A167786 Triangle of z Transform coefficients from General Pascal [1,8,1} A142458 polynomials multiplied by factor 3^Floor[(2*k - 1)/3].

0, 3, 3, 6, 9, 45, 45, 27, 234, 540, 360, 27, 315, 1305, 1980, 990, 81, 1026, 6750, 18360, 20790, 8316, 243, 3807, 26379, 115830, 234630, 212058, 70686, 243, 7938, 37800, 177660, 582120, 939708, 706860, 201960, 729, 26001, 280827, 873180, 3087315

Offset: 0

Author

Roger L. Bagula, Nov 12 2009

Keywords

• nonn
• uned

Comments

Row sums are:

{0, 3, 9, 99, 1161, 4617, 55323, 663633, 2654289, 31850739, 382206681...}

These are a sequence of Infinite sums that give A142458.

Even terms are factored by (1+2*n) which is the MacMahon (1+2*n)^k,but the polynomials seem fundamental

other than that.

A060187 MacMahon gives A013609 Triangle of coefficients in expansion of (1 + 2x)^n.

I looked for a simple infinite sum for the {1,8,1} and failed.

This reasoning comes from finding that the general z Transform polynomials are

related to the Eulerian: in fact this type of Eulerian polynomials A008292 gives (1+n)^k binomial.

The polynomials given here form a set of infinite sum sequences.

Examples

			{0},
{3},
{3, 6},
{9, 45, 45},
{27, 234, 540, 360},
{27, 315, 1305, 1980, 990},
{81, 1026, 6750, 18360, 20790, 8316},
{243, 3807, 26379, 115830, 234630, 212058, 70686},
{243, 7938, 37800, 177660, 582120, 939708, 706860, 201960},
{729, 26001, 280827, 873180, 3087315, 8058204, 10814958, 6967620, 1741905},
{2187, -308610, 1076490, 7334820, 17120565, 48411594, 104968710, 120570120, 67934295, 15096510}
		

Crossrefs

A142458, A060187, A008292

Programs

Mathematica

m = 3 A[n_, 1] := 1 A[n_, n_] := 1
A[n_, k_] := (m*n - m*k + 1)A[n - 1, k - 1] + (m*k - (m - 1))A[n - 1, k]
a = Table[A[n, k], {n, 10}, {k, n}]
p[x_, n_] = x*Sum[a[[n, k]]*x^(k - 1), {k, 1, n}]/(x - 1)
b = Table[p[x, n], {n, 0, 10}]
Table[3^Floor[(2*k - 1)/3]*CoefficientList[ExpandAll[ InverseZTransform[b[[k]], x, n] /. UnitStep[ -1 + n] -> 1], n], {k, 1, Length[b]}]

Formula

m=3;

A(n,k)= (m*n - m*k + 1)A(n - 1, k - 1} + (m*k - (m - 1))A(n - 1, k)

q(n,k)=InverseZTransform[x*Sum[a[[n, k]]*x^(k - 1), {k, 1, n}]/(x - 1)^n, x, k]

out_n,k=3^Floor[(2*k - 1)/3]*coefficients(q[n,k])