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 - a(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
Also, the identity (49*n + 1)^2 - (49*n^2 + 2*n)*7^2 = 1 can be written as A158066(n)^2 - a(n)*7^2 = 1 (see Barbeau's paper in link). - Vincenzo Librandi, Feb 11 2012
The continued fraction expansion of sqrt(4*a(n)) is [14n; {3, 1, 1, 7n-1, 1, 1, 3, 28n}]. - Magus K. Chu, Sep 17 2022
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
E. J. Barbeau, Polynomial Excursions, Chapter 10: Diophantine equations (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(7^2*t+2)).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
G.f.: x*(51+47*x)/(1-x)^3.
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
E.g.f.: (51*x + 49*x^2)*exp(x). - G. C. Greubel, Feb 02 2018
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {51, 200, 447}, 50]
PROG
(Magma) I:=[51, 200, 447]; [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