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.

A353225 Expansion of e.g.f. (1 - x^4)^(-1/x^3).

Original entry on oeis.org

1, 1, 1, 1, 1, 61, 361, 1261, 3361, 128521, 1678321, 11670121, 56596321, 1773048421, 37020623641, 410615985781, 3056256665281, 88439609228881, 2516514283997281, 39513591769228561, 409546654143301441, 11679302565962651341, 413008783534735181641
Offset: 0

Views

Author

Seiichi Manyama, May 01 2022

Keywords

Crossrefs

Programs

  • Mathematica
    With[{nn=30},CoefficientList[Series[(1-x^4)^(-1/x^3),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Sep 17 2024 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace((1-x^4)^(-1/x^3)))
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(-log(1-x^4)/x^3)))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=1, (i+3)\4, (4*j-3)/j*v[i-4*j+4]/(i-4*j+3)!)); v;
    
  • PARI
    a(n) = n!*sum(k=0, n\4, abs(stirling(n-3*k, n-4*k, 1))/(n-3*k)!);

Formula

a(0) = 1; a(n) = (n-1)! * Sum_{k=1..floor((n+3)/4)} (4*k-3)/k * a(n-4*k+3)/(n-4*k+3)!.
a(n) = n! * Sum_{k=0..floor(n/4)} |Stirling1(n-3*k,n-4*k)|/(n-3*k)!.
a(n) ~ sqrt(2*Pi) * n^(n + 1/2) / (4*exp(n)). - Vaclav Kotesovec, May 04 2022

A353222 Expansion of e.g.f. (1 - x^3)^(-1/x).

Original entry on oeis.org

1, 0, 2, 0, 12, 60, 120, 2520, 15120, 90720, 1693440, 13305600, 140374080, 2724321600, 27744837120, 414096883200, 8689288608000, 111399326438400, 2114134793971200, 48501156601497600, 759659036405068800, 17279306372135808000, 434100706059205785600
Offset: 0

Views

Author

Seiichi Manyama, May 01 2022

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace((1-x^3)^(-1/x)))
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(-log(1-x^3)/x)))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=1, (i+1)\3, (3*j-1)/j*v[i-3*j+2]/(i-3*j+1)!)); v;

Formula

a(0) = 1; a(n) = (n-1)! * Sum_{k=1..floor((n+1)/3)} (3*k-1)/k * a(n-3*k+1)/(n-3*k+1)!.
a(n) ~ sqrt(2*Pi) * n^(n + 1/2) / (3*exp(n)). - Vaclav Kotesovec, May 04 2022
Showing 1-2 of 2 results.