1, 1, 2, 3, 7, ... Multiplication dual of Fibonacci?

[Bhushit Joshipura]

Fibonacci series is:
F(0) = 1; F(1) = 1; F(n) = F(n-1) + F(n-2) for n > 1

That is,
F(n) = [ F(n-1) + F(n-2) ] * 1 for n > 1.

What is the series in which the operations are just reversed:

G(0) = 1; G(1) = 1; G(n) = [ G(n-1) * G(n-2) ] + 1 for n > 1 ?

Initial a few numbers could be:

1, 1, 2, 3, 7, 22, 155, ...

Thanks in advance,
-Bhushit (not a mathematician)

[Richard Tobin]

In article <1188314627.8...@z24g2000prh.googlegroups.com>,
Bhushit Joshipura <josh...@gmail.com> wrote:

>Initial a few numbers could be:
>
>1, 1, 2, 3, 7, 22, 155, ...

See http://www.research.att.com/~njas/sequences/A007660 for
some information about this sequence.

-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

[Bhushit Joshipura]

Thanks. The page refers to a way of writing continued fractions - I
did not understand it :-(

Also, I have some interesting continued fraction (and similar
formations). Is there a general place where I can search and
contribute?

-Bhushit

On Aug 28, 8:43 pm, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:
> In article <1188314627.825148.109...@z24g2000prh.googlegroups.com>,

[Bhushit Joshipura]

Looking again at my posting, we might have hit a way of generating
"series of series" by swapping + and *.

Base Series - Fibonacci: Let F0(0) = 0; F0(1) = 1; F0(n) = [ F(n-1)*1
+ F(n-2)*1 ] * 1

There are 4 symbols to be flipped. This way we can generate 15 (=2^4
-1) number of series.

For example, another such series could be:
F15(n) = (F(n-1)+1)*(F(n-2)+1) + 1; every sign flipped

This must be a well-known technique to mathematicians. What do you
call it?

-Bhushit

[Proginoskes]

On Aug 28, 8:23 am, Bhushit Joshipura <joship...@gmail.com> wrote:
> Fibonacci series is:
> F(0) = 1; F(1) = 1; F(n) = F(n-1) + F(n-2) for n > 1
>
> That is,
> F(n) = [ F(n-1) + F(n-2) ] * 1 for n > 1.
>
> What is the series in which the operations are just reversed:
>
> G(0) = 1; G(1) = 1; G(n) = [ G(n-1) * G(n-2) ] + 1 for n > 1 ?

Related sequence:

G(0) = 1: G(n) = G(0) * ... * G(n-1) + 1, for n >= 1.

The G(i)'s are pairwise coprime. Taking a prime factor of each G(i)
produces an infinite set of primes. (Not my proof; possibly from the
American Mathematical Monthly.)

--- Christopher Heckman