

A158482


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


5



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
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OFFSET

1,1


COMMENTS

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


FORMULA

a(n) = 3*a(n1) 3*a(n2) +a(n3).
G.f: x*(15+12*x+x^2)/(1x)^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)


MATHEMATICA

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


PROG

(Magma) I:=[15, 57, 127]; [n le 3 select I[n] else 3*Self(n1)3*Self(n2)+1*Self(n3): n in [1..40]];
(PARI) a(n) = 14*n^2+1;


CROSSREFS

Cf. A005843, A158481.
Sequence in context: A020222 A256867 A336251 * A184223 A084815 A240152
Adjacent sequences: A158479 A158480 A158481 * A158483 A158484 A158485


KEYWORD

nonn,easy


AUTHOR

Vincenzo Librandi, Mar 20 2009


STATUS

approved



