T(n, k) = T(n-1, k)*n + binomial(n, k)*(-1)^(n-k) = T(n, k-1)/k + binomial(n, k)*(-1)^(n-k)/(n-k+1) = T(n-1, k-1)*n/k = T(n-k, 0)*binomial(n, k) =
A000166(n-k)*binomial(n,k) [with T(0, 0) = 1]; so T(n, n) = 1, T(n, n-1) = 0, T(n, n-2) = n*(n-1)/2 for n >= 0.
Sum_{k=0..n} T(n, k) = Sum_{k=0..n} k * T(n, k) = n! for all n > 0, n, k integers. -
Wouter Meeussen, May 29 2001
O.g.f. for k-th column: (1/k!)*Sum_{i>=k} i!*x^i/(1+x)^(i+1).
O.g.f. for k-th row: k!*Sum_{i=0..k} (-1)^i/i!*(1-x)^i. (End)
E.g.f. for number of permutations with exactly k fixed points is x^k/(k!*exp(x)*(1-x)). -
Vladeta Jovovic, Aug 25 2002
Sum_{k=0..n} T(n, k)*x^k is the permanent of the n X n matrix with x's on the diagonal and 1's elsewhere; for x = 0, 1, 2, 3, 4, 5, 6 see
A000166,
A000142,
A000522,
A010842,
A053486,
A053487,
A080954. -
Philippe Deléham, Dec 12 2003; for x = 1+i see
A009551 and
A009102. -
John M. Campbell, Oct 11 2011
T(n,k) = Sum_{j=0..n} (-1)^(j-k)*binomial(j,k)*n!/j!. -
Paul Barry, May 25 2006
T(n,k) = (n!/k!)*Sum_{j=0..n-k} ((-1)^j)/j!, 0 <= k <= n. From the Appell type of the triangle and the subfactorial formula.
T(n,0) = n*Sum_{j=0..n-1} (j/(j+1))*T(n-1,j), T(0,0)=1. From the z-sequence of this Sheffer triangle z(j)=j/(j+1) with e.g.f. (1-exp(x)*(1-x))/x. See the W. Lang link under
A006232 for Sheffer a- and z-sequences. -
Wolfdieter Lang, Jan 21 2008
T(n,k) = (n/k)*T(n-1,k-1) for k >= 1. See above. From the a-sequence of this Sheffer triangle a(0)=1, a(n)=0, n >= 1 with e.g.f. 1. See the W. Lang link under
A006232 for Sheffer a- and z-sequences. -
Wolfdieter Lang, Jan 21 2008
T(n,k) = T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k=0 and
T(n,k) = T(n-1,k-1) + T(n-1,k)*(n-1-k) + T(n-1,k+1)*(k+1) for k>=1.
(End)
Sum_{k=0..n} k^2 * T(n, k) = 2*n! if n > 1. -
Michael Somos, Jun 06 2017
The lowering and raising operators of this Appell sequence of polynomials P(n,x) are L = d/dx and R = x + d/dL log[exp(-L)/(1-L)] = x-1 + 1/(1-L) = x + L + L^2 - ... such that L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
P(n,x) = (1-L)^(-1) exp(-L) x^n = (1+L+L^2+...)(x-1)^n = n! Sum_{k=0..n} (x-1)^k / k!.
The polynomials of this pair P_n(x) and Q_n(x) are umbral compositional inverses; i.e., P_n(Q.(x)) = x^n = Q_n(P.(x)), where, e.g., (Q.(x))^n = Q_n(x).
For more on the infinitesimal generator, noted by Bala below, see
A238385. (End)
Sum_{k=0..n} (-1)^k * T(n,k) =
A000023(n).
Sum_{k=0..n} (-1)^k * k * T(n,k) =
A335111(n). (End)
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