OFFSET
1,1
COMMENTS
The identity (128*n^2+32*n+1)^2-(4*n^2+n)*(64*n+8)^2=1 can be written as A157337(n)^2-A007742(n)*a(n)^2=1. This is the case s=2 of the identity (8*n^2*s^4+8*n*s^2+1)^2-(n^2*s^2+n)*(8*n*s^3+4*s)^2=1. - Vincenzo Librandi, Jan 29 2012
Likewise, the immediate identity (a(n)^2+1)^2-(a(n)^2+2)*a(n)^2 = 1 can be rewritten as A158686(8*n+1)^2-(A158686(8*n+1)+1)*a(n)^2=1. - Bruno Berselli, Feb 13 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
From Vincenzo Librandi, Jan 29 2012: (Start)
G.f.: 8*(1+7*x)/(x-1)^2. [corrected by Georg Fischer, May 12 2019]
a(n) = 2*a(n-1)-a(n-2). (End)
E.g.f.: 8*(1+8*x)*exp(x). - G. C. Greubel, Feb 01 2018
MATHEMATICA
Range[72, 5000, 64] (* Vladimir Joseph Stephan Orlovsky, Jul 16 2011 *)
LinearRecurrence[{2, -1}, {72, 136}, 50] (* Vincenzo Librandi, Jan 29 2012 *)
PROG
(Magma) I:=[72, 136]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Jan 29 2012
(PARI) for(n=1, 40, print1(64*n + 8", ")); \\ Vincenzo Librandi, Jan 29 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 27 2009
STATUS
approved