Variants of variations (A007526)

Hugo Pfoertner all at
Tue Feb 27 12:30:20 CET 2007

Hugo Pfoertner wrote:
> SeqFans,
> Peter Luschny proposed an interesting generalization of
> Peter has compiled the material on a temporary webpage

This link is now replaced by

and it is definitely worthwhile to throw a glance at Peter's compilation
on this subject, providing a good example for an exploratory use of the


Hello sequences,

I tried to look at recursive sequences and their properties. As a result I created a webpage about them:

Right now I have the following types of sequences there:

I discuss the properties of these sequences and have references to all such sequences in OEIS. 

Editors of OEIS,
I made exact copies of definitions of these sequences from OEIS into my page. If you are interested you may check for mistakes and consistency.

Best, Tanya

Need personalized email and website? Look no further. It's easy
with Doteasy $0 Web Hosting! Learn more at

I have a couple of related things to discuss.

First, form the recursive sequence of rationals:

b(1) = 1. b(n) = 1 + 1/b(n-1)^2.

1, 2, 5/4, 41/25, 2306/1681,...

The numerators of this sequence are already in the OEIS (twice): A076725 

So, according to a comment at A076725, b(n) (<- the notation used in the 
comment) equals A076725(n)/A076725(n-1)^2.

This is easily shown to be the case.

Next, form the continued fraction of rational terms:

r(n) is such that the continued fraction [r(1);r(2),...,r(n)] = b(n), for 
every positive integer n.

So, we have {r(n)}: 1, 1, 1/3, 9/13, 91/289,...

(I don't know if the numerators or denominators is in the OEIS.)

So, for example:

b(4) = 41/25 = 1 + 1/(1 + 1/(1/3 + 13/9)).
And b(5) = 2306/1681 = 1 + 1/(1 + 1/(1/3 + 1/(9/13 + 289/91))).

Now, according to a result of mine published as a problem in the American 
Mathematical Monthly*,...

*(The problem as printed contained errors and omitted key information. 
The solution was later printed with corrections. Unfortunately, I don't 
know which issues the problem/solution were printed in, for I do not 

...If c(n) is the nth convergent of a continued fraction, where c(n) = 
f(c(n-1)) for every positive integer, where f(x) is a continuous function 
with a finite nonzero derivative at x = limit(m -> infinity) c(m),*...

*(There are other conditions on f and c(1) too. I know that c(1) had to 
be such that {c(k)} approached a limit. I can't remember the rest. Sorry.)

then, if c(n) = the continued fraction of real terms [r(1),r(2),...r(n)], 
for every positive integer n, then limit{n -> infinity} r(n)* r(n+1) = - 
1/f'(x) - 2 - f'(x), where f'(x) is the derivative of f at x = limit{n-> 
infinity} c(n).

So, if x = the root of x^3 - x -1 = 0,* (I am assuming that b(n) -> x.)...

*(What is this root? I have no access to a computer algebra program, and 
I do not know how to solve the cubic off the top of my head. I am 
embarrassed that I don't know the root to even this simple cubic.)

...then limit(n -> infinity} r(n)*r(n+1) = (x^3 /2) - 2 + (2 /x^3).

A quick check on my calculator shows that indeed the above limit is 
probably in the general vicinity of the limit. What is the closed form 
for this limit of {r(n)*r(n+1)}?

I planned on submitting the above r-sequence (as numerator and 
denominator sequences) to the OEIS, but I have too many loose ends to tie 

Leroy Quet

More information about the SeqFan mailing list