OFFSET
1,1
COMMENTS
The identity (36*n + 1)^2 - (36*n^2 + 2*n)*6^2 = 1 can be written as a(n)^2 - A158064(n)*6^2 = 1. - Vincenzo Librandi, Feb 11 2012
Parametrize Pythagorean triangles with parameters a and b and side lengths x = b^2 - a^2, y = 2*a*b and z = a^2 + b^2. Generate one Pythagorean triangle with a=n-1 and b=n and side lengths (x1, y1, z1), and another one with a=n, b=n+1 and side lengths (x2, y2, z2). Then 2*a(n) = x2 - x1 + 12*(y2-y1) + 6*(z2-z1). - J. M. Bergot, Jul 16 2013
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*(6^2*t+2)).
Index entries for linear recurrences with constant coefficients, signature (2, -1).
FORMULA
G.f.: x*(37-x)/(1-x)^2. - Vincenzo Librandi, Feb 11 2012
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Feb 11 2012
a(n) = 12*A008585(n) + 1 (see illustration in links). - John Elias, Jun 29 2021
MATHEMATICA
Range[37, 2000, 36] (* Vladimir Joseph Stephan Orlovsky, Jun 15 2011 *)
LinearRecurrence[{2, -1}, {37, 73}, 50] (* Vincenzo Librandi, Feb 11 2012 *)
PROG
(Magma) I:=[37, 73]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Feb 11 2012
(PARI) for(n=1, 40, print1(36*n + 1", ")); \\ Vincenzo Librandi, Feb 11 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 12 2009
STATUS
approved