A325873 - OEIS

%I #9 Oct 30 2019 13:27:32

%S 1,0,1,0,0,1,0,1,0,1,0,1,4,0,1,0,8,5,10,0,1,0,26,58,15,20,0,1,0,194,

%T 217,238,35,35,0,1,0,1142,2035,1008,728,70,56,0,1,0,9736,13470,11611,

%U 3444,1848,126,84,0,1

%N T(n, k) = [x^k] Sum_{k=0..n} |Stirling1(n, k)|*FallingFactorial(x, k), triangle read by rows, for n >= 0 and 0 <= k <= n.

%e Triangle starts:

%e [0] [1]

%e [1] [0, 1]

%e [2] [0, 0,    1]

%e [3] [0, 1,    0,     1]

%e [4] [0, 1,    4,     0,     1]

%e [5] [0, 8,    5,     10,    0,    1]

%e [6] [0, 26,   58,    15,    20,   0,    1]

%e [7] [0, 194,  217,   238,   35,   35,   0,   1]

%e [8] [0, 1142, 2035,  1008,  728,  70,   56,  0,  1]

%e [9] [0, 9736, 13470, 11611, 3444, 1848, 126, 84, 0, 1]

%t p[n_] := Sum[Abs[StirlingS1[n, k]] FactorialPower[x, k], {k, 0, n}];

%t Table[CoefficientList[FunctionExpand[p[n]], x], {n, 0, 9}] // Flatten

%o (Sage)

%o def a_row(n):

%o     s = sum(stirling_number1(n,k)*falling_factorial(x,k) for k in (0..n))

%o     return expand(s).list()

%o [a_row(n) for n in (0..9)]

%Y Cf. A079642 (variant), A129062, A325872.

%K nonn,tabl

%O 0,13

%A _Peter Luschny_, Jun 27 2019