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)^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).
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)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {15, 57, 127}, 50]
PROG
(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]];
(PARI) a(n) = 14*n^2+1;
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 20 2009
STATUS
approved