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A158556
a(n) = 28*n^2 + 1.
3
1, 29, 113, 253, 449, 701, 1009, 1373, 1793, 2269, 2801, 3389, 4033, 4733, 5489, 6301, 7169, 8093, 9073, 10109, 11201, 12349, 13553, 14813, 16129, 17501, 18929, 20413, 21953, 23549, 25201, 26909, 28673, 30493, 32369, 34301, 36289, 38333, 40433, 42589, 44801, 47069
OFFSET
0,2
COMMENTS
The identity (28*n^2 + 1)^2 - (196*n^2 + 14) * (2*n)^2 = 1 can be written as a(n)^2 - A158555(n)*A005843(n)^2 = 1.
FORMULA
G.f.: (1 + 26*x + 29*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 09 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(28))*Pi/sqrt(28) + 1)/2.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(28))*Pi/sqrt(28) + 1)/2. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: exp(x)*(1 + 28*x + 28*x^2).
a(n) = A247541(2*n). (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1, 29, 113}, 50] (* Vincenzo Librandi, Mar 02 2012 *)
28*Range[0, 40]^2+1 (* Harvey P. Dale, Jun 30 2022 *)
PROG
(Magma) I:=[1, 29, 113]; [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, Mar 02 2012
(PARI) for(n=0, 40, print1(28*n^2+1", ")); \\ Vincenzo Librandi, Mar 02 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 21 2009
EXTENSIONS
Comment rewritten, a(0) added by R. J. Mathar, Oct 16 2009
STATUS
approved