OFFSET
0,2
COMMENTS
Sequence found by reading the segment (1, 11) together with the line from 11, in the direction 11, 41, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Sep 10 2011
The identity (10n^2 + 1)^2 - (25n^2 + 5)*(2n)^2 = 1 can be written as a(n)^2 - A158445(n)*A005843(n)^2 = 1. - Vincenzo Librandi, Jan 03 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) = A033583(n) + 1.
For n > 0: a(n) = A010010(n)/2.
From Vincenzo Librandi, Jan 03 2012: (Start)
G.f: x*(11 + 8*x + x^2)/(1-x)^3.
a(n) = a(-n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(10))*coth(Pi/sqrt(10)))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(10))*csch(Pi/sqrt(10)))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(10))*sinh(Pi/sqrt(5)).
Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(10))*csch(Pi/sqrt(10)). (End)
E.g.f.: exp(x)*(1 + 10*x + 10*x^2). - Stefano Spezia, Feb 05 2021
MATHEMATICA
Table[10*n^2+1, {n, 0, 50}] (* Vincenzo Librandi, Jan 03 2012 *)
PROG
(PARI) a(n)=10*n^2+1 \\ Charles R Greathouse IV, Oct 16 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, Mar 13 2009
STATUS
approved