A158537 a(n) = 22*n^2 + 1.

1, 23, 89, 199, 353, 551, 793, 1079, 1409, 1783, 2201, 2663, 3169, 3719, 4313, 4951, 5633, 6359, 7129, 7943, 8801, 9703, 10649, 11639, 12673, 13751, 14873, 16039, 17249, 18503, 19801, 21143, 22529, 23959, 25433, 26951, 28513, 30119, 31769, 33463, 35201, 36983

Offset

0,2

Comments

From the identity (22*n^2 + 1)^2 - (121*n^2 + 11)*(2*n)^2 = 1 we derive a(n)^2 - A158536(n) * A005843(n)^2 = 1.

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

G.f.: (1 + 20*x + 23*x^2)/(1-x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

From Amiram Eldar, Mar 06 2023: (Start)

Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(22))*Pi/sqrt(22) + 1)/2.

Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(22))*Pi/sqrt(22) + 1)/2. (End)

Mathematica

LinearRecurrence[{3, -3, 1}, {1, 23, 89}, 50] (* Vincenzo Librandi, Feb 12 2012 *)
22*Range[0, 40]^2+1 (* Harvey P. Dale, May 04 2019 *)

Prog

(Magma) I:=[1, 23, 89]; [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 10 2012
(PARI) for(n=1, 40, print1(22*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 10 2012

Crossrefs

Cf. A005843, A158536.

Sequence in context: A050255 A014088 A244453 * A117049 A142062 A050529

Adjacent sequences: A158534 A158535 A158536 * A158538 A158539 A158540

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 21 2009

Extensions

Comment rewritten, a(0) added by R. J. Mathar, Oct 16 2009

Status

approved