A158482 a(n) = 14*n^2 + 1.

15, 57, 127, 225, 351, 505, 687, 897, 1135, 1401, 1695, 2017, 2367, 2745, 3151, 3585, 4047, 4537, 5055, 5601, 6175, 6777, 7407, 8065, 8751, 9465, 10207, 10977, 11775, 12601, 13455, 14337, 15247, 16185, 17151, 18145, 19167, 20217, 21295, 22401

Offset: 1

Author

Vincenzo Librandi, Mar 20 2009

Keywords

• nonn
• easy

Comments

The identity (14*n^2 + 1)^2 - (49*n^2 + 7)*(2*n)^2 = 1 can be written as a(n)^2 - A158481(n)*A005843(n)^2 = 1.

Sequence found by reading the line from 15, in the direction 15, 57, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277. Also sequence found by reading the same line in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 13 2011

Links

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

Crossrefs

Cf. A005843, A118277, A158481, A195019, A195020.

Programs

Magma

I:=[15, 57, 127]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];

Mathematica

LinearRecurrence[{3,-3,1},{15,57,127},50]

PARI

a(n) = 14*n^2+1;

Formula

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

G.f: x*(15+12*x+x^2)/(1-x)^3.

From Amiram Eldar, Feb 05 2021: (Start)

Sum_{n>=0} 1/a(n) = (1 - (Pi/sqrt(14))*coth(Pi/sqrt(14)))/2.

Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(14))*csch(Pi/sqrt(14)))/2.

Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(14))*sinh(Pi/sqrt(7)).

Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(14))*csch(Pi/sqrt(14)). (End)