OFFSET
0,2
COMMENTS
The identity (8*n^2 +1)^2 - (64*n^2 +16)*n^2 = 1 can be written as a(n)^2 -A157912(n)*n^2 = 1 for n>0. - Vincenzo Librandi, Feb 09 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 8*n^2 + 1.
G.f.: (1+3*x)^2/(1-x)^3.
a(n) = a(n-1) + 16*n - 8 with a(0)=1. - Vincenzo Librandi, Aug 08 2010
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(8))*coth(Pi/sqrt(8)))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(8))*csch(Pi/sqrt(8)))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(8))*sinh(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(8))*csch(Pi/sqrt(8)). (End)
E.g.f.: (1 +8*x +8*x^2)*exp(x). - G. C. Greubel, May 26 2021
MAPLE
seq(1+8*n^2, n=0..100); # Robert Israel, Sep 04 2015
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1, 9, 33}, 40] (* Vincenzo Librandi, Feb 09 2012 *)
PROG
(Magma) I:=[1, 9, 33]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 09 2012
(PARI) for(n=0, 50, print1(8*n^2+1", ")); \\ Vincenzo Librandi, Feb 09 2012
(Sage) [8*n^2 +1 for n in (0..40)] # G. C. Greubel, May 26 2021
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Mar 23 2003
STATUS
approved