A157337 a(n) = 128*n^2 + 32*n + 1.

161, 577, 1249, 2177, 3361, 4801, 6497, 8449, 10657, 13121, 15841, 18817, 22049, 25537, 29281, 33281, 37537, 42049, 46817, 51841, 57121, 62657, 68449, 74497, 80801, 87361, 94177, 101249, 108577, 116161, 124001, 132097, 140449, 149057

Offset: 1

Author

Vincenzo Librandi, Feb 27 2009

Keywords

• nonn
• easy

Comments

The identity (128*n^2+32*n+1)^2 - (4*n^2+n)*(64*n+8)^2 = 1 can be written as a(n)^2 - A007742(n)*A157336(n)^2 = 1 (see also second part of the comment in A157336). - Vincenzo Librandi, Jan 29 2012

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..10000
• Vincenzo Librandi, X^2-AY^2=1
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A007742, A157336.

Programs

Magma

I:=[161, 577, 1249]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // Vincenzo Librandi, Jan 29 2012

Mathematica

LinearRecurrence[{3,-3,1},{161,577,1249},50] (* Vincenzo Librandi, Jan 29 2012 *)

PARI

for(n=1, 40, print1(128*n^2 + 32*n + 1", ")); \\ Vincenzo Librandi, Jan 29 2012

Formula

G.f.: x*(x^2 + 94*x + 161)/(1-x)^3. - Vincenzo Librandi, Jan 29 2012

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jan 29 2012

a(n) = 2*A017077(n)^2 - 1. - Bruno Berselli, Jan 29 2012

E.g.f.: (1 + 160*x + 128*x^2)*exp(x) - 1. - G. C. Greubel, Feb 01 2018