Variants of variations (A007526)
Hugo Pfoertner
all at abouthugo.de
Tue Feb 27 12:30:20 CET 2007
Hugo Pfoertner wrote:
>
> SeqFans,
>
[...]
> Peter Luschny proposed an interesting generalization of
> http://www.research.att.com/~njas/sequences/A007526
>
> Peter has compiled the material on a temporary webpage
> http://www.luschny.de/math/temp/variationen.html
This link is now replaced by
http://www.luschny.de/math/seq/variations.html
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
OEIS.
Hugo
Hello sequences,
I tried to look at recursive sequences and their properties. As a result I created a webpage about them:
http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html
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
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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
up.
Thanks,
Leroy Quet
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