OFFSET
0,2
COMMENTS
a(n)/a(n-1) tends to (11+sqrt(21))/2 = 7.79128784...
For n>=2, a(n) equals 5^n times the permanent of the (2n-2)X(2n-2) tridiagonal matrix with 1/sqrt(5)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..1119
Index entries for linear recurrences with constant coefficients, signature (11,-25).
FORMULA
G.f.: (1-6x)/(1-11x+25x^2).
a(n) = Sum_{k=0..n} A165253(n,k)*5^(n-k).
a(n) = ((21-sqrt(21))*(11+sqrt(21))^n+(21+sqrt(21))*(11-sqrt(21))^n )/(42*2^n). [Klaus Brockhaus, Sep 26 2009]
MATHEMATICA
LinearRecurrence[{11, -25}, {1, 5}, 30] (* Harvey P. Dale, Oct 02 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Philippe Deléham, Sep 14 2009
STATUS
approved