A123185 Triangular array from a zero coefficient sum of recursive polynomials: p(k, x) = x*p(k - 1, x) + p(k - 2, x).

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, -1, 7, -12, 14, -13, 7, -3, 1, 2, -6, 16, -22, 20, -16, 8, -3, 1, -1, 9, -18, 30, -35, 27, -19, 9, -3, 1, 2, -7, 25, -40, 50, -51, 35, -22, 10, -3, 1

Offset

0,4

Comments

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}

Links

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)

Formula

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)

G.f.: g(t,x) = (1 - t + (1-2x) t^2)/(1 - t x - t^2). - Robert Israel, Jul 12 2016

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

Example

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

Maple

S:= series((1 - t + (1-2*x)*t^2)/(1 - t*x - t^2), t, 21):
R:= [seq(coeff(S, t, n), n=0..19)]:
seq(seq(coeff(R[n], x, j), j=0..n-1), n=1..20); # Robert Israel, Jul 12 2016

Mathematica

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]

Crossrefs

Cf. A114525, A168561.

Sequence in context: A077233 A282290 A178795 * A133569 A141071 A004648

Adjacent sequences: A123182 A123183 A123184 * A123186 A123187 A123188

Keyword

tabl,sign

Author

Roger L. Bagula, Oct 02 2006

Extensions

Edited by Robert Israel, Jul 13 2016

Status

approved