OFFSET
0,2
COMMENTS
Together with A002310 these are the two sequences satisfying the requirement that (a(n)^2 + a(n-1)^2)/(1 - a(n)*a(n-1)) be an integer; in both cases this integer is -5. - Floor van Lamoen, Oct 26 2001
REFERENCES
From a posting to Netnews group sci.math by ksbrown(AT)seanet.com (K. S. Brown) on Aug 15 1996.
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See p. 44.
Tanya Khovanova, Recursive Sequences
MathPages, N = (x^2 + y^2)/(1+xy) is a Square
Index entries for linear recurrences with constant coefficients, signature (5,-1).
FORMULA
Sequences A002310, A002320 and A049685 have this in common: each one satisfies a(n+1) = (a(n)^2+5)/a(n-1) - Graeme McRae, Jan 30 2005
G.f.: (1-2x)/(1-5x+x^2). - Philippe Deléham, Nov 16 2008
a(n) = Sum_{k = 0..n} A238731(n,k)*2^k. - _Philippe Deléham, Mar 05 2014
E.g.f.: exp(5*x/2)*(sqrt(21)*cosh(sqrt(21)*x/2) + sinh(sqrt(21)*x/2))/sqrt(21). - Stefano Spezia, Jul 07 2025
From Peter Bala, Jul 07 2025: (Start)
a(n) = ( (4 + sqrt(21))*(5 - sqrt(21))^(n+1) - (4 - sqrt(21))*(5 + sqrt(21))^(n+1) )/(2^(n+1)*sqrt(21)).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n) + 5/a(2*n)) = 1/15, since 5/(a(2*n) + 5/a(2*n)) = 1/a(2*n-1) + 1/a(2*n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n-1) + 5/a(2*n-1)) = 1/5, since 5/(a(2*n-1) + 5/a(2*n-1)) = 1/a(2*n-2) + 1/a(2*n). (End)
MATHEMATICA
LinearRecurrence[{5, -1}, {1, 3}, 30] (* Harvey P. Dale, Nov 13 2014 *)
PROG
(Haskell)
a002320 n = a002320_list !! n
a002320_list = 1 : 3 :
(zipWith (-) (map (* 5) (tail a002320_list)) a002320_list)
-- Reinhard Zumkeller, Oct 16 2011
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Joe Keane (jgk(AT)jgk.org)
STATUS
approved