A001946 a(n) = 11*a(n-1) + a(n-2).

2, 11, 123, 1364, 15127, 167761, 1860498, 20633239, 228826127, 2537720636, 28143753123, 312119004989, 3461452808002, 38388099893011, 425730551631123, 4721424167835364, 52361396397820127, 580696784543856761, 6440026026380244498, 71420983074726546239

Offset: 0

Author

N. J. A. Sloane, Simon Plouffe

Keywords

• easy
• nonn

Comments

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

References

• J. 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).

Links

• 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 recurrences a(n) = k*a(n - 1) +/- a(n - 2)
• Index entries for linear recurrences with constant coefficients, signature (11, 1).

Crossrefs

Cf. A000032, A000045, A121171, A124296, A124297.

Programs

Magma

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

Maple

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

Mathematica

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 *)

Formula

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

Contribution 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(a(n+k)/a(k), k=infinity) = (A001946(n) + A049666(n)*sqrt(125))/2.

Limit(A001946(n)/A049666(n), n=infinity) = 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