login
This site is supported by donations to The OEIS Foundation.

 

Logo

Many excellent designs for a new banner were submitted. We will use the best of them in rotation.

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A145502 a(n+1)=a(n)^2+2*a(n)-2 and a(1)=2 12
2, 6, 46, 2206, 4870846, 23725150497406, 562882766124611619513723646, 316837008400094222150776738483768236006420971486980606 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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))

From Peter Bala, Nov 12 2012: (Start)

The present sequence corresponds to the case x = 3 of the following general remarks. Sequences A145503 through A145510 correspond to the cases x = 4 through x = 11 respectively.

Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1. Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^2 + 2*a(n) - 2 with the initial condition a(1) = x - 1.

A second recurrence is a(n) = (a(1) + 2)*{product {k = 1..n-1} a(k)} - 2.

The following algebraic identity is valid for x > 2:

(x + 1)/sqrt(x^2 - 4) = (1 + 1/(x - 1))*(y + 1)/sqrt(y^2 - 4), where y - 1 = (x - 1)^2 + 2*(x - 1) - 2. Iterating the identity yields the product expansion (x + 1)/sqrt(x^2 - 4) = product {n = 1..inf} (1 + 1/a(n)).

A second expansion is product {n = 1..inf} (1 + 2/(a(n) + 1) = sqrt((x + 2)/(x - 2)). For an alternative approach to these identities see the Bala link.

(End)

LINKS

Table of n, a(n) for n=1..8.

Peter Bala, Notes on A145502-A145510

FORMULA

From Peter Bala, Nov 12 2012: (Start)

a(n) = phi^(2^n) + (1/phi)^(2^n) - 1, where phi := (1 + sqrt(5))/2 is the golden ratio.

a(n) = A001566(n-1) - 1.

Recurrence: a(n) = 4*{product {k = 1..n-1} a(k)} - 2 with a(1) = 2.

Product {n = 1..inf} (1 + 1/a(n)) = 4/sqrt(5).

Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(5).

(End)

MATHEMATICA

aa = {}; k = 2; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa

or

k = 1; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)

CROSSREFS

A145502-A145510. A001566

Sequence in context: A078603 A001587 A078537 * A072444 A052596 A098710

Adjacent sequences:  A145499 A145500 A145501 * A145503 A145504 A145505

KEYWORD

nonn,easy

AUTHOR

Artur Jasinski, Oct 11 2008

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement .

Last modified April 21 12:08 EDT 2014. Contains 240824 sequences.