A162514 Triangle of coefficients of polynomials defined by the Binet form P(n,x) = U^n + L^n, where U = (x + d)/2, L = (x - d)/2, d = (4 + x^2)^(1/2). Decreasing powers of x.
2, 1, 0, 1, 0, 2, 1, 0, 3, 0, 1, 0, 4, 0, 2, 1, 0, 5, 0, 5, 0, 1, 0, 6, 0, 9, 0, 2, 1, 0, 7, 0, 14, 0, 7, 0, 1, 0, 8, 0, 20, 0, 16, 0, 2, 1, 0, 9, 0, 27, 0, 30, 0, 9, 0, 1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2, 1, 0, 11, 0, 44, 0, 77, 0, 55, 0, 11, 0, 1, 0, 12, 0, 54, 0, 112, 0, 105, 0, 36, 0, 2, 1, 0, 13, 0
Offset: 0
Examples
Triangle begins 2; == 2 1, 0; == x + 0 1, 0, 2; == x^2 + 2 1, 0, 3, 0; == x^3 + 3*x + 0 1, 0, 4, 0, 2; 1, 0, 5, 0, 5, 0; 1, 0, 6, 0, 9, 0, 2; 1, 0, 7, 0, 14, 0, 7, 0; 1, 0, 8, 0, 20, 0, 16, 0, 2; 1, 0, 9, 0, 27, 0, 30, 0, 9, 0; 1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2; ... From _Wolfdieter Lang_, Aug 07 2014: (Start) The row polynomials R(n, x) are: R(0, x) = 2, R(1, x) = 1 = x*P(1,1/x), R(2, x) = 1 + 2*x^2 = x^2*P(2,1/x), R(3, x) = 1 + 3*x^2 = x^3*P(3,1/x), ... (End)
Links
- G. C. Greubel, Rows n=0..100 of triangle, flattened
Programs
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Mathematica
Table[Reverse[CoefficientList[LucasL[n, x], x]], {n, 0, 12}]//Flatten (* G. C. Greubel, Nov 05 2018 *)
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PARI
P(n)= { local(U, L, d, r, x); if ( n<0, return(0) ); x = 'x+O('x^(n+1)); d=(4 + x^2)^(1/2); U=(x+d)/2; L=(x-d)/2; r = U^n+L^n; r = truncate(r); return( r ); } for (n=0, 10, print(Vec(P(n))) ); /* show triangle */ /* Joerg Arndt, Jul 24 2011 */
Formula
P(n,x) = x*P(n-1,x) + P(n-2,x) for n >= 2, P(0,x) = 2, P(1,x) = x.
From Wolfdieter Lang, Aug 07 2014: (Start)
T(n,m) = [x^(n-m)] P(n,x), m = 0, 1, ..., n and n >= 0.
G.f. of polynomials P(n,x): (2 - x*z)/(1 - x*z - z^2).
G.f. of row polynomials R(n,x) = Sum_{m=0..n} T(n,m)*x^m: (2 - z)/(1 - z - (x*z)^2) (rows for P(n,x) reversed).
(End)
For n > 0, T(n,2*m+1) = 0, T(n,2*m) = A034807(n,m). - Paolo Bonzini, Jun 23 2016
Extensions
Name clarified by Wolfdieter Lang, Aug 07 2014
Comments