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.

A355164 a(n) = exp(-1/3) * Sum_{k>=0} (3*k + 2)^n / (3^k * k!).

Original entry on oeis.org

1, 3, 12, 63, 405, 3024, 25515, 239355, 2465478, 27600669, 333051669, 4303119330, 59202612693, 863285928327, 13288589222508, 215177742933579, 3654114236490393, 64902307993517160, 1202782377224829015, 23207417212751493327, 465302639045308247262, 9677171073270491712513, 208434297638273958963225
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 22 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[Exp[2 x + (Exp[3 x] - 1)/3], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = 2 a[n - 1] + Sum[Binomial[n - 1, k - 1] 3^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 22}]
    Table[Sum[Binomial[n, k] 2^(n - k) 3^k BellB[k, 1/3], {k, 0, n}], {n, 0, 22}]

Formula

E.g.f.: exp(2*x + (exp(3*x) - 1) / 3).
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 3^(k-1) * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * A004212(k).
a(n) ~ 3^(n + 2/3) * n^(n + 2/3) * exp(n/LambertW(3*n) - n - 1/3) / (sqrt(1 + LambertW(3*n)) * LambertW(3*n)^(n + 2/3)). - Vaclav Kotesovec, Jun 27 2022

A355167 a(n) = exp(-1/4) * Sum_{k>=0} (4*k + 3)^n / (4^k * k!).

Original entry on oeis.org

1, 4, 20, 128, 1008, 9280, 96704, 1120768, 14274816, 197833728, 2958521344, 47415508992, 809838505984, 14670950907904, 280760761434112, 5655835404271616, 119561580162646016, 2645030742360588288, 61087848487323959296, 1469652941137655103488, 36758243982057508175872, 954111239026567129595904
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 22 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 21; CoefficientList[Series[Exp[3 x + (Exp[4 x] - 1)/4], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = 3 a[n - 1] + Sum[Binomial[n - 1, k - 1] 4^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 21}]
    Table[Sum[Binomial[n, k] 3^(n - k) 4^k BellB[k, 1/4], {k, 0, n}], {n, 0, 21}]

Formula

E.g.f.: exp(3*x + (exp(4*x) - 1) / 4).
a(0) = 1; a(n) = 3 * a(n-1) + Sum_{k=1..n} binomial(n-1,k-1) * 4^(k-1) * a(n-k).
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * A004213(k).
a(n) ~ 2^(2*n + 3/2) * n^(n + 3/4) * exp(n/LambertW(4*n) - n - 1/4) / (sqrt(1 + LambertW(4*n)) * LambertW(4*n)^(n + 3/4)). - Vaclav Kotesovec, Jun 27 2022
Showing 1-2 of 2 results.