OFFSET
1,1
COMMENTS
The identity (8*n^2 + 1)^2 - (16*n^2 + 4)*(2*n)^2 = 1 can be written as A081585(n)^2 - a(n)*A005843(n)^2 = 1.
Sequence found by reading the line from 20, in the direction 20, 68, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 02 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
From Bruno Berselli, Sep 06 2011: (Start)
G.f.: 4*x*(5 + 2*x + x^2)/(1-x)^3.
a(n) = 4*A053755(n). (End)
From Amiram Eldar, Mar 05 2023: (Start)
Sum_{n>=1} 1/a(n) = (coth(Pi/2)*Pi/2 - 1)/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/2)*Pi/2)/8. (End)
MATHEMATICA
a[n_] := 16*n^2 + 4; Array[a, 50] (* Amiram Eldar, Mar 05 2023 *)
PROG
(Magma) [16*n^2+4: n in [1..50]];
(PARI) a(n)=16*n^2+4 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 19 2009
EXTENSIONS
Comment rewritten by Bruno Berselli, Sep 06 2011
STATUS
approved