A002320 a(n) = 5*a(n-1) - a(n-2).

1, 3, 14, 67, 321, 1538, 7369, 35307, 169166, 810523, 3883449, 18606722, 89150161, 427144083, 2046570254, 9805707187, 46981965681, 225104121218, 1078538640409, 5167589080827, 24759406763726, 118629444737803

Offset: 0

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A054477.

Programs

Haskell

a002320 n = a002320_list !! n
a002320_list = 1 : 3 :
   (zipWith (-) (map (* 5) (tail a002320_list)) a002320_list)
-- Reinhard Zumkeller, Oct 16 2011

Mathematica

LinearRecurrence[{5,-1},{1,3},30] (* Harvey P. Dale, Nov 13 2014 *)

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)