cp's OEIS Frontend

A105794 Inverse of a generalized Stirling number triangle of first kind.

%I A105794 #68 Feb 16 2025 02:28:00
%S A105794 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 A105794 0,21,70,56,14,1,1,-1,1,42,231,294,126,20,1,-1,1,0,85,735,1407,924,
%U A105794 246,27,1,1,-1,1,170,2290,6363,6027,2400,435,35,1
%N A105794 Inverse of a generalized Stirling number triangle of first kind.
%C A105794 Inverse of number triangle A105793. Row sums are the generalized Bell numbers A000296.
%C A105794 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 A105794 From _Peter Bala_, Jul 10 2013: (Start)
%C A105794 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 A105794 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 A105794 The row polynomials are a particular case of the Actuarial polynomials - see Roman 4.3.4.  (End)
%D A105794 S. Roman, The umbral calculus, Pure and Applied Mathematics 111, Academic Press Inc., New York, 1984. Reprinted by Dover in 2005.
%H A105794 Robert Israel, <a href="/A105794/b105794.txt">Table of n, a(n) for n = 0..9869</a>
%H A105794 Peter Bala, <a href="/A105794/a105794.pdf">Notes on A105794</a>
%H A105794 Peter Bala, <a href="/A143395/a143395.pdf">A 3 parameter family of generalized Stirling numbers</a>
%H A105794 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 A105794 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 A105794 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 A105794 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 A105794 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 A105794 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 A105794 Recurrence relation: T(n+1,k) = T(n,k-1) + (k-1)*T(n,k).
%F A105794 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 A105794 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 A105794 From _Peter Bala_, Jul 10 2013: (Start)
%F A105794 T(n,k) = Sum_{i = 0..n-1} (-1)^i*Stirling2(n-1-i,k-1), for n >= 1, k >= 1.
%F A105794 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 A105794 Dobinski-type formula for the row polynomials: R(n,x) = exp(-x)*Sum_{k >= 0} (k-1)^n*x^k/k!.
%F A105794 Sum_{k = 0..n} binomial(n,k)*R(k,x) = n-th Bell polynomial (n-th row polynomial of A048993). (End)
%F A105794 From _Peter Bala_, Jan 13 2014: (Start)
%F A105794 T(n,k) = Sum_{i = 0..n} (-1)^(n-i)*binomial(n,i)*Stirling2(i,k).
%F A105794 T(n,k) = Sum_{i = 0..n} (-2)^(n-i)*binomial(n,i)*Stirling2(i+1,k+1).
%F A105794 Matrix product P^(-1) * S = P^(-2) * S1, where P = A007318, S = A048993 and S1 = A008277. (End)
%F A105794 From _Werner Schulte_, Feb 15 2025: (Start)
%F A105794 Conjecture 1: Sum_{k=0..n} (-1)^(n-k) * T(n, k) * (k+m)! / m! = (m+2)^n for m >= 0.
%F A105794 Conjecture 2: (-1)^n - B(n) = Sum_{k=1..n} (-1)^k * T(n, k) * (k-1)! / (k+1) where B(n) = B(n, 0) is n-th Bernoulli number. (End)
%e A105794 The triangle starts with
%e A105794   n=0:  1;
%e A105794   n=1: -1,  1;
%e A105794   n=2:  1, -1, 1;
%e A105794   n=3: -1,  1, 0, 1;
%e A105794   n=4:  1, -1, 1, 2, 1;
%e A105794   n=5: -1,  1, 0, 5, 5, 1;
%e A105794   ... - _Wolfdieter Lang_, Jun 20 2011
%p A105794 B:= Matrix(12,12,shape=triangular[lower],(n,k) -> combinat:-stirling1(n-1,k-1)+(n-1)*combinat:-stirling1(n-2,k-1)):
%p A105794 A:= B^(-1):
%p A105794 seq(seq(A[i,j],j=1..i),i=1..12); # _Robert Israel_, Jan 19 2015
%p A105794 T := (n, k) -> add((-1)^(n - i)*binomial(n, i)*Stirling2(i, k), i=0..n):
%p A105794 seq(seq(T(n, k), k=0..n), n=0..9);  # _Peter Luschny_, Feb 15 2025
%t A105794 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 A105794 Cf. A000296 (row sums), A105793 (matrix inverse), A227341.
%Y A105794 Cf. A007318, A008277, A048993.
%K A105794 easy,sign,tabl
%O A105794 0,14
%A A105794 _Paul Barry_, Apr 20 2005