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.

A333881 Expansion of e.g.f. exp(Sum_{k>=0} x^(3*k + 1) / (3*k + 1)!).

Original entry on oeis.org

1, 1, 1, 1, 2, 6, 16, 37, 114, 478, 1907, 6777, 28414, 148579, 758916, 3580189, 18981485, 117883917, 720627553, 4193077474, 26795418840, 191751387094, 1352954503595, 9301704998742, 69285817230370, 559142785301527, 4453089770243547, 35182348161102172
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 08 2020

Keywords

Comments

Number of partitions of n-set into blocks congruent to 1 mod 3.

Crossrefs

Programs

  • Mathematica
    nmax = 27; CoefficientList[Series[Exp[Sum[x^(3 k + 1)/(3 k + 1)!, {k, 0, nmax}]], {x, 0, nmax}], x] Range[0, nmax]!
    a[n_] := a[n] = Sum[Boole[MemberQ[{1}, Mod[k, 3]]] Binomial[n - 1, k - 1] a[n - k], {k, 1, n}]; a[0] = 1; Table[a[n], {n, 0, 27}]
    nmax = 30; CoefficientList[Series[Exp[Exp[x]/3 - 2*Sin[Pi/6 - Sqrt[3]*x/2] / (3*Exp[x/2])], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Apr 15 2020 *)

Formula

E.g.f.: exp(exp(x)/3 - 2*sin(Pi/6 - sqrt(3)*x/2) / (3*exp(x/2))). - Vaclav Kotesovec, Apr 15 2020
a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/3)} binomial(n-1,3*k) * a(n-3*k-1). - Seiichi Manyama, Sep 22 2023

A352430 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/5)} binomial(n,5*k+1) * a(n-5*k-1).

Original entry on oeis.org

1, 1, 2, 6, 24, 120, 721, 5054, 40488, 364896, 3654000, 40249441, 483659508, 6296246424, 88269037584, 1325861901000, 21243052172161, 361630022931666, 6518319228715302, 124018898163736536, 2483799332459535000, 52231733840672804881, 1150683180739820615582, 26502219276887376327696
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 16 2022

Keywords

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, 5 k + 1] a[n - 5 k - 1], {k, 0, Floor[(n - 1)/5]}]; Table[a[n], {n, 0, 23}]
    nmax = 23; CoefficientList[Series[1/(1 - Sum[x^(5 k + 1)/(5 k + 1)!, {k, 0, nmax}]), {x, 0, nmax}], x] Range[0, nmax]!
  • PARI
    my(N=40, x='x+O('x^N)); Vec(serlaplace(1/(1-sum(k=0, N\5, x^(5*k+1)/(5*k+1)!)))) \\ Seiichi Manyama, Mar 23 2022

Formula

E.g.f.: 1 / (1 - Sum_{k>=0} x^(5*k+1) / (5*k+1)!).

A333883 Expansion of e.g.f. exp(Sum_{k>=0} x^(6*k + 1) / (6*k + 1)!).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 9, 37, 121, 331, 793, 1718, 5163, 32281, 217921, 1188709, 5291353, 20031170, 66744741, 267996541, 2030569465, 18368560519, 138812739409, 853152218102, 4409607501927, 19826125988257, 99717123889777, 871344991322017, 9658479225877057
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 08 2020

Keywords

Comments

Number of partitions of n-set into blocks congruent to 1 mod 6.

Crossrefs

Programs

  • Mathematica
    nmax = 30; CoefficientList[Series[Exp[Sum[x^(6 k + 1)/(6 k + 1)!, {k, 0, nmax}]], {x, 0, nmax}], x] Range[0, nmax]!
    a[n_] := a[n] = Sum[Boole[MemberQ[{1}, Mod[k, 6]]] Binomial[n - 1, k - 1] a[n - k], {k, 1, n}]; a[0] = 1; Table[a[n], {n, 0, 30}]
    nmax = 30; CoefficientList[Series[Exp[x*HypergeometricPFQ[{}, {1/3, 1/2, 2/3, 5/6, 7/6}, x^6/46656]], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Apr 15 2020 *)

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/6)} binomial(n-1,6*k) * a(n-6*k-1). - Seiichi Manyama, Sep 22 2023

A365969 Expansion of e.g.f. exp( Sum_{k>=0} x^(5*k+1) / (5*k+1) ).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 121, 841, 3361, 10081, 25201, 3684241, 50309281, 369738721, 1926648361, 7980936601, 1335634023361, 27705746752321, 302258931418081, 2283161710263841, 13419441405835201, 2498339829188508481, 70152448708746111961, 1025314852704395518441
Offset: 0

Views

Author

Seiichi Manyama, Sep 23 2023

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(sum(k=0, N\5, x^(5*k+1)/(5*k+1)))))

Formula

a(0) = 1; a(n) = (n-1)! * Sum_{k=0..floor((n-1)/5)} a(n-5*k-1)/(n-5*k-1)!.
a(0) = a(1) = ... = a(4) = 1; a(n) = a(n-1) + 120 * binomial(n-1,5) * a(n-5).
Showing 1-4 of 4 results.