cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A145503 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=3.

Original entry on oeis.org

3, 13, 193, 37633, 1416317953, 2005956546822746113, 4023861667741036022825635656102100993
Offset: 1

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Author

Artur Jasinski, Oct 11 2008

Keywords

Comments

General formula for a(n+1)=a(n)^2+2*a(n)-2 and a(1)=k+1 is a(n)=Floor[((k + Sqrt[k^2 + 4])/2)^(2^((n+1) - 1)).
Essentially the same as A110407. [R. J. Mathar, Mar 18 2009]

Crossrefs

Programs

  • Mathematica
    aa = {}; k = 3; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
    (* or *)
    k = 2; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}]
    NestList[#^2+2#-2&,3,10] (* Harvey P. Dale, Feb 01 2018 *)

Formula

From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 2 + sqrt(3).
a(n) = A003010(n-1) - 1. a(n) = 2*A002812(n-1) - 1.
Recurrence: a(n) = 5*(Product {k = 1..n-1} a(k)) - 2 with a(1) = 3.
Product_{n >= 1} (1 + 1/a(n)) = 5/6*sqrt(3).
Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt(3).
(End)