/5)). - Vaclav Kotesovec, Aug 14 2021
a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k-1) * (5 - 4*k/n) * (k-1)! * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 11 2023

A347021 Expansion of e.g.f. 1 / (1 - 4 * log(1 + x))^(1/4).

Original entry on oeis.org

1, 1, 4, 32, 364, 5444, 100520, 2210760, 56406240, 1637877600, 53327583360, 1924096475520, 76198487927040, 3285955396558080, 153273199794071040, 7689131281851770880, 412809183978447306240, 23616192920003184176640, 1434201753814306170808320
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 11 2021

Keywords

Crossrefs

Cf. A006252, A007696, A320343, A346983, A347016, A347020, A347022, A347023.

Programs

  • Mathematica
    nmax = 18; CoefficientList[Series[1/(1 - 4 Log[1 + x])^(1/4), {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[StirlingS1[n, k] 4^k Pochhammer[1/4, k], {k, 0, n}], {n, 0, 18}]

Formula

a(n) = Sum_{k=0..n} Stirling1(n,k) * A007696(k).
a(n) ~ n! * exp(1/16) / (Gamma(1/4) * 2^(1/2) * n^(3/4) * (exp(1/4) - 1)^(n + 1/4)). - Vaclav Kotesovec, Aug 14 2021
a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k-1) * (4 - 3*k/n) * (k-1)! * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 11 2023

A347023 E.g.f.: 1 / (1 - 6 * log(1 + x))^(1/6).

Original entry on oeis.org

1, 1, 6, 72, 1254, 28794, 819888, 27869316, 1101032100, 49570797780, 2505156062472, 140417898936336, 8644973807845368, 579908437058338920, 42098286646367326368, 3288252917244250703664, 274974019392668843164176, 24510436934573885695407504, 2319947117871178825560902112
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 11 2021

Keywords

Comments

In general, for k > 1, if e.g.f. = 1 / (1 - k*log(1 + x))^(1/k), then a(n) ~ n! * exp(1/k^2) / (Gamma(1/k) * k^(1/k) * n^(1 - 1/k) * (exp(1/k) - 1)^(n + 1/k)). - Vaclav Kotesovec, Aug 14 2021

Crossrefs

Cf. A006252, A008542, A320343, A346985, A347019, A347020, A347021, A347022.

Programs

  • Mathematica
    nmax = 18; CoefficientList[Series[1/(1 - 6 Log[1 + x])^(1/6), {x, 0, nmax}], x] Range[0, nmax]!
Table[Sum[StirlingS1[n, k] 6^k Pochhammer[1/6, k], {k, 0, n}], {n, 0, 18}]

Formula

a(n) = Sum_{k=0..n} Stirling1(n,k) * A008542(k).
a(n) ~ n! * exp(1/36) / (Gamma(1/6) * 6^(1/6) * n^(5/6) * (exp(1/6) - 1)^(n + 1/6)). - Vaclav Kotesovec, Aug 14 2021

A365599 Expansion of e.g.f. 1 / (1 - 3 * log(1 + x))^(2/3).

Original entry on oeis.org

1, 2, 8, 54, 498, 5868, 83940, 1413480, 27375240, 599437440, 14641665120, 394657325280, 11635613604000, 372469741813440, 12864889063033920, 476870475257550720, 18882021780125953920, 795381867831610978560, 35515223076159203880960
Offset: 0

Views

Author

Seiichi Manyama, Sep 11 2023

Keywords

Crossrefs

Cf. A335531, A347020.
Cf. A365575.

Programs

  • Mathematica
    a[n_] := Sum[Product[3*j + 2, {j, 0, k - 1}] * StirlingS1[n, k], {k, 0, n}]; Array[a, 19, 0] (* Amiram Eldar, Sep 13 2023 *)
  • PARI
    a(n) = sum(k=0, n, prod(j=0, k-1, 3*j+2)*stirling(n, k, 1));

Formula

a(n) = Sum_{k=0..n} (Product_{j=0..k-1} (3*j+2)) * Stirling1(n,k).
a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k-1) * (3 - k/n) * (k-1)! * binomial(n,k) * a(n-k).
a(n) ~ Gamma(1/3) * n^(n + 1/6) / (3^(1/6) * sqrt(2*Pi) * (exp(1/3) - 1)^(n + 2/3) * exp(n - 2/9)). - Vaclav Kotesovec, Nov 11 2023
Showing 1-4 of 4 results.

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