A123361 Triangle read by rows: T(n,k) = coefficient of x^k in the polynomial p[n,x] defined by p[0,x]=1, p[1,x]=1+x and p[n,x]=(1+x)(2-x)(3-x)...(n-x) for n >= 2 (0 <= k <= n).

1, 1, 1, 2, 1, -1, 6, 1, -4, 1, 24, -2, -17, 8, -1, 120, -34, -83, 57, -13, 1, 720, -324, -464, 425, -135, 19, -1, 5040, -2988, -2924, 3439, -1370, 268, -26, 1, 40320, -28944, -20404, 30436, -14399, 3514, -476, 34, -1, 362880, -300816, -154692, 294328, -160027, 46025, -7798, 782, -43, 1

Offset: 0

Author

Roger L. Bagula, Nov 09 2006

Keywords

• sign
• tabl

Comments

Changing the initial conditions in the recursion produces a different triangular sequence. The result here is a variation of Stirling's numbers of the first kind. The Chang and Sederberg version of this recursion produces an even function in sections.

Examples

			Triangle begins:
     1;
     1,     1;
     2,     1,    -1;
     6,     1,    -4,     1;
    24,    -2,   -17,     8,    -1;
   120,   -34,   -83,    57,   -13,     1;
   720,  -324,  -464,   425,  -135,    19,    -1;
  5040, -2988, -2924,  3439, -1370,   268,   -26,     1;
		

References

• Chang and Sederberg, Over and Over Again, MAA, 1997, page 209 (Moving Averages).

Links

• G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Crossrefs

Cf. A008275.

Programs

Maple

p[0]:=1: p[1]:=1+x: for n from 2 to 10 do p[n]:=sort(expand((n-x)*p[n-1])) od: for n from 0 to 10 do seq(coeff(p[n],x,k),k=0..n) od; # yields sequence in triangular form

Mathematica

p[ -1, x] = 1; p[0, x] = x + 1; p[k_, x_] := p[k, x] = (-x + k + 1)*p[k - 1, x] w = Table[CoefficientList[p[n, x], x], {n, -1, 10}]; Flatten[w]

Extensions

Edited by N. J. A. Sloane, Nov 24 2006, Jun 17 2007