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A001946 a(n) = 11*a(n-1) + a(n-2).
(Formerly M2009 N0794)
26
2, 11, 123, 1364, 15127, 167761, 1860498, 20633239, 228826127, 2537720636, 28143753123, 312119004989, 3461452808002, 38388099893011, 425730551631123, 4721424167835364, 52361396397820127, 580696784543856761, 6440026026380244498, 71420983074726546239 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
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
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
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
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 *)
PROG
(Magma) [ Lucas(5*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011
CROSSREFS
Sequence in context: A354978 A247736 A155928 * A121337 A269069 A361036
KEYWORD
easy,nonn
AUTHOR
STATUS
approved

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