

A123221


Bezier transform of Mahonian numbers triangle A008302; p(k, x) = Sum[x^m, {m, 0, k}]*p(k  1, x).


0



1, 1, 1, 0, 1, 1, 1, 0, 2, 3, 5, 3, 1, 1, 0, 3, 5, 11, 22, 20, 15, 9, 4, 1, 1, 0, 4, 7, 18, 41, 90, 101, 101, 90, 71, 49, 29, 14, 5, 1, 1, 0, 5, 9, 26, 64, 154, 359, 455, 531, 573, 573, 531, 455, 359, 259, 169, 98, 49, 20, 6, 1, 1, 0, 6, 11, 35, 91, 234, 583, 1415, 1940, 2493
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OFFSET

1,9


COMMENTS

This method of recursive polynomials is a better method of obtaining these polynomials than that given originally in A008302. The jump in polynomial power levels in Mahonian numbers is very like that in Poncelet transforms.


LINKS

Table of n, a(n) for n=1..73.


FORMULA

p(k, x) = Sum[x^m, {m, 0, k}]*p(k  1, x)>t(n,m) Coefficient Bezier transform : t'(n,m)=t(n,m)*x^m*(1x)^(nm)


EXAMPLE

{1},
{1},
{1, 0, 1, 1},
{1, 0, 2, 3, 5, 3, 1},
{1, 0, 3, 5, 11, 22, 20, 15, 9, 4, 1},
{1, 0, 4, 7, 18, 41, 90, 101, 101, 90, 71, 49, 29, 14, 5, 1},
{1,0, 5, 9, 26, 64, 154, 359, 455, 531, 573, 573, 531, 455, 359, 259, 169, 98,
49, 20, 6, 1},


MATHEMATICA

(* Mahonian Polynomials*) p[0, x] = 1; p[1, x] = x + 1; p[k_, x_] := p[k, x] = Sum[x^m, {m, 0, k}]*p[k  1, x]; w = Table[CoefficientList[p[n, x], x], {n, 0, 10}]; (* Bezier transform*) v = Table[CoefficientList[Sum[w[[n + 1]][[m + 1]]*x^m*(1  x)^(n  m), {m, 0, Length[w[[n + 1]]]  1}], x], {n, 0, 10}]; Flatten[v]


CROSSREFS

Cf. A008302.
Sequence in context: A069110 A238684 A202694 * A197032 A254862 A172984
Adjacent sequences: A123218 A123219 A123220 * A123222 A123223 A123224


KEYWORD

nonn,uned,tabf


AUTHOR

Roger L. Bagula, Oct 05 2006


STATUS

approved



