A158574 a(n) = 256*n^2 + 16.

16, 272, 1040, 2320, 4112, 6416, 9232, 12560, 16400, 20752, 25616, 30992, 36880, 43280, 50192, 57616, 65552, 74000, 82960, 92432, 102416, 112912, 123920, 135440, 147472, 160016, 173072, 186640, 200720, 215312, 230416, 246032, 262160, 278800, 295952, 313616, 331792

Offset: 0

Author

Vincenzo Librandi, Mar 21 2009

Keywords

• nonn
• easy

Comments

The identity (32*n^2 + 1)^2 - (256*n^2 + 16)*(2*n)^2 = 1 can be written as A158575(n)^2 - a(n)*A005843(n)^2 = 1.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A005843, A158575.

Programs

Magma

I:=[16, 272, 1040]; [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, Feb 15 2012

Mathematica

LinearRecurrence[{3, -3, 1}, {16, 272, 1040}, 50] (* Vincenzo Librandi, Feb 15 2012 *)

PARI

for(n=0, 50, print1(256*n+16", ")); \\ Vincenzo Librandi, Feb 15 2012

Formula

G.f.: 16*(1 + 14*x + 17*x^2)/(1-x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

From Amiram Eldar, Mar 09 2023: (Start)

Sum_{n>=0} 1/a(n) = (coth(Pi/4)*Pi/4 + 1)/32.

Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/4)*Pi/4 + 1)/32. (End)

Extensions

Comment rewritten, a(0) added by R. J. Mathar, Oct 16 2009