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A176663 T(n, k) = [x^k] Sum_{j=0..n} j!*binomial(x, j), for 0 <= k <= n, triangle read by rows.

%I A176663 #29 Feb 08 2025 23:35:31
%S A176663 1,1,1,1,0,1,1,2,-2,1,1,-4,9,-5,1,1,20,-41,30,-9,1,1,-100,233,-195,76,
%T A176663 -14,1,1,620,-1531,1429,-659,161,-20,1,1,-4420,11537,-11703,6110,
%U A176663 -1799,302,-27,1,1,35900,-98047,106421,-61174,20650,-4234,519,-35,1
%N A176663 T(n, k) = [x^k] Sum_{j=0..n} j!*binomial(x, j), for 0 <= k <= n, triangle read by rows.
%F A176663 From _Peter Luschny_, Jul 02 2019: (Start)
%F A176663 Sum_{k=0..n} T(n, k)*x^k = Sum_{k=0..n} (x)_k, where (x)_k denotes the falling factorial.
%F A176663 Let T be the lower triangular matrix associated to the T(n, k) and S the lower triangular matrix associated to the Stirling set numbers S2(n, k). Then S*T = A186020 (seen as a matrix) and T*S = A000012 (seen as a matrix). (End)
%F A176663 T(n, k) = Sum_{i=0..n-k} Stirling1(i+k, k). - _Igor Victorovich Statsenko_, May 25 2024
%e A176663 Triangle starts:
%e A176663   {1},
%e A176663   {1,       1},
%e A176663   {1,       0,      1},
%e A176663   {1,       2,     -2,        1},
%e A176663   {1,      -4,      9,       -5,      1},
%e A176663   {1,      20,    -41,       30,     -9,       1},
%e A176663   {1,    -100,    233,     -195,     76,     -14,     1},
%e A176663   {1,     620,  -1531,     1429,   -659,     161,   -20,     1},
%e A176663   {1,   -4420,  11537,   -11703,   6110,   -1799,   302,   -27,   1},
%e A176663   {1,   35900, -98047,   106421, -61174,   20650, -4234,   519, -35,   1},
%e A176663   {1, -326980, 928529, -1066279, 662506, -248675, 59039, -8931, 835, -44, 1}
%p A176663 with(PolynomialTools):
%p A176663 T_row := n -> CoefficientList(expand(add(k!*binomial(x, k), k=0..n)), x):
%p A176663 ListTools:-Flatten([seq(T_row(n), n=0..9)]); # _Peter Luschny_, Jul 02 2019
%t A176663 p[x_, n_] := Sum[k! Binomial[x, k], {k, 0, n}];
%t A176663 Table[CoefficientList[FunctionExpand[p[x, n]], x], {n, 0, 10}] // Flatten
%t A176663 (* Alternative: *)
%t A176663 Table[CoefficientList[FunctionExpand[Sum[FactorialPower[x, k], {k, 0, n}]], x], {n, 0, 10}] // Flatten (* _Peter Luschny_, Jul 02 2019 *)
%Y A176663 Row sums are A040000. Alternating row sums are A058006, which are also T(n,1).
%Y A176663 Cf. A186020.
%K A176663 sign,tabl
%O A176663 0,8
%A A176663 _Roger L. Bagula_, Apr 23 2010
%E A176663 Edited by _Peter Luschny_, Jul 02 2019