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

0,4

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.

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

Example

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;

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]

Crossrefs

Cf. A008275.

Sequence in context: A181538 A322128 A125731 * A265315 A179380 A107106

Adjacent sequences: A123358 A123359 A123360 * A123362 A123363 A123364

Keyword

sign,tabl

Author

Roger L. Bagula, Nov 09 2006

Extensions

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

Status

approved