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 1-3 of 3 results.

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

A355165 a(n) = exp(-1/4) * Sum_{k>=0} (4*k + 2)^n / (4^k * k!).

Original entry on oeis.org

1, 3, 13, 79, 601, 5339, 53861, 607527, 7560625, 102637235, 1506225085, 23726435583, 398852249097, 7120170905995, 134408217821205, 2673140092099543, 55832167947587425, 1221199519275467107, 27902127744298836845, 664446811342185649583, 16457968670922936733113, 423242969435491221774907
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 22 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 21; CoefficientList[Series[Exp[2 x + (Exp[4 x] - 1)/4], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = 2 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] 2^(n + k) BellB[k, 1/4], {k, 0, n}], {n, 0, 21}]

Formula

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

A363908 a(n) = exp(-1/5) * Sum_{k>=0} (5*k + 4)^n / (5^k * k!).

Original entry on oeis.org

1, 5, 30, 225, 2075, 22500, 276875, 3790625, 57050000, 934984375, 16549046875, 314146406250, 6358972578125, 136603266015625, 3101556258593750, 74164388642578125, 1861859526474609375, 48936176077929687500, 1343302192888037109375, 38425435693841064453125, 1143143051078878906250000
Offset: 0

Views

Author

Michael De Vlieger, Jun 27 2023

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Exp[4 x + (Exp[5 x] - 1)/5], {x, 0, #}], x]* Range[0, #]! &[21]
Showing 1-3 of 3 results.