A137381 Triangle: p(x) = (t/log(1 + t))^a0*(1 + t)^x; a0=2; weights (n+1)!*n!.

1, 2, 2, 1, 6, 6, 0, -12, 0, 24, -12, 120, 0, -240, 120, 360, -2280, 0, 4800, -3600, 720, -13260, 68040, 0, -151200, 138600, -45360, 5040, 638400, -2899680, 0, 6773760, -7056000, 2963520, -564480, 40320, -39630528, 166320000, 0, -406425600, 464002560, -228614400, 57576960, -7257600, 362880

Offset

1,2

Comments

p(n,x,alpha) = sum(i=0..n, (sum(k=1..i, binomial(k+alpha-1,alpha-1) *sum(j=0..k, ((-1)^j*j!*stirling1(j+i,j) *binomial(k,j))/(j+i)!))) *binomial(x,n-i)). - Vladimir Kruchinin, Jan 12 2012

Links

Table of n, a(n) for n=1..45.

V. Kruchinin, D. Kruchinin, Application of a composition of generating functions for obtaining explicit formulas of polynomials, arXiv: 1211.0099

Formula

p(x) = (t/Log[1 + t])^a0*(1 + t)^x; a0=2;weights (n+1)!*n!;

T(n,r) = n!*(n+1)!*sum(i=0..n,((sum(k=1..i, (k+1)*sum(j=0..k,((-1)^j*j! * stirling1(j+i,j)* C(k,j))/(j+i)!) ))*stirling1(n-i,k))/(n-i)!). - Vladimir Kruchinin, Jan 12 2012

Example

{1},
{2, 2},
{1, 6, 6},
{0, -12, 0, 24},
{-12, 120, 0, -240, 120},
{360, -2280, 0, 4800, -3600, 720},
{-13260, 68040, 0, -151200, 138600, -45360, 5040},
{638400, -2899680, 0, 6773760, -7056000, 2963520, -564480, 40320}

Mathematica

a0 = 2;
p[t_] = (t/Log[1 + t])^a0*(1 + t)^x;
Table[ ExpandAll[(n!*(n + 1)!)*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}];
a = Table[ CoefficientList[(n!*(n + 1)!)*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}];
Flatten[a]

Prog

(Maxima) T(n, r):=n!*(n+1)!*sum(((sum((k+1)*sum(((-1)^j*j!*stirling1(j+i, j)* binomial(k, j))/(j+i)!, j, 0, k) , k, 1, i))*stirling1(n-i, k))/(n-i)!, i, 0, n); /* Vladimir Kruchinin, Jan 12 2012 */

Crossrefs

Sequence in context: A134558 A344639 A230420 * A109316 A162980 A162979

Adjacent sequences: A137378 A137379 A137380 * A137382 A137383 A137384

Keyword

uned,tabl,sign,less

Author

Roger L. Bagula, Apr 09 2008

Status

approved