OFFSET
0,2
COMMENTS
11 = Lucas(5) divides a(1+10k), a(2+10k), and a(9+10k). Last digit of a(n) is 1, or a(n) mod 10 = 1. For odd n there exists the so-called Aurifeuillian factorization A001946(n) = Lucas(5n) = Lucas(n)*A(n)*B(n) = A000032(n)*A124296(n)*A124297(n), where A(n) = A124296(n) = 5*F(n)^2 - 5*F(n) + 1 and B(n) = A124297(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n).
LINKS
John Cerkan, Table of n, a(n) for n = 0..2373
Eric Weisstein's World of Mathematics, Aurifeuillean Factorization
Index entries for linear recurrences with constant coefficients, signature (4,-2,-6,4,2,-1).
FORMULA
a(n) = 5*Fibonacci(n)^2 + 5*Fibonacci(n) + 1.
G.f.: -(11*x^5-21*x^4-15*x^3+31*x^2-7*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). [Colin Barker, Jan 03 2013]
MATHEMATICA
Table[5*Fibonacci[n]^2+5*Fibonacci[n]+1, {n, 0, 50}]
LinearRecurrence[{4, -2, -6, 4, 2, -1}, {1, 11, 11, 31, 61, 151}, 30] (* Harvey P. Dale, Feb 23 2023 *)
PROG
(PARI) a(n)=subst(5*t*(t+1)+1, t, fibonacci(n)) \\ Charles R Greathouse IV, Jan 03 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Alexander Adamchuk, Oct 25 2006
STATUS
approved