cp's OEIS Frontend

Row sums = A001787: (1, 4, 12, 32, 80, 192, ...).

A134083 * [1,2,3,...] = A084850: (1, 5, 20, 68, 208, 592, ...).

Row sums = A052905: (1, 5, 10, 16, 23, 31, 40, ...).

Row sums = A034856: (1, 4, 8, 13, 19, 26, ...).

Row sums = A014480: (1, 6, 20, 56, 144, 352, ...).

Also 2*A134083-A007318 as infinite lower triangular matrices. - Philippe Deléham, Oct 15 2007

Row sums = A052905: (1, 5, 10, 16, 23, 31, 40, ...).

Equals (1, 2, 3, ...) convolved with (1, 2, 4, 4, 4, ...). a(3) = 22 = (1, 2, 3, 4) dot (4, 4, 2, 1) = (4 + 8 + 6 + 4). - Gary W. Adamson, May 01 2009

a(n) is also the number of ways to place 2 nonattacking bishops on a 2 X (n+1) board. - Vaclav Kotesovec, Jan 29 2010

Partial sums are A174723. - Wesley Ivan Hurt, Apr 16 2016

Also the number of irredundant sets in the n-cocktail party graph. - Eric W. Weisstein, Aug 09 2017

Let b(n) = LPT[ A001147 ] = -A001147(n-1) for n > 0 and 1 for n=0, where LPT represents the action of the list partition transform described in A133314.

Then T(n,k) = binomial(n,k) * b(n-k) .

Form the matrix of polynomials TB(n,k,t) = T(n,k) * t^(n-k) = binomial(n,k) * b(n-k) * t^(n-k) = binomial(n,k) * Pb(n-k,t),

beginning as

1;

-1, 1;

-1*t, -2, 1;

-3*t^2, -3*t, -3, 1;

-15*t^3, -12*t^2, -6*t, -4, 1;

-105*t^4, -75*t^3, -30*t^2, -10*t, -5, 1;

Let Pc(n,t) = LPT(Pb(.,t)).

Then [TB(t)]^(-1) = TC(t) = [ binomial(n,k) * Pc(n-k,t) ] = LPT(TB),

whose first column is

Pc(0,t) = 1

Pc(1,t) = 1

Pc(2,t) = 2 + t

Pc(3,t) = 6 + 6*t + 3*t^2

Pc(4,t) = 24 + 36*t + 30*t^2 + 15*t^3

Pc(5,t) = 120 + 240*t + 270*t^2 + 210*t^3 + 105*t^4.

The coefficients of these polynomials are given by the reverse of A102625 with the highest order coefficients given by A001147 with an additional leading 1.

Note this is not the complete matrix TC. The complete matrix is formed by multiplying along the diagonal of the lower triangular Pascal matrix by these polynomials, embedding trees of coefficients in the matrix.

exp[Pb(.,t)*x] = 1 + [(1-2t*x)^(1/2) - 1] / (t-0) = [1 + a finite diff. of [(1-2t*x)^(1/2)] with step t] = e.g.f. of the first column of TB.

exp[Pc(.,t)*x] = 1 / { 1 + [(1-2t*x)^(1/2) - 1] / t } = 1 / exp[Pb(.,t)*x) = e.g.f. of the first column of TC.

TB(t) and TC(t), being inverse to each other, are the generators of an Abelian group.

TB(0) and TC(0) are generators for a subgroup representing the iterated Laguerre operator described in A132013 and A132014.

Let sb(t,m) and sc(t,m) be the associated sequences under the LPT to TB(t)^m = B(t,m) and TC(t)^m = C(t,m).

Let Esb(t,m) and Esc(t,m) be e.g.f.'s for sb(t,m) and sc(t,m), rB(t,m) and rC(t,m) be the row sums of B(t,m) and C(t,m) and aB(t,m) and aC(t,m) be the alternating row sums.

Then B(t,m) is the inverse of C(t,m), Esb(t,m) is the reciprocal of Esc(t,m) and sb(t,m) and sc(t,m) form a reciprocal pair under the LPT. Similar relations hold among the row sums and the alternating sign row sums and associated quantities.

All the group members have the form B(t,m) * C(u,p) = TB(t)^m * TC(u)^p = [ binomial(n,k) * s(n-k) ]

with associated e.g.f. Es(x) = exp[m * Pb(.,t) * x] * exp[p * Pc(.,u) * x] for the first column of the matrix, with terms s(n), so group multiplication is isomorphic to matrix multiplication and to multiplication of the e.g.f.'s for the associated sequences (see examples).

These results can be extended to other groups of integer-valued arrays by replacing the 2 by any natural number in the expression for exp[Pb(.,t)*x].

More generally,

[ G.f. for M = Product_{i=0..j} B[s(i),m(i)] * C[t(i),n(i)] ]

= exp(u*x) * Product_{i=0..j} { exp[m(i) * Pb(.,s(i)) * x] * exp[n(i) * Pc(.,t(i)) * x] }

= exp(u*x) * Product_{i=0..j} { 1 + [ (1 - 2*s(i)*x)^(1/2) - 1 ] / s(i) }^m(i) / { 1 + [ (1 - 2*t(i)*x)^(1/2) - 1 ] / t(i) }^n(i)

= exp(u*x) * H(x)

[ E.g.f. for M ] = I_o[2*(u*x)^(1/2)] * H(x).

M is an integer-valued matrix for m(i) and n(i) positive integers and s(i) and t(i) integers. To invert M, change B to C in Product for M.

H(x) is the e.g.f. for the first column of M and diagonally multiplying the Pascal matrix by the terms of this column generates M. See examples.

The G.f. for M, i.e., the e.g.f. for the row polynomials of M, implies that the row polynomials form an Appell sequence (see Wikipedia and Mathworld). - Tom Copeland, Dec 03 2013