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 A157364(n)^2-a(n)*A157363(n)^2=1.
The continued fraction expansion of sqrt(4*a(n)) is [14n-1; {1, 2, 2, 7n-1, 2, 2, 1, 28n-2}]. - Magus K. Chu, Sep 17 2022
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
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
G.f.: x*(47+51*x)/(1-x)^3.
E.g.f. x*(47 + 49*x)*exp(x). - G. C. Greubel, Feb 02 2018
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {47, 192, 435}, 50]
Table[49n^2-2n, {n, 40}] (* Harvey P. Dale, Jun 10 2019 *)
PROG
(Magma) I:=[47, 192, 435]; [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)=49*n^2-2*n \\ Charles R Greathouse IV, Dec 23 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Feb 28 2009
STATUS
approved