A158575 a(n) = 32*n^2 + 1.

1, 33, 129, 289, 513, 801, 1153, 1569, 2049, 2593, 3201, 3873, 4609, 5409, 6273, 7201, 8193, 9249, 10369, 11553, 12801, 14113, 15489, 16929, 18433, 20001, 21633, 23329, 25089, 26913, 28801, 30753, 32769, 34849, 36993, 39201, 41473, 43809, 46209, 48673, 51201, 53793

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 a(n)^2-A158574(n)*A005843(n)^2 = 1. - Comment rewritten by R. J. Mathar, Oct 16 2009

Sequence found by reading the line segment from 1 to 33 together with the line from 33, in the direction 33, 129, ..., in the square spiral whose vertices are the generalized 18-gonal numbers A274979. - Omar E. Pol, Apr 21 2021

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000217, A005843, A010021, A081585, A158563, A158574, A244082.

Cf. A274979 (generalized 18-gonal numbers).

Programs

Magma

I:=[1, 33, 129]; [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}, {1, 33, 129}, 50] (* Vincenzo Librandi, Feb 15 2012 *)
32*Range[0,40]^2+1 (* Harvey P. Dale, Jul 20 2021 *)

PARI

for(n=0, 50, print1(32*n^2+1", ")); \\ Vincenzo Librandi, Feb 15 2012

Formula

G.f.: (1+30*x+33*x^2)/(1-x)^3.

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

For n > 0 a(n) = sqrt(8*(A000217(4*n-1)^2 + A000217(4*n)^2) + 1). - J. M. Bergot, Sep 03 2015

a(n) = A244082(n) + 1. - Omar E. Pol, Apr 21 2021

From Amiram Eldar, Mar 09 2023: (Start)

Sum_{n>=0} 1/a(n) = (1 + coth(Pi/(4*sqrt(2)))*Pi/(4*sqrt(2)))/2.

Sum_{n>=0} (-1)^n/a(n) = (1 + cosech(Pi/(4*sqrt(2)))*Pi/(4*sqrt(2)))/2. (End)

From Elmo R. Oliveira, Jan 16 2025: (Start)

E.g.f.: exp(x)*(1 + 32*x + 32*x^2).

a(n) = A081585(2*n). (End)

Extensions

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