0,1

For odd n there is the Aurifeuillian factorization a(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]. The largest prime divisors of a(n) for n>0 are listed in A121171[n] = {11, 41, 31, 2161, 151, 2521, 911, ...}. - Alexander Adamchuk, Oct 25 2006

For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

John Riordan, Combinatorial Identities, Wiley, 1968, p. 139.

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200

Tanya Khovanova, Recursive Sequences.

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.

Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.

Index entries for linear recurrences with constant coefficients, signature (11, 1).

Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).

a(n) = Lucas(5n) = Fibonacci(5n-1) + Fibonacci(5n+1). - Alexander Adamchuk, Oct 25 2006

a(n) = ((11 + 5*sqrt(5))/2)^n + ((11 - 5*sqrt(5))/2)^n. - Tanya Khovanova, Feb 06 2007

From Johannes W. Meijer, Jun 12 2010: (Start)

a(2n+1) = 11*A097842(n), a(2n) = A065705(n).

a(3n+1) = A041226(5n), a(3n+2) = A041226(5n+3), a(3n+3) = 2* A041226(5n+4).

Limit_{k->oo} a(n+k)/a(k) = (a(n) + A049666(n)*sqrt(125))/2.

Limit_{n->oo} a(n)/A049666(n) = sqrt(125). (End)

From Peter Bala, Mar 22 2015: (Start)

a(n) = Fibonacci(10*n)/Fibonacci(5*n) for n >= 1.

a(n) = ( Fibonacci(5*n + 2*k) - F(5*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.

a(n) = ( Fibonacci(5*n + 2*k + 1) + F(5*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k.

a(n) = Sum_{k = 0..2*n} binomial(2*n,k)*Lucas(n + k). (End)

a(n) = [x^n] ( (1 + 11*x + sqrt(1 + 22*x + 125*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 26 2015

E.g.f.: 2*cosh(5*sqrt(5)*x/2)*(cosh(11*x/2) + sinh(11*x/2)). - Stefano Spezia, Jan 18 2025

Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = 1/(10*sqrt(5)). - Amiram Eldar, Apr 06 2026

A001946:=(-2+11*z)/(-1+11*z+z**2); # Conjectured by Simon Plouffe in his 1992 dissertation

Table[Fibonacci[5n-1]+Fibonacci[5n+1], {n, 0, 30}] (* Alexander Adamchuk, Oct 25 2006 *)

LinearRecurrence[{11, 1}, {2, 11}, 20] (* Harvey P. Dale, Jan 25 2024 *)

(Magma) [ Lucas(5*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

Cf. A000032, A000045, A041226, A049666, A065705, A097842, A121171, A124296, A124297.

Sequence in context: A354978 A247736 A155928 * A121337 A269069 A361036

Adjacent sequences: A001943 A001944 A001945 * A001947 A001948 A001949

easy,nonn,changed

N. J. A. Sloane, Simon Plouffe

approved