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).
1, 24, 924, 26432, 705320, 18858840, 520059540, 14980405440, 453247114320, 14433720701400, 483908513388300, 17068210823664000, 632607429473019000, 24602295329058447000, 1002393959071727722500, 42720592574082543120000
Offset: 0
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).
Links
- A. E. Fekete, Apropos two notes on notation, Amer. Math. Monthly, 101 (1994), 771-778.
- H. W. Gould, Harris Kwong, and Jocelyn Quaintance, On Certain Sums of Stirling Numbers with Binomial Coefficients, J. Integer Sequences, 18 (2015), #15.9.6.
- C. Jordan, On Stirling's Numbers, Tohoku Math. J., 37 (1933), 254-278.
Programs
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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
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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 *)
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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