A157916 a(n) = 50*n^2 + 1.

51, 201, 451, 801, 1251, 1801, 2451, 3201, 4051, 5001, 6051, 7201, 8451, 9801, 11251, 12801, 14451, 16201, 18051, 20001, 22051, 24201, 26451, 28801, 31251, 33801, 36451, 39201, 42051, 45001, 48051, 51201, 54451, 57801, 61251, 64801, 68451, 72201, 76051, 80001

Offset

1,1

Comments

The identity (50*n^2+1)^2 - (625*n^2+25)*(2*n)^2 = 1 can be written as a(n)^2 - A157915(n)*A005843(n)^2 = 1. - Vincenzo Librandi, Feb 10 2012

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

From Vincenzo Librandi, Feb 10 2012: (Start)

G.f.: x*(51+48*x+x^2)/(1-x)^3.

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

From Amiram Eldar, Mar 07 2023: (Start)

Sum_{n>=1} 1/a(n) = (coth(Pi/(5*sqrt(2)))*Pi/(5*sqrt(2)) - 1)/2.

Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/(5*sqrt(2)))*Pi/(5*sqrt(2)))/2. (End)

Maple

A157916:=n->50*n^2+1: seq(A157916(n), n=1..60); # Wesley Ivan Hurt, Jan 27 2017

Mathematica

LinearRecurrence[{3, -3, 1}, {51, 201, 451}, 40] (* Vincenzo Librandi, Feb 10 2012 *)

Prog

(Magma) I:=[51, 201, 451]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 10 2012
(PARI) for(n=1, 40, print1(50*n^2 + 1", ")); \\ Vincenzo Librandi, Feb 10 2012

Crossrefs

Cf. A005843, A157915.

Sequence in context: A369922 A031431 A157365 * A007264 A158640 A107253

Adjacent sequences: A157913 A157914 A157915 * A157917 A157918 A157919

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 09 2009

Status

approved