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
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 09 2009
STATUS
approved