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%I #15 Oct 13 2017 05:17:49
%S 1,1,1,2,1,-1,6,1,-4,1,24,-2,-17,8,-1,120,-34,-83,57,-13,1,720,-324,
%T -464,425,-135,19,-1,5040,-2988,-2924,3439,-1370,268,-26,1,40320,
%U -28944,-20404,30436,-14399,3514,-476,34,-1,362880,-300816,-154692,294328,-160027,46025,-7798,782,-43,1
%N 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).
%C 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.
%D Chang and Sederberg, Over and Over Again, MAA, 1997, page 209 (Moving Averages).
%H G. C. Greubel, <a href="/A123361/b123361.txt">Table of n, a(n) for the first 100 rows, flattened</a>
%e Triangle begins:
%e 1;
%e 1, 1;
%e 2, 1, -1;
%e 6, 1, -4, 1;
%e 24, -2, -17, 8, -1;
%e 120, -34, -83, 57, -13, 1;
%e 720, -324, -464, 425, -135, 19, -1;
%e 5040, -2988, -2924, 3439, -1370, 268, -26, 1;
%p 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
%t 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]
%Y Cf. A008275.
%K sign,tabl
%O 0,4
%A _Roger L. Bagula_, Nov 09 2006
%E Edited by _N. J. A. Sloane_, Nov 24 2006, Jun 17 2007
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