A157919 a(n) = 50*n^2 - 1.

49, 199, 449, 799, 1249, 1799, 2449, 3199, 4049, 4999, 6049, 7199, 8449, 9799, 11249, 12799, 14449, 16199, 18049, 19999, 22049, 24199, 26449, 28799, 31249, 33799, 36449, 39199, 42049, 44999, 48049, 51199, 54449, 57799, 61249, 64799, 68449, 72199, 76049, 79999

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 - A157918(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*(49 + 52*x - x^2)/(x-1)^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) = (1 - cot(Pi/(5*sqrt(2)))*Pi/(5*sqrt(2)))/2.

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

Mathematica

LinearRecurrence[{3, -3, 1}, {49, 199, 449}, 50] (* Vincenzo Librandi, Feb 10 2012 *)
50 Range[40]^2-1 (* Harvey P. Dale, Dec 15 2018 *)

Prog

(Magma) I:=[49, 199, 449]; [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(50*n^2 - 1", ")); \\ Vincenzo Librandi, Feb 10 2012

Crossrefs

Cf. A157918, A005843.

Sequence in context: A198386 A351663 A016982 * A297895 A204709 A192359

Adjacent sequences: A157916 A157917 A157918 * A157920 A157921 A157922

Keyword

nonn,easy

Author

Vincenzo Librandi, Mar 09 2009

Status

approved