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A001785 Second-order reciprocal Stirling number (Fekete) a(n) = [[2n+4, n]]. The number of n-orbit permutations of a (2n+4)-set with at least 2 elements in each orbit. Also known as associated Stirling numbers of the first kind (e.g., Comtet).

%I A001785 M5382 N2338 #64 Aug 05 2023 13:08:47
%S A001785 1,120,7308,303660,11098780,389449060,13642629000,486591585480,
%T A001785 17856935296200,678103775949600,26726282654771700,1094862336960892500,
%U A001785 46641683693715610500,2066075391660447667500,95122549872697437090000
%N A001785 Second-order reciprocal Stirling number (Fekete) a(n) = [[2n+4, n]]. The number of n-orbit permutations of a (2n+4)-set with at least 2 elements in each orbit. Also known as associated Stirling numbers of the first kind (e.g., Comtet).
%D A001785 L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 256.
%D A001785 C. Jordan, Calculus of Finite Differences. Budapest, 1939, p. 152.
%D A001785 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A001785 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A001785 A. E. Fekete, <a href="http://www.jstor.org/stable/2974533">Apropos two notes on notation</a>, Amer. Math. Monthly, 101 (1994), 771-778.
%H A001785 H. W. Gould, Harris Kwong, and Jocelyn Quaintance, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Kwong/kwong9.html">On Certain Sums of Stirling Numbers with Binomial Coefficients</a>, J. Integer Sequences, 18 (2015), #15.9.6.
%H A001785 C. Jordan, <a href="https://www.jstage.jst.go.jp/article/tmj1911/37/0/37_0_254/_pdf">On Stirling's Numbers</a>, Tohoku Math. J., 37 (1933), 254-278.
%F A001785 a(n) = [[2n+4, n]] = Sum_{i=0..n} (-1)^i*binomial(2n+4, 2n+4-i)*[2n+4-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
%F A001785 Recurrence: 30*(n-1)*(116*n+75)*a(n) + (-6960*n^3-49760*n^2-112691*n-80787)*a(n-1) + (n+1)*(2*n+1)*(20*n+21)*a(n-2) = 0. - _R. J. Mathar_, Jul 18 2015
%F A001785 For n>0, a(n) = (1113 + 1447*n + 600*n^2 + 80*n^3)*(2*n+4)!/(1215*2^(n+3)*(n-1)!). - _Vaclav Kotesovec_, Jan 17 2016
%F A001785 Recurrence (for n>1): (n-1)*(80*n^3 + 360*n^2 + 487*n + 186)*a(n) = (n+2)*(2*n+3)*(80*n^3 + 600*n^2 + 1447*n + 1113)*a(n-1). - _Vaclav Kotesovec_, Jan 18 2016
%p A001785 with(combinat):s1 := (n,k)->sum((-1)^i*binomial(n,i)*abs(stirling1(n-i,k-i)),i=0..n); for j from 1 to 20 do s1(2*j+4,j); od; # Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
%t A001785 Prepend[Table[Sum[(-1)^i Binomial[2 n + 4, 2 n + 4 - i] Abs@ StirlingS1[2 n + 4 - i, n - i], {i, 0, n}], {n, 14}] , 1] (* _Michael De Vlieger_, Jan 04 2016 *)
%o A001785 (PARI) a(n) = if (!n, 1, sum(i=0, n, (-1)^i*binomial(2*n+4, 2*n+4-i)*abs(stirling(2*n+4-i, n-i, 1)))); \\ _Michel Marcus_, Jan 04 2016
%o A001785 (Magma) [1] cat [(1113+1447*n+600*n^2+80*n^3)*Factorial(2*n+4)/(1215*2^(n+ 3)*Factorial(n-1)): n in [1..15]]; // _Vincenzo Librandi_, Jan 18 2016
%Y A001785 Cf. A000907, A000483, A001784.
%K A001785 nonn
%O A001785 0,2
%A A001785 _N. J. A. Sloane_
%E A001785 More terms from Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
%E A001785 Offset changed to 0 by _Michel Marcus_, Jan 04 2016