A124296 a(n) = 5*F(n)^2 - 5*F(n) + 1, where F(n) = Fibonacci(n).

1, 1, 1, 11, 31, 101, 281, 781, 2101, 5611, 14851, 39161, 102961, 270281, 708761, 1857451, 4865911, 12744061, 33372361, 87382901, 228792301, 599019851, 1568309051, 4105974961, 10749725281, 28143378001, 73680695281, 192899171531

Offset

0,4

Comments

11 = Lucas(5) divides a(3+10k), a(7+10k), and a(8+10k). The last digit of a(n) is 1, so 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..2375

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.: -(x^5+9*x^4-15*x^3+x^2+3*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}]
5#^2-5#+1&/@Fibonacci[Range[0, 30]] (* Harvey P. Dale, Nov 29 2011 *)

Prog

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

Crossrefs

Cf. A000032, A000045, A121171, A001946, A124297.

Bisections: A001604, A156094.

Sequence in context: A068841 A316982 A192246 * A223388 A152220 A272263

Adjacent sequences: A124293 A124294 A124295 * A124297 A124298 A124299

Keyword

nonn,easy

Author

Alexander Adamchuk, Oct 25 2006

Status

approved