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Triangle: p(x) = (t/log(1 + t))^a0*(1 + t)^x; a0=2; weights (n+1)!*n!.
0

%I #23 Feb 18 2024 05:15:01

%S 1,2,2,1,6,6,0,-12,0,24,-12,120,0,-240,120,360,-2280,0,4800,-3600,720,

%T -13260,68040,0,-151200,138600,-45360,5040,638400,-2899680,0,6773760,

%U -7056000,2963520,-564480,40320,-39630528,166320000,0,-406425600,464002560,-228614400,57576960,-7257600,362880

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

%C 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

%H V. Kruchinin, D. Kruchinin, <a href="http://arxiv.org/abs/1211.0099">Application of a composition of generating functions for obtaining explicit formulas of polynomials</a>, arXiv: 1211.0099

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

%F 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

%e {1},

%e {2, 2},

%e {1, 6, 6},

%e {0, -12, 0, 24},

%e {-12, 120, 0, -240, 120},

%e {360, -2280, 0, 4800, -3600, 720},

%e {-13260, 68040, 0, -151200, 138600, -45360, 5040},

%e {638400, -2899680, 0, 6773760, -7056000, 2963520, -564480, 40320}

%t a0 = 2;

%t p[t_] = (t/Log[1 + t])^a0*(1 + t)^x;

%t Table[ ExpandAll[(n!*(n + 1)!)*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}];

%t a = Table[ CoefficientList[(n!*(n + 1)!)*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}];

%t Flatten[a]

%o (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 */

%K uned,tabl,sign,less

%O 1,2

%A _Roger L. Bagula_, Apr 09 2008