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A158601
a(n) = 400*n^2 + 20.
2
20, 420, 1620, 3620, 6420, 10020, 14420, 19620, 25620, 32420, 40020, 48420, 57620, 67620, 78420, 90020, 102420, 115620, 129620, 144420, 160020, 176420, 193620, 211620, 230420, 250020, 270420, 291620, 313620, 336420, 360020, 384420, 409620, 435620, 462420, 490020
OFFSET
0,1
COMMENTS
The identity (40*n^2 + 1)^2 - (400*n^2 + 20)*(2*n)^2 = 1 can be written as A158602(n)^2 - a(n)*A005843(n)^2 = 1.
LINKS
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
FORMULA
G.f.: -20*(1 + 18*x + 21*x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 16 2023: (Start)
Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/40.
Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/40. (End)
From Elmo R. Oliveira, Jan 15 2025: (Start)
E.g.f.: 20*exp(x)*(1 + 20*x + 20*x^2).
a(n) = 20*A158493(n). (End)
MATHEMATICA
400 Range[0, 40]^2+20 (* Harvey P. Dale, Feb 05 2011 *)
LinearRecurrence[{3, -3, 1}, {20, 420, 1620}, 50] (* Vincenzo Librandi, Feb 16 2012 *)
PROG
(Magma) I:=[20, 420, 1620]; [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 16 2012
(PARI) for(n=0, 40, print1(400*n^2 + 20", ")); \\ Vincenzo Librandi, Feb 16 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Mar 22 2009
EXTENSIONS
Comment rewritten, formula replaced by R. J. Mathar, Oct 28 2009
STATUS
approved