A157914 a(n) = 8*n^2 - 1.

7, 31, 71, 127, 199, 287, 391, 511, 647, 799, 967, 1151, 1351, 1567, 1799, 2047, 2311, 2591, 2887, 3199, 3527, 3871, 4231, 4607, 4999, 5407, 5831, 6271, 6727, 7199, 7687, 8191, 8711, 9247, 9799, 10367, 10951, 11551, 12167, 12799, 13447, 14111, 14791

Offset

1,1

Comments

The identity (8*n^2 - 1)^2 - (16*n^2 - 4)*(2*n)^2 = 1 can be written as a(n)^2 - A158443(n)*A005843(n)^2 = 1.

Sequence found by reading the line from 7, in the direction 7, 31, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 03 2011

Bisection of A195605 (odd part). - Bruno Berselli, Sep 21 2011

The identity (8*n^2 - 1)^2 - (64*n^2 - 16)*(n)^2 = 1 can be written as a(n)^2 - A157913(n)*(n)^2 = 1. - Vincenzo Librandi, Feb 09 2012

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

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

G.f: x*(7+10*x-x^2)/(1-x)^3.

a(n) = A139098(n) - 1. - Omar E. Pol, Sep 03 2011

E.g.f.: (8*x^2 + 8*x - 1)*exp(x) + 1. - G. C. Greubel, Jul 15 2017

From Amiram Eldar, Feb 04 2021: (Start)

Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(8))*cot(Pi/sqrt(8)))/2.

Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(8))*csc(Pi/sqrt(8)) - 1)/2.

Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(8))*csc(Pi/sqrt(8)).

Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(8))/sqrt(2). (End)

Mathematica

Table[8n^2-1, {n, 50}]

Prog

(Magma) I:=[7, 31, 71]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]];
(PARI) a(n)=8*n^2-1 \\ Charles R Greathouse IV, Sep 03 2011

Crossrefs

Cf. A157913, A158443, A005843, A139098, A195605.

Sequence in context: A163354 A105428 A050547 * A090684 A033199 A304163

Adjacent sequences: A157911 A157912 A157913 * A157915 A157916 A157917

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 09 2009

Status

approved