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 a(n)^2-A157365(n)*A157366(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 (3,-3,1).
FORMULA
G.f: x*(4999 + 4604*x + x^2)/(1-x)^3.
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
E.g.f.: (1 + 4998*x + 4802*x^2)*exp(x) - 1. - G. C. Greubel, Feb 02 2018
EXAMPLE
For n=1, a(1)=4999; n=2, a(2)=19601; n=3, a(3)=43807
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {4999, 19601, 43807}, 50]
PROG
(Magma) I:=[4999, 19601, 43807]; [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) = 4802*n^2+196*n+1.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 28 2009
STATUS
approved