G.f.: 1/(1 - 5*x - x^2).
a(n) = Sum_{alpha=RootOf(-1+5*z+z^2)} (1/29)*(5+2*alpha)*alpha^(-1-n).
a(n-1) = (((5 + sqrt(29))/2)^n - ((5 - sqrt(29))/2)^n)/sqrt(29). -
Gary W. Adamson, Jul 01 2003
a(n) = U(n, 5*i/2)*(-i)^n with i^2 = -1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See triangle
A049310.
Let M = {{0, 1}, {1, 5}}, then a(n) is the lower-right term of M^n. -
Roger L. Bagula, May 29 2005
a(n) = F(n, 5), the n-th Fibonacci polynomial evaluated at x = 5. -
T. D. Noe, Jan 19 2006
a(n) = denominator of n-th convergent to [1, 4, 5, 5, 5, ...], for n > 0. Continued fraction [1, 4, 5, 5, 5, ...] = 0.807417596..., the inradius of a right triangle with legs 2 and 5. n-th convergent =
A100237(n)/
A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135, 566/701, ... -
Gary W. Adamson, Dec 21 2007
Limit_{k->oo} a(n+k)/a(k) = (
A087130(n) + a(n-1)*sqrt(29))/2.
Limit_{n->oo}
A087130(n)/a(n-1) = sqrt(29). (End)
Define the 2 X 2 matrix A = {{1, 1}, {5, 4}}. Then:
a(n) is the upper-left term of (1/5)*(A^(n+2) - A^(n+1));
a(n) is the upper-right term of A^(n+1);
a(n) is the lower-left term of (1/5)*A^(n+1);
a(n) is the lower-right term of (Sum_{k=0..n} A^k). (End)
G.f.: x/(1 - 5*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 5 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). -
Peter Bala, May 08 2024
E.g.f.: exp(5*x/2)*(29*cosh(sqrt(29)*x/2) + 5*sqrt(29)*sinh(sqrt(29)*x/2))/29. -
Stefano Spezia, Nov 20 2025
