T(n, k) = T(n-1, k-1) - (n+1)*T(n-1, k), n >= k >= 0; T(n, k) = 0, n < k; T(n, -1) = 0, T(0, 0) = 1.
E.g.f. for k-th column of signed triangle: ((log(1+x))^k)/(k!*(1+x)^2).
Triangle (signed) = [-2, -1, -3, -2, -4, -3, -5, -4, -6, -5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; triangle (unsigned) = [2, 1, 3, 2, 4, 3, 5, 4, 6, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...], where DELTA is Deléham's operator defined in
A084938 (unsigned version in
A143491).
E.g.f.: (1 + x)^(y-2). -
Vladeta Jovovic, May 17 2004 [For row polynomials s(n, y)]
With P(n, t) = Sum_{j=0..n-2} T(n-2,j) * t^j and P(1, t) = -1 and P(0, t) = 1, then G(x, t) = -1 + exp[P(.,t)*x] = [(1+x)^t - 1 - t^2 * x] / [t(t-1)], whose compositional inverse in x about 0 is given in
A074060. G(x, 0) = -log(1+x) and G(x, 1) = (1+x) log(1+x) - 2x. G(x, q^2) occurs in formulas on pages 194-196 of the Manin reference. -
Tom Copeland, Feb 17 2008
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then T(n,i) = f(n,i,2), for n=1,2,...; i=0..n. -
Milan Janjic, Dec 21 2008
T(n, k) = Sum_{j=0..n} (-1)^(n-j)*(n-j+1)!*binomial(n, j)*Stirling1(j, k). -
Mélika Tebni, May 02 2022
Recurrence for row polynomials {s(n, x)}_{n>=0}: s(0, x) = 1, s(n, x) = (x - 2)*exp(-(d/dx)) s(n-1, x), for n >= 1. This is adapted from the general Sheffer result given by S. Roman, Corollary 3.7.2., p. 50.
Recurrence for column sequence {T(n, k)}_{n>=k}: T(n, n) = 1, T(n, k) = (n!/(n-k))*Sum_{j=k..n-1} (1/j!)*(a(n-1-j) + k*beta(n-1-j))*T(n-1, k), for k >= 0, where alpha = repeat(-2, 2) and beta(n) = [x^n] (d/dx)log(log(x)/x) = (-1)^(n+1)*
A002208(n+1)/
A002209(n+1), for n >= 0. This is the adapted Boas-Buck recurrence, also given in Rainville, Theorem 50., p. 141, For the references and a comment see
A046521. (End)
