OFFSET
1,1
COMMENTS
The identity (98*n+1)^2-(49*n^2+n)*(14)^2 = 1 can be written as A157947(n)^2-a(n)*14^2 = 1. - Vincenzo Librandi, Feb 10 2012
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 14 in the first table at p. 85, case d(t) = t*(7^2*t+1)).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
G.f.: x*(-50-48*x)/(x-1)^3. - Vincenzo Librandi, Feb 10 2012
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Vincenzo Librandi, Feb 10 2012
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {50, 198, 444}, 50] (* Vincenzo Librandi, Feb 10 2012 *)
Table[49n^2+n, {n, 50}] (* Harvey P. Dale, Aug 22 2015 *)
PROG
(Magma) [ 49*n^2+n: n in [1..50] ];
(PARI) for(n=1, 50, print1(49*n^2 + n", ")); \\ Vincenzo Librandi, Feb 10 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Nov 22 2010
STATUS
approved