cp's OEIS Frontend

Coefficients of terms in the series reversion of (1-k*x-(k+1)*x^2)/(1+x). - Paul Barry, May 21 2005

Equals A131427 * A007318 as infinite lower triangular matrices. [Philippe Deléham, Sep 15 2008]

Sum_{k=0..n} T(n,k)*x^k = A168491(n), A000007(n), A000108(n), A151374(n), A005159(n), A151403(n), A156058(n), A156128(n), A156266(n), A156270(n), A156273(n), A156275(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Nov 15 2013

Diagonal sums are A052709(n+1). - Philippe Deléham, Nov 15 2013

By rows, equals partial sums of A053121 reversed rows. Example: Row 4 of A053121 = (2, 0, 3, 0, 1) -> (1, 0, 3, 0, 2) -> (1, 1, 4, 4, 6). - Gary W. Adamson, Dec 28 2008, edited by Michel Marcus, Sep 22 2015

Row sums = A097813: (1, 2, 6, 16, 38, 84, 178, ...).

b(n)= A006519(n), with b(n) defined in the formula. For every odd n b(n)=1.

The row polynomials Po(n,x) := 2*b(n)*sum(a(n,m)*x^m,m=0..n-1), n>=1, appear as numerators of the generating functions for the odd numbered column sequences of array A034870. b(n) is defined in the formula below.

Row sums = (1, 2, 4, 8, ...). A131047 * (1,2,3, ...) = A087447 starting (1, 4, 10, 24, 56, ...). A generalized set of analogous triangles: (1/(Q+1)) * (P^Q - 1/P), Q an integer, generates triangles with row sums = powers of (Q+1). Cf. A131048, A131049, A131050, A131051 for triangles having Q = 2,3,4 and 5, respectively.

A007318, Pascal's triangle, = this triangle + A119467, since one triangle = the zeros or masks of the other. - Gary W. Adamson, Jun 12 2007

Row n gives the coefficients in the expansion of (1/4)*(1 + x)^n + (9/4)*2^n*(1 - x)^(1 + n)*Phi(x, -n, 1/2) - (3/2)*(1 - x)^(n + 2)*Phi(x, -1 - n, 1), where Phi is the Lerch transcendant.

T is the transpose of A132440.

Let M(t) = exp(t*T) = limit [1 + t*T/n]^n as n tends to infinity.

Then M(1) = the transpose of the lower triangular Pascal matrix A007318, with inverse M(-1).

Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L P_n(x) = n * P_(n-1)(x) and

R P_n(x) = P_(n+1)(x), the matrix T represents the action of L in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x) = x^n/n!, L = DxD and R = D^(-1).

See A132440 as an analog and more general discussion.

Sum_{n>=0} c_n T^n / n! = e^(c.T) gives the Maurer-Cartan form matrix for the one-dimensional Leibniz group defined by multiplication of a Taylor series by the formal Taylor series e^(c.x) (cf. Olver). - Tom Copeland, Nov 05 2015

From Tom Copeland, Jul 02 2018: (Start)

The transpose Psc^Trn of the lower triangular Pascal matrix Psc = A007318 gives the numerical coefficients of the Maurer-Cartan form matrix M of the Leibniz group Leibniz(n)(1,1) presented on p. 9 of the Olver paper. M = exp[c. * T] with (c.)^n = c_n and T the Lie infinitesimal generator of this entry. The columns e^T are the rows of the Pascal matrix A007318.

M can be obtained by multiplying each n-th column vector of Psc by c_n and then transposing the result; i.e., with the diagonal matrix H = Diag(c_0, c_1, c_2, ...), M = (Psc * H)^Trn = H * Psc^Trn.

M is a matrix representation of the differential operator S = e^{c.*D} with D = d/dx, which acting on x^m gives the Appell polynomial p_m(x) = (c. + x)^m, with (c.)^k = c_k an arbitrary indeterminate except for c_0 = 1. For example, S x^2 = (c. + x)^2 = c_0*x^2 + 2*c_1*x + c_2, and M * (0,0,1,0,0,...)^Trn = (c_2,2*c_1,c_0,0,0,...)^Trn = V, so V^Trn = (0,0,1,0,...) * M^Trn = (0,0,1,0,...) * Psc * H = (c_2,2*c_1,c_0,0,...).

The differential lowering and raising operators for the Appell sequence are given by L = D and R = x + dlog(S)/dD, with L p_n(x = n * p_(n-1)(x) and R p_n(x) = p_(n+1)(x).

(End)

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is 1/((1-z)*(1-x*z*(1-z)/(1-2*z))).

Column m (without leading zeros) is obtained from convolution of A000012 (powers of 1) with m-fold convoluted A011782.

Central binomial coefficients c = A000984(n) > 1 appear once in the middle column C(2n, n), and thereafter in one or more later rows to the left as C(r,k) and to the right as C(r, r-k), k < r/2; the last time in row r = c = C(c,1) = C(c,c-1). For these, a(n) = (A003016(n)+1)/2. For all other numbers n > 1, a(n) = A003016(n)/2. - M. F. Hasler, Mar 01 2023

Column k has e.g.f. Bessel_I(k,2x). - Paul Barry, Mar 10 2010

T(n,k) is the number of binary sequences of length n in which the number of ones minus the number of zeros is k. If 2 divides(n+k), such a sequence will have (n+k)/2 ones and (n-k)/2 zeros. Since there are C(n,(n+k)/2) ways to choose the sequence entries that get a one, T(n,k)=binomial(n,(n+k)/2) whenever (n+k) is even and T(n,k)= 0 otherwise. See the example below in the example section. - Dennis P. Walsh, Apr 11 2012

T(n,k) is the number of walks on the semi-infinite integer line with n steps that end at k. The walks start at 0, move at each step either one to the left or one to the right, and never enter the region of negative k. [Walks with impenetrable wall at -1/2. Dyck excursions of n steps that end at level k.] The variant without the restriction of negative positions is A053121. - R. J. Mathar, Nov 02 2023