A176665 Triangle of polynomial coefficients of p(x,n) = Sum_{k=0..n} (k + 1)^n * k! * binomial(x, k), read by rows.

1, 1, 2, 1, -5, 9, 1, 109, -165, 64, 1, -3303, 6188, -3494, 625, 1, 169711, -357254, 254434, -74635, 7776, 1, -13084359, 30063342, -24927719, 9549230, -1718079, 117649, 1, 1417404703, -3486909736, 3229823067, -1474126800, 354928391, -43216649, 2097152

Offset

0,3

Comments

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

Links

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Formula

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) ).

Example

Triangle begins as:
  1;
  1,         2;
  1,        -5,        9;
  1,       109,     -165,        64;
  1,     -3303,     6188,     -3494,     625;
  1,    169711,  -357254,    254434,  -74635,     7776;
  1, -13084359, 30063342, -24927719, 9549230, -1718079, 117649;

Mathematica

(* First program *)
p[x_, n_]:= Sum[(k+1)^n*k!*Binomial[x, k], {k, 0, n}];
Table[CoefficientList[ExpandAll[p[x, n]], x], {n, 0, 10}]//Flatten
(* Second program *)
f[n_]:= CoefficientList[Sum[(k+1)^n*Product[x-j, {j, 0, k-1}], {k, 0, n}], x];
Table[f[n], {n, 0, 10}] (* G. C. Greubel, Feb 07 2021 *)

Prog

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

Crossrefs

Cf. A083318.

Sequence in context: A318052 A026078 A349635 * A199050 A306539 A193629

Adjacent sequences: A176662 A176663 A176664 * A176666 A176667 A176668

Keyword

sign,tabl

Author

Roger L. Bagula, Apr 23 2010

Extensions

Edited by G. C. Greubel, Feb 07 2021

Status

approved