A124297 a(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n).

1, 11, 11, 31, 61, 151, 361, 911, 2311, 5951, 15401, 40051, 104401, 272611, 712531, 1863551, 4875781, 12760031, 33398201, 87424711, 228859951, 599129311, 1568486161, 4106261531, 10750188961, 28144128251, 73681909211, 192901135711

Offset: 0

Author

Alexander Adamchuk, Oct 25 2006

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A000032, A000045, A121171, A001946, A124296.

Bisections: A001604, A156095.

Programs

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 *)

PARI

a(n)=subst(5*t*(t+1)+1,t,fibonacci(n)) \\ Charles R Greathouse IV, Jan 03 2013

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]