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%I #28 Jan 04 2024 06:52:47
%S 2,1,2,3,2,2,4,9,3,2,7,16,18,4,2,11,35,40,30,5,2,18,66,105,80,45,6,2,
%T 29,126,231,245,140,63,7,2,47,232,504,616,490,224,84,8,2,76,423,1044,
%U 1512,1386,882,336,108,9,2,123,760,2115,3480,3780,2772,1470,480,135,10,2
%N Triangular array T(n,k) = C(n,k)*Lucas(n-k), 0 <= k <= n.
%F G.f.: (2 - (2x + 1)*t)/(1 - (2x + 1)*t + (x^2 + x - 1)*t^2) = 2 + (1 + 2*x)*t + (3 + 2*x + 2*x^2)*t^2 + (4 + 9*x + 3*x^2 + 2*x^3)*t^3 + ... .
%F The row polynomials L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k satisfy L(n,x)*F(n,x) = F(2*n,x), where F(n,x) = Sum_{k = 0..n} C(n,k)*Fibonacci(n-k)*x^k.
%F Other identities and formulas include:
%F L(n+1,x)^2 - L(n,x)*L(n+2,x) = -5*(x^2 + x - 1)^n;
%F L(n+1,x) - (x^2 + x - 1)*L(n-1,x) = 5*F(n,x) for n >= 1;
%F L(2*n,x) - 2*(x^2 + x - 1)^n = 5*F(n,x)^2;
%F L(n,2*x) = Sum_{ k = 0..n} C(n,k)*L(n-k,x)*x^k;
%F L(n,3*x) = Sum_{ k = 0..n} C(n,k)*L(n-k,2*x)*x^k etc.
%F Sum_{k = 0..n} C(n,k)*L(k,x)*F(n-k,x) = 2^n*F(n,x).
%F Row sums: L(n,1) = Lucas(2*n); Alternating row sums: L(n,-1) = (-1)^n*Lucas(n); L(n,1/phi) = (-1)^n*L(n,-phi) = sqrt(5)^n for n >= 1, where phi = (1 + sqrt(5))/2.
%F The polynomials L(n,-x) satisfy a Riemann hypothesis: the zeros of L(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
%F From _Peter Bala_, Jun 29 2016:
%F L(n,x) = (x + phi)^n + (x - 1/phi)^n, where phi = (1 + sqrt(5))/2. The zeros of L(n,x) are given by -1/2 - i*sqrt(5)/2*cot( (2*k - 1)*Pi/(2*n) ) for k = 1..n.
%F d/dx(L(n,x)) = n*L(n-1,x).
%F L(-n,x) = L(n,x)/(x^2 + x - 1)^n.
%F L(n,x - 1) = (-1)^n*L(n,-x).
%F L(n,x)^2 - L(2*n,x) = 2*(x^2 + x - 1)^n.
%F L(n,x)^3 - L(3*n,x) = 3*(x^2 + x - 1)^n*L(n,x).
%F L(n,x)^4 - L(4*n,x) = 4*(x^2 + x - 1)^n*L(n,x)^2 - 2*(x^2 + x - 1)^(2*n).
%F If n divides m and m/n is odd then L(n,x) divides L(m,x) in the polynomial ring Z[x].
%F L(n,x) = F(n+1,x) - (x^2 + x - 1)*F(n-1,x) = 2*F(n+1,x) - (2*x + 1)*F(n,x), where F(n,x) = Sum_{k = 0..n} binomial(n,k)*Fibonacci(n-k)*x^k denotes the n-th row polynomial of A094440 (taken with an offset of 0).
%F exp( Sum_{n >= 1} L(n,x)*z^n/n ) = Sum_{n >= 0} F(n+1,x)*z^n.
%F exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*z^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)* F(n+3,x)*z^n. (End)
%F From _Peter Bala_, Dec 31 2023: (Start)
%F For n >= 1, the n-th row polynomial L(n,x) is the numerator of 1/(n-1)! * (d/dx)^(n-1) (2*x + 1)/(1 - x - x^2).
%F Recurrence: for n >= 1, n*L(n+1,x) = n*(2*x + 1)*L(n,x) + (1 - x - x^2)* d/dx(L(n,x)) with L(1,x) = 2*x + 1. (End)
%e Triangle starts
%e 2;
%e 1, 2;
%e 3, 2, 2;
%e 4, 9, 3, 2;
%e 7, 16, 18, 4, 2;
%e 11, 35, 40, 30, 5, 2;
%e 18, 66, 105, 80, 45, 6, 2;
%e ...
%p with(combinat): lucas := n -> fibonacci(n-1) + fibonacci(n+1): T := (n,k) -> binomial(n,k)*lucas(n-k): for n from 0 to 10 do seq( T(n,k), k = 0..n) od;
%t Flatten[Table[Binomial[n,k]LucasL[n-k],{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, Nov 06 2011 *)
%Y Cf. A000032 (T(n,0)), A000045, A005248 (row sums), A061084 (alter. row sums), A094440.
%K easy,nonn,tabl
%O 0,1
%A _Peter Bala_, Aug 17 2007
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