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A162515 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = (U^n-L^n)/d, where U=(x+d)/2, L=(x-d)/2, d=(4 + x^2)^(1/2). 5

%I

%S 0,1,1,0,1,0,1,1,0,2,0,1,0,3,0,1,1,0,4,0,3,0,1,0,5,0,6,0,1,1,0,6,0,10,

%T 0,4,0,1,0,7,0,15,0,10,0,1,1,0,8,0,21,0,20,0,5,0,1,0,9,0,28,0,35,0,15,

%U 0,1,1,0,10,0,36,0,56,0,35,0,6,0,1,0,11,0,45,0,84,0,70,0,21,0,1,1,0,12,0,55

%N Triangle of coefficients of polynomials defined by Binet form: P(n,x) = (U^n-L^n)/d, where U=(x+d)/2, L=(x-d)/2, d=(4 + x^2)^(1/2).

%C Row sums 0,1,1,2,3,5,... are the Fibonacci numbers, A000045.

%C Note that the coefficients are given in decreasing order. - _M. F. Hasler_, Dec 07 2011

%C Essentially a mirror image of A168561. - _Philippe Deléham_, Dec 08 2013

%H T. Copeland, <a href="https://tcjpn.wordpress.com/2015/10/12/the-elliptic-lie-triad-kdv-and-ricattt-equations-infinigens-and-elliptic-genera/">Addendum to Elliptic Lie Triad</a>

%F P(n,x) = x*P(n-1,x)+P(n-2,x), where P(0,x)=0 and P(1,x)=1.

%F T(n,k) = T(n-1,k)+T(n-2,k-2) for n>=2. - _Philippe Deléham_, Dec 08 2013

%e First rows:

%e 0 (row 0)

%e 1

%e 1...0

%e 1...0...1

%e 1...0...2...0

%e 1...0...3...0...1

%e 1...0...4...0...3...0

%e Row 6 matches P(6,x)=x^5 + 4*x^3 + 3*x.

%t d = (4 + x^2)^(1/2); u = (x + d)/2; l = (x - d)/2;

%t f[n_] := (u^n - l^n)/d;

%t Factor[f[1]] (* P(1,x) *)

%t Factor[f[5]] (* P(5,x) *)

%t Factor[f[8]]

%t Factor[f[21]]

%t Factor[f[144]]

%o (PARI) P(n) = my( d=(4 + x^2)^(1/2), U=(x+d)/2, L=(x-d)/2); Pol((U^n-L^n)/d) \\ _M. F. Hasler_, Dec 07 2011

%Y Cf. A000045, A162514, A162516, A162517, A053119, A049310.

%K nonn,tabf

%O 0,10

%A _Clark Kimberling_, Jul 05 2009

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Last modified January 17 19:58 EST 2019. Contains 319251 sequences. (Running on oeis4.)