OFFSET
1,1
COMMENTS
The identity (4802*n^2+196*n+1)^2-(49*n^2+2*n)*(686*n+14)^2=1 can be written as A157367(n)^2-A157365(n)*a(n)^2=1.
This formula is the case s=7 of the identity (2*s^4*n^2+4*s^2*n+1)^2-(s^2*n^2+2*n)*(2*s^3*n+2*s)^2=1. - Bruno Berselli, Feb 11 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
G.f.: 14*x*(50-x)/(1-x)^2.
a(n) = 2*a(n-1)-a(n-2).
E.g.f.: 14*((1 + 49*x)*exp(x) - 1). - G. C. Greubel, Feb 02 2018
MATHEMATICA
686*Range[36]+14 (* or *) LinearRecurrence[{2, -1}, {700, 1386}, 50]
PROG
(Magma) I:=[700, 1386, 2072]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];
(PARI) a(n) = 686*n+14.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 28 2009
STATUS
approved