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 1-2 of 2 results.

A337552 a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * (3*k-2) * a(n-k).

Original entry on oeis.org

1, 1, 6, 37, 330, 3613, 47652, 732625, 12875118, 254540413, 5591435136, 135108218353, 3561467337546, 101704047315037, 3127751183515020, 103059820083026449, 3622223857996975110, 135266462416766669917, 5348457650664454581240, 223227700948792985989777
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 31 2020

Keywords

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] (3 k - 2) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
    nmax = 19; CoefficientList[Series[1/(Exp[x] (2 - 3 x) - 1), {x, 0, nmax}], x] Range[0, nmax]!
  • PARI
    seq(n)={Vec(serlaplace(1 / (exp(x + O(x*x^n)) * (2 - 3*x) - 1)))} \\ Andrew Howroyd, Aug 31 2020

Formula

E.g.f.: 1 / (exp(x) * (2 - 3*x) - 1).
a(n) ~ n! * c / ((1-c) * (2/3 - c)^(n+1)), where c = -LambertW(-exp(-2/3)/3). - Vaclav Kotesovec, Aug 31 2020

A337554 a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * (5*k-4) * a(n-k).

Original entry on oeis.org

1, 1, 8, 53, 560, 6961, 105898, 1867393, 37713620, 856269401, 21606253238, 599664843433, 18156702186880, 595557844417441, 21037627605306578, 796218790808110673, 32143778726932363340, 1378765268603813275081, 62619174356163136219918, 3001963660666272082265113
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 31 2020

Keywords

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] (5 k - 4) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 19}]
    nmax = 19; CoefficientList[Series[1/(Exp[x] (4 - 5 x) - 3), {x, 0, nmax}], x] Range[0, nmax]!
  • PARI
    seq(n)={Vec(serlaplace(1 / (exp(x + O(x*x^n)) * (4 - 5*x) - 3)))} \\ Andrew Howroyd, Aug 31 2020

Formula

E.g.f.: 1 / (exp(x) * (4 - 5*x) - 3).
a(n) ~ n! * c / (3*(1-c) * (4/5 - c)^(n+1)), where c = -LambertW(-3*exp(-4/5)/5). - Vaclav Kotesovec, Aug 31 2020
Showing 1-2 of 2 results.