A157336 a(n) = 8*(8*n + 1).

72, 136, 200, 264, 328, 392, 456, 520, 584, 648, 712, 776, 840, 904, 968, 1032, 1096, 1160, 1224, 1288, 1352, 1416, 1480, 1544, 1608, 1672, 1736, 1800, 1864, 1928, 1992, 2056, 2120, 2184, 2248, 2312, 2376, 2440, 2504, 2568, 2632, 2696, 2760, 2824, 2888, 2952

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 A157337(n)^2-A007742(n)*a(n)^2=1. This is the case s=2 of the identity (8*n^2*s^4+8*n*s^2+1)^2-(n^2*s^2+n)*(8*n*s^3+4*s)^2=1. - Vincenzo Librandi, Jan 29 2012

Likewise, the immediate identity (a(n)^2+1)^2-(a(n)^2+2)*a(n)^2 = 1 can be rewritten as A158686(8*n+1)^2-(A158686(8*n+1)+1)*a(n)^2=1. - Bruno Berselli, Feb 13 2012

Links

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

Crossrefs

Cf. A007742, A157337.

Programs

Magma

I:=[72, 136]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Jan 29 2012

Mathematica

Range[72, 5000, 64] (* Vladimir Joseph Stephan Orlovsky, Jul 16 2011 *)
LinearRecurrence[{2, -1}, {72, 136}, 50] (* Vincenzo Librandi, Jan 29 2012 *)

PARI

for(n=1, 40, print1(64*n + 8", ")); \\ Vincenzo Librandi, Jan 29 2012

Formula

From Vincenzo Librandi, Jan 29 2012: (Start)

G.f.: 8*(1+7*x)/(x-1)^2. [corrected by Georg Fischer, May 12 2019]

a(n) = 2*a(n-1)-a(n-2). (End)

E.g.f.: 8*(1+8*x)*exp(x). - G. C. Greubel, Feb 01 2018