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.

A001784 Second-order reciprocal Stirling number (Fekete) a(n) = [[2n+3, n]]. The number of n-orbit permutations of a (2n+3)-set with at least 2 elements in each orbit. Also known as associated Stirling numbers of the first kind (e.g., Comtet).

Original entry on oeis.org

1, 24, 924, 26432, 705320, 18858840, 520059540, 14980405440, 453247114320, 14433720701400, 483908513388300, 17068210823664000, 632607429473019000, 24602295329058447000, 1002393959071727722500, 42720592574082543120000
Offset: 0

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Author

Keywords

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 256.
  • C. Jordan, Calculus of Finite Differences. Budapest, 1939, p. 152.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    with(combinat):s1 := (n,k)->sum((-1)^i*binomial(n,i)*abs(stirling1(n-i,k-i)),i=0..n); 1; for j from 1 to 20 do s1(2*j+3,j); od; # Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
  • Mathematica
    Prepend[Table[Sum[(-1)^i Binomial[2 n + 3, 2 n + 3 - i] Abs@ StirlingS1[2 n + 3 - i, n - i], {i, 0, n}], {n, 15}] , 1] (* Michael De Vlieger, Jan 04 2016 *)
  • PARI
    a(n) = if (!n, 1, sum(i=0, n, (-1)^i*binomial(2*n+3, 2*n+3-i)*abs(stirling(2*n+3-i, n-i, 1)))); \\ Michel Marcus, Jan 04 2016

Formula

a(n) = [[2n+3, n]] = Sum_{i=0..n} (-1)^i*binomial(2n+3, 2n+3-i)*[2n+3-i, n-i] where [n, k] is the unsigned Stirling number of the first kind. - Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
Conjecture: 480*(n+1)*a(n) +30*(-32*n^2-14821*n+42287)*a(n-1) +(878700*n^2-403433*n+5134227)*a(n-2) +(911423*n-656446)*(2*n-3)*a(n-3)=0. - R. J. Mathar, Jul 18 2015
Conjecture: (n-2)*(20*n^2-5*n-3)*a(n) -n*(2*n+1)*(20*n^2+35*n+12)*a(n-1)=0. - R. J. Mathar, Jul 18 2015
For n>0, a(n) = (67 + 75*n + 20*n^2)*(2*n+3)!/(405*2^n*(n-1)!). - Vaclav Kotesovec, Jan 17 2016

Extensions

More terms from Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
Offset changed to 0 by Michel Marcus, Jan 04 2016