A176665 - OEIS

%I #6 Feb 07 2021 19:42:39

%S 1,1,2,1,-5,9,1,109,-165,64,1,-3303,6188,-3494,625,1,169711,-357254,

%T 254434,-74635,7776,1,-13084359,30063342,-24927719,9549230,-1718079,

%U 117649,1,1417404703,-3486909736,3229823067,-1474126800,354928391,-43216649,2097152

%N Triangle of polynomial coefficients of p(x,n) = Sum_{k=0..n} (k + 1)^n * k! * binomial(x, k), read by rows.

%C Row sums are: A083318 = {1, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025, ...}.

%H G. C. Greubel, <a href="/A176665/b176665.txt">Rows n = 0..100 of the triangle, flattened</a>

%F Let p(x,n) = Sum_{k=0..n} (k + 1)^n * k! * binomial(x, k) then the number triangle is given by T(n, m) = coefficients( p(x,n) ).

%e Triangle begins as:

%e   1;

%e   1,         2;

%e   1,        -5,        9;

%e   1,       109,     -165,        64;

%e   1,     -3303,     6188,     -3494,     625;

%e   1,    169711,  -357254,    254434,  -74635,     7776;

%e   1, -13084359, 30063342, -24927719, 9549230, -1718079, 117649;

%t (* First program *)

%t p[x_, n_]:= Sum[(k+1)^n*k!*Binomial[x, k], {k, 0, n}];

%t Table[CoefficientList[ExpandAll[p[x, n]], x], {n, 0, 10}]//Flatten

%t (* Second program *)

%t f[n_]:= CoefficientList[Sum[(k+1)^n*Product[x-j, {j,0,k-1}], {k,0,n}], x];

%t Table[f[n], {n, 0, 10}] (* _G. C. Greubel_, Feb 07 2021 *)

%o (Sage)

%o def p(n, x): return sum( (k+1)^n*factorial(k)*binomial(x, k) for k in (0..n))

%o flatten([[( p(n, x) ).series(x, n+1).list()[k] for k in (0..n)] for n in (0..10)]) # _G. C. Greubel_, Feb 07 2021

%Y Cf. A083318.

%K sign,tabl

%O 0,3

%A _Roger L. Bagula_, Apr 23 2010

%E Edited by _G. C. Greubel_, Feb 07 2021