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%I #57 Dec 18 2019 00:53:43
%S 1,-1,1,1,-1,1,-1,1,0,1,1,-1,1,2,1,-1,1,0,5,5,1,1,-1,1,10,20,9,1,-1,1,
%T 0,21,70,56,14,1,1,-1,1,42,231,294,126,20,1,-1,1,0,85,735,1407,924,
%U 246,27,1,1,-1,1,170,2290,6363,6027,2400,435,35,1
%N Inverse of a generalized Stirling number triangle of first kind.
%C Inverse of number triangle A105793. Row sums are the generalized Bell numbers A000296.
%C T(n,k)*(-1)^(n-k) gives the inverse Sheffer matrix of A094645. In the umbral notation (cf. Roman, p. 17, quoted in A094645) this is Sheffer for (1-t,-log(1-t)). - _Wolfdieter Lang_, Jun 20 2011
%C From _Peter Bala_, Jul 10 2013: (Start)
%C For a graph G and a positive integer k, the graphical Stirling number S(G;k) is the number of partitions of the vertex set of G into k nonempty independent sets (an independent set in G is a subset of the vertices, no two elements of which are adjacent). Omitting the first two rows and columns of this triangle produces the triangle of graphical Stirling numbers of cycles on n vertices (interpreting a 2-cycle as a single edge). In comparison, the classical Stirling numbers of the second kind, A008277, are the graphical Stirling numbers of path graphs on n vertices. See Galvin and Than. See also A227341.
%C This is the triangle of connection constants for expressing the polynomial sequence (x-1)^n in terms of the falling factorial polynomials, that is, (x-1)^n = Sum_{k = 0..n} T(n,k)*x_(k), where x_(k) = x*(x-1)*...*(x-k+1) denotes the falling factorial.
%C The row polynomials are a particular case of the Actuarial polynomials - see Roman 4.3.4. (End)
%D S. Roman, The umbral calculus, Pure and Applied Mathematics 111, Academic Press Inc., New York, 1984. Reprinted by Dover in 2005.
%H Robert Israel, <a href="/A105794/b105794.txt">Table of n, a(n) for n = 0..9869</a>
%H P. Bala, <a href="/A105794/a105794.pdf">Notes on A105794</a>
%H P. Bala, <a href="/A143395/a143395.pdf">A 3 parameter family of generalized Stirling numbers</a>
%H B. Duncan and R. Peele, <a href="http://www.emis.de/journals/JIS/VOL12/Peele/peele5.html">Bell and Stirling Numbers for Graphs</a>, Journal of Integer Sequences 12 (2009), article 09.7.1.
%H D. Galvin and D. T. Thanh, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v20i1p73">Stirling numbers of forests and cycles</a>, Electr. J. Comb. Vol. 20(1): P73 (2013).
%H Sophie Morier-Genoud, <a href="https://arxiv.org/abs/1907.12790">Counting Coxeter's friezes over a finite field via moduli spaces</a>, arXiv:1907.12790 [math.CO], 2019.
%F Term k in row n is given by {(-1)^(k+n) * (Sum_{j=0..k} (-1)^j * binomial(k,j) * (1-j)^n) / k! }; i.e., a finite difference. - _Tom Copeland_, Jun 05 2006
%F O.g.f. for row n = n-th finite difference of the Touchard (Bell) polynomials, T(x,j) and so the e.g.f. for these finite differences and therefore the sequence = exp{x*[exp(t)-1]-t} = exp{t*[T(x,.)-1]} umbrally. - _Tom Copeland_, Jun 05 2006
%F The e.g.f. A(x,t) = exp(x*(exp(t)-1)-t) satisfies the partial differential equation x*dA/dx - dA/dt = (1-x)*A.
%F Recurrence relation: T(n+1,k) = T(n,k-1) + (k-1)*T(n,k).
%F Let f(x) = ((x-1)/x)*exp(x). For n >= 1, the n-th row polynomial R(n,x) = x*exp(-x)*(x*d/dx)^(n-1)(f(x)) and satisfies the recurrence equation R(n+1,x) = (x-1)*R(n,x) + x*R'(n,x). - _Peter Bala_, Oct 28 2011
%F Let f(x) = exp(exp(x)-x). Then R(n,exp(x)) = 1/f(x)*(d/dx)^n(f(x)). Similar formulas hold for A008277, A039755, A111577, A143494 and A154537. - _Peter Bala_, Mar 01 2012
%F From _Peter Bala_, Jul 10 2013: (Start)
%F T(n,k) = Sum_{i = 0..n-1} (-1)^i*Stirling2(n-1-i,k-1), for n >= 1, k >= 1.
%F The k-th column o.g.f. is 1/(1+x)*x^k/((1-x)*(1-2*x)*...*(1-(k-1)*x) (the empty product occurring in the denominator when k = 0 and k = 1 is taken equal to 1).
%F Dobinski-type formula for the row polynomials: R(n,x) = exp(-x)*Sum_{k >= 0} (k-1)^n*x^k/k!.
%F Sum_{k = 0..n} binomial(n,k)*R(k,x) = n-th Bell polynomial (n-th row polynomial of A048993). (End)
%F From _Peter Bala_, Jan 13 2014: (Start)
%F T(n,k) = Sum_{i = 0..n} (-1)^(n-i)*binomial(n,i)*Stirling2(i,k).
%F T(n,k) = Sum_{i = 0..n} (-2)^(n-i)*binomial(n,i)*Stirling2(i+1,k+1).
%F Matrix product P^(-1) * S = P^(-2) * S1, where P = A007318, S = A048993 and S1 = A008277. (End)
%e The triangle starts with
%e n=0: 1;
%e n=1: -1, 1;
%e n=2: 1, -1, 1;
%e n=3: -1, 1, 0, 1;
%e n=4: 1, -1, 1, 2, 1;
%e n=5: -1, 1, 0, 5, 5, 1;
%e ... - _Wolfdieter Lang_, Jun 20 2011
%p B:= Matrix(12,12,shape=triangular[lower],(n,k) -> combinat:-stirling1(n-1,k-1)+(n-1)*combinat:-stirling1(n-2,k-1)):
%p A:= B^(-1):
%p seq(seq(A[i,j],j=1..i),i=1..12); # _Robert Israel_, Jan 19 2015
%t Table[Sum[(-1)^(n - i)*Binomial[n, i] StirlingS2[i, k], {i, 0, n}], {n, 0, 10}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Oct 14 2019 *)
%Y Cf. A000296 (row sums), A105793 (matrix inverse), A227341.
%Y Cf. A007318, A008277, A048993.
%K easy,sign,tabl
%O 0,14
%A _Paul Barry_, Apr 20 2005
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