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%I #12 Jul 14 2016 00:18:39
%S 1,-1,1,2,-3,1,-1,3,-3,1,2,-4,4,-3,1,-1,5,-7,5,-3,1,2,-5,9,-10,6,-3,1,
%T -1,7,-12,14,-13,7,-3,1,2,-6,16,-22,20,-16,8,-3,1,-1,9,-18,30,-35,27,
%U -19,9,-3,1,2,-7,25,-40,50,-51,35,-22,10,-3,1
%N Triangular array from a zero coefficient sum of recursive polynomials: p(k, x) = x*p(k - 1, x) + p(k - 2, x).
%C Except for p(0,x)=1 all sum to zero: Table[Sum[CoefficientList[p[n, x], x][[m]], {m, 1, n + 1}], {n, 0, 12}] {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
%H Robert Israel, <a href="/A123185/b123185.txt">Table of n, a(n) for n = 0..10010</a> (rows 0 to 140, flattened)
%F p(k, x) = x*p(k - 1, x) + p(k - 2, x) for k >= 3 with p(0,x) = 1, p(1,x) = x-1 and p(2,x) = x^2-3x+2. (Corrected by _Robert Israel_, Jul 12 2016)
%F G.f.: g(t,x) = (1 - t + (1-2x) t^2)/(1 - t x - t^2). - _Robert Israel_, Jul 12 2016
%F For n >= 1, T(n,k) = A114525(n,k) + 2*A114525(n-1,k) - 5*A168561(n-1,k), where we take A114525(n-1,n) = A168561(n-1,n) = 0. - _Robert Israel_, Jul 13 2016
%e 1
%e -1, 1
%e 2, -3, 1
%e -1, 3, -3, 1
%e 2, -4, 4, -3, 1
%e -1, 5, -7, 5, -3, 1
%e 2, -5, 9, -10, 6, -3, 1
%p S:= series((1 - t + (1-2*x)*t^2)/(1 - t*x - t^2), t, 21):
%p R:= [seq(coeff(S,t,n),n=0..19)]:
%p seq(seq(coeff(R[n],x,j),j=0..n-1),n=1..20); # _Robert Israel_, Jul 12 2016
%t p[0, x] = 1; p[1, x] = x - 1; p[2, x] = x^2 - 3x + 2; p[k_, x_] := p[k, x] = x*p[k - 1, x] + p[k - 2, x] ; w = Table[CoefficientList[p[n, x], x], {n, 0, 10}]; Flatten[w]
%Y Cf. A114525, A168561.
%K tabl,sign
%O 0,4
%A _Roger L. Bagula_, Oct 02 2006
%E Edited by _Robert Israel_, Jul 13 2016
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