A136487 Coefficients of polynomial recursion p(n,x) = (x-1)*(p(n-1,x) - (x+1)*p(n-2,x)), p(0,x) = 1, p(1,x) = x+1.
1, 1, 1, 1, 1, -1, -1, -1, 0, 2, 0, -1, 2, 0, -4, 0, 2, -3, 2, 7, -4, -5, 2, 1, 5, -5, -11, 11, 7, -7, -1, 1, -8, 12, 16, -28, -8, 20, 0, -4, 13, -25, -20, 60, -2, -46, 12, 12, -3, -1, -21, 50, 19, -120, 38, 92, -50, -24, 15, 2, -1
Offset: 0
Examples
First few rows are: 1; 1, 1; {}; 1, 1, -1, -1; -1, 0, 2, 0, -1; 2, 0, -4, 0, 2; -3, 2, 7, -4, -5, 2, 1; 5, -5, -11, 11, 7, -7, -1, 1; -8, 12, 16, -28, -8, 20, 0, -4; 13, -25, -20, 60, -2, -46, 12, 12, -3, -1; -21, 50, 19, -120, 38, 92, -50, -24, 15, 2, -1;
Links
- Robert Israel, Table of n, a(n) for n = 0..10103(rows 0 to 141, flattened)
Programs
-
Magma
m:=12; function p(n,x) if n le 1 then return (x+1)^n; else return (x-1)*(p(n-1,x) - (x+1)*p(n-2,x)); end if; end function; R
:=PowerSeriesRing(Integers(), m+2); T:= func< n,k | Coefficient(R!( p(n,x) ), k) >; [1,1,1] cat [T(n,k): k in [0..n], n in [3..m]]; // G. C. Greubel, Jul 31 2023 -
Maple
F:= proc(n) option remember; expand((1-x)*procname(n-1)+(1-x^2)*procname(n-2)) end proc: F(0):= 1: F(1):= 1+x: R:=proc(n) local V,j; V:= F(n); seq(coeff(V,x,j),j=0..degree(V)) end proc: for i from 0 to 20 do R(i) od; # Robert Israel, Dec 03 2018
-
Mathematica
P[x,0]= 1; P[x,1]= x+1; P[x_,n_]:= P[x,n]= (x-1)*(P[x,n-1] - (x+1)*P[x,n-2]); Table[CoefficientList[P[x,n],x],{n,0,10}]//Flatten
-
SageMath
def p(n,x): if n<2: return (x+1)^n else: return (x-1)*(p(n-1,x) - (x+1)*p(n-2,x)) def T(n): P.
= PowerSeriesRing(QQ) return P( p(n,x) ).list() flatten([T(n) for n in range(13)]) # G. C. Greubel, Jul 31 2023
Formula
T(n, k) = coefficient [x^k] ( p(x, n) ), where p(x,n) = (x-1)*p(x,n-1) - (x^2-1)*p(x,n-2), p(x,0) = 1, p(x,1) = x+1.
Sum_{k >= 0} T(n, k) = A130706(n).
From Robert Israel, Dec 03 2018: (Start)
T(n,k) = T(n-1,k-1) - T(n-1,k) - T(n-2,k-2) + T(n-2,k).
G.f. as array: (1-2*x)/(1 + x*(y-1)+x^2*(1-y^2)).
T(n,0) = (-1)^(n+1)*A000045(n-2) for n >= 3. (End)
Extensions
Edited by Robert Israel, Dec 03 2018
Comments