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%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
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