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 1-5 of 5 results.

A346987 Expansion of e.g.f. 1 / (1 + 5 * log(1 - x))^(1/5).

Original entry on oeis.org

1, 1, 7, 86, 1524, 35370, 1015590, 34757400, 1381147440, 62498177880, 3172764322680, 178566159846480, 11034757650750960, 742773843654742080, 54094804600076176320, 4238009228531321452800, 355400361455423327193600, 31764402860426288679456000, 3014207878695233997923193600
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 11 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 18; CoefficientList[Series[1/(1 + 5 Log[1 - x])^(1/5), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[Abs[StirlingS1[n, k]] 5^k Pochhammer[1/5, k], {k, 0, n}], {n, 0, 18}]
  • Maxima
    a[n]:=if n=0 then 1 else sum(n!/(n-k)!*(5/k-4/n)*a[n-k],k,1,n);
    makelist(a[n],n,0,50); /* Tani Akinari, Aug 27 2023 */

Formula

a(n) = Sum_{k=0..n} |Stirling1(n,k)| * A008548(k).
a(n) ~ n! * exp(n/5) / (Gamma(1/5) * 5^(1/5) * n^(4/5) * (exp(1/5) - 1)^(n + 1/5)). - Vaclav Kotesovec, Aug 14 2021
For n > 0, a(n) = Sum_{k=1..n} (n!/(n-k)!)*(5/k-4/n)*a(n-k). - Tani Akinari, Aug 27 2023

A347015 Expansion of e.g.f. 1 / (1 + 3 * log(1 - x))^(1/3).

Original entry on oeis.org

1, 1, 5, 42, 498, 7644, 144156, 3225648, 83536008, 2457701928, 80970232104, 2953056534768, 118112744060208, 5140622709134496, 241863782829704928, 12232551538417012992, 661818290353375962240, 38140594162828447248000, 2332567001993176540206720, 150880256846462633823648000
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 10 2021

Keywords

Crossrefs

Programs

  • Maple
    g:= proc(n) option remember; `if`(n<2, 1, (3*n-2)*g(n-1)) end:
    a:= n-> add(abs(Stirling1(n, k))*g(k), k=0..n):
    seq(a(n), n=0..19);  # Alois P. Heinz, Aug 10 2021
  • Mathematica
    nmax = 19; CoefficientList[Series[1/(1 + 3 Log[1 - x])^(1/3), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[Abs[StirlingS1[n, k]] 3^k Pochhammer[1/3, k], {k, 0, n}], {n, 0, 19}]

Formula

a(n) = Sum_{k=0..n} |Stirling1(n,k)| * A007559(k).
a(n) ~ n! * exp(n/3) / (Gamma(1/3) * 3^(1/3) * n^(2/3) * (exp(1/3) - 1)^(n + 1/3)). - Vaclav Kotesovec, Aug 14 2021
a(0) = 1; a(n) = Sum_{k=1..n} (3 - 2*k/n) * (k-1)! * binomial(n,k) * a(n-k). - Seiichi Manyama, Sep 09 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

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

A347019 E.g.f.: 1 / (1 + 6 * log(1 - x))^(1/6).

Original entry on oeis.org

1, 1, 8, 114, 2358, 64074, 2157828, 86714592, 4049302404, 215458069428, 12867377875632, 852254389954296, 61998666080311800, 4914000741835488744, 421488717980664846960, 38897664480760253351904, 3843081247426270376211216, 404727487161912602921083536
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(n/k) / (Gamma(1/k) * k^(1/k) * n^(1 - 1/k) * (exp(1/k) - 1)^(n + 1/k)). - Vaclav Kotesovec, Aug 14 2021

Crossrefs

Programs

  • Mathematica
    nmax = 17; CoefficientList[Series[1/(1 + 6 Log[1 - x])^(1/6), {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[Abs[StirlingS1[n, k]] 6^k Pochhammer[1/6, k], {k, 0, n}], {n, 0, 17}]

Formula

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

A365584 Expansion of e.g.f. 1 / (1 + 4 * log(1-x))^(3/4).

Original entry on oeis.org

1, 3, 24, 300, 5100, 109692, 2854344, 87164088, 3055516800, 120916282368, 5331444120576, 259168711406976, 13769882994784896, 793844510730348672, 49353915922852214016, 3291455140392403401984, 234388011123877880424960, 17750517946502792294592000
Offset: 0

Views

Author

Seiichi Manyama, Sep 10 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[Product[4*j + 3, {j, 0, k - 1}] * Abs[StirlingS1[n, k]], {k, 0, n}]; Array[a, 18, 0] (* Amiram Eldar, Sep 10 2023 *)
  • PARI
    a(n) = sum(k=0, n, prod(j=0, k-1, 4*j+3)*abs(stirling(n, k, 1)));

Formula

a(n) = Sum_{k=0..n} (Product_{j=0..k-1} (4*j+3)) * |Stirling1(n,k)|.
a(0) = 1; a(n) = Sum_{k=1..n} (4 - k/n) * (k-1)! * binomial(n,k) * a(n-k).
a(n) ~ Gamma(1/4) * n^(n + 1/4) / (2^(3/2) * sqrt(Pi) * (exp(1/4) - 1)^(n + 3/4) * exp(3*n/4)). - Vaclav Kotesovec, Nov 11 2023
Showing 1-5 of 5 results.