In
http://openstudy.com/study#/updates/4fcb99aee4b0c6963ad51730
@asnaseer showed that every element of the sequence$$
a_{0}=1,\; a_{1}=2,\; a_{2}=24,\; a_{n}=\frac{ 6a_{n-1}^{2}a_{n-3}-8a_{n-1}a_{n-2}^{2}}{a_{n-2}a_{n-3}}.

$$
is an integer.
Show now that every n divides $a_n$

Question

In 
http://openstudy.com/study#/updates/4fcb99aee4b0c6963ad51730
@asnaseer showed that every element of the sequence$$
a_{0}=1,\; a_{1}=2,\; a_{2}=24,\; a_{n}=\frac{ 6a_{n-1}^{2}a_{n-3}-8a_{n-1}a_{n-2}^{2}}{a_{n-2}a_{n-3}}.

$$
is an integer.
Show now that every n divides $a_n$

asnaseer · Answer

I believe I also showed that every $a_n$ is divisible by $a_{n-1}$ and hence every $a_n$ is divisible by all $a_m$ m=n-1,n-2,...,1

asnaseer · Answer

I don't know if that helps here?

anonymous · Answer

Since you were able to create a linear recurrence:$$r_n=6r_{n-1}-8r_{n-2}$$$$r_1 = 2,r_2=12$$we should solve it.

asnaseer · Answer

what do you mean by "solve it" joemath?

anonymous · Answer

Coming up with a closed formula, not a recursive one.

asnaseer · Answer

is there a way of transforming this into an expression for $r_n$ in terms of n?

anonymous · Answer

yes, using linear algebra, i'll post the short version.

asnaseer · Answer

BTW: @eliassaab with all these interesting properties, does this sequence have any particular significance?

anonymous · Answer

$$r_n-6r_{n-1}+8r_{n-2}=0$$has the characteristic equation:$$r^2-6r+8=0\Longrightarrow (r-2)(r-4)=0\Longrightarrow r=2,r=4$$So a closed formula for rn would be of the form:$$r_n=c_12^n+c_24^n$$now we use the initial conditions to see what the constants are.

anonymous · Answer

@asnaseer, I do not think it does.

anonymous · Answer

$$1=2c_1+4c_2$$$$12=4c_1+16c_2$$This system tells us that:$$c_1=-1,c_2=1$$So the closed formula for r_n is:$$r_n=4^n-2^n$$Now maybe we can figure out a closed formula for a_n using this.

asnaseer · Answer

would it be:$$a_n=(4^n-2^n)(4^{n-1}-2^{n-1})...(4-2)$$

anonymous · Answer

yeah, it seems that:$$r_nr_{n-1}r_{n-2}\ldots r_1=\frac{a_n}{a_{n-1}}\cdot \frac{a_{n-1}}{a_{n-2}}\cdots \frac{a_1}{a_0}=a_n$$

asnaseer · Answer

which simplifies to:$$a_n=2^n*2^{n-1}*2^{n-2}*...*2(2^n-1)(2^{n-1}-1)...(2-1)$$

anonymous · Answer

Maybe now that we have a closed formula for a_n, we should proceed by induction? or try to show we can factor an n out of that expression.

asnaseer · Answer

$$a_n=\prod_{k=1}^{n}2^k*\prod_{k=1}^{n}(2^k-1)$$

asnaseer · Answer

wolframs solution to this does not seem to factor out an n :( http://www.wolframalpha.com/input/?i=product+%282^k%29%282^k-1%29+k%3D1+to+n

asnaseer · Answer

so maybe it needs induction as you suggested joemath?

anonymous · Answer

maybe modular arithmetic? for example, if n was a prime, say p, then the term:$$4^{p-1}-2^{p-1}$$would be divisible by p, which can be shown using Fermat's Little Theorem.  Since one of the terms of that product is divisible by p, the whole thing would be divisible by p.

anonymous · Answer

The induction was turning up bust =/

anonymous · Answer

I'll think about this more while im eating dinner lol.  Im thinking modular arithmetic is the way to go.  If we can show that for any prime to a power, if:$$p^k\mid n\Longrightarrow p^k\mid a_n$$i believe we would be done.

asnaseer · Answer

this looks too much like number theory to me - which (unfortunately) is beyond my grasp at the moment. but I look forward to the result :)

anonymous · Answer

@asnaseer with
$$
a_n=2^n*2^{n-1}*2^{n-2}*...*2(2^n-1)(2^{n-1}-1)...(2-1)
$$
You can use Euler totient Theorem whics says that
If a and n is coprime, then 
$$
a^{\phi(n)} \equiv 1\text{  ( mod n)  }
$$

anonymous · Answer

@joemath314159, you  argument of using Fermat's Little Theorem is a step in the right
direction, you need Euler totient theorem  which is a generalization of  Fermat's Little Theorem

anonymous · Answer

Here is how you finish the problem.
Case 1, n is odd, then
n is coprime with 2 so
n divides $ 2^{\phi(n)} -1 $ which is one of the factors 
$$
(2^n-1)(2^{n-1}-1)...(2-1)
$$
so  n divides a_n

anonymous · Answer

Case2. 
If n is even the $n=2^k\,b$, with k<n and b is odd. 
Now $ 2^k $ is one of the factors
$$2^n*2^{n-1}*2^{n-2}*...*2$$
and b is coprime with 2 so b divides one the factors
$$  (2^n-1)(2^{n-1}-1)...(2-1)$$
by case 1.
Now, we can deduce that n divides $ a_n$

asnaseer · Answer

I won't even pretend to understand this proof :)
A bit beyond my capabilities I'm afraid :(

anonymous · Answer

You know what $ \phi (n) $ is?

asnaseer · Answer

afraid not

anonymous · Answer

It is the number of integers m  less than n and such the GCD(m,n)=1.
$$
\phi(5) = 4 \
\phi ( 6)=2\
\phi (10) =4 \
\phi(p) = p-1 \text { if }  p \text { is a prime}
$$

anonymous · Answer

If we take 10, the numbers m less than 10 that have GCD(m,10)=1 aew
1,3,7,9

asnaseer · Answer

ok - in case 1, I understand n is coprime with 2, but I then don't understand why this implies:
n divides $2^{\phi(n)} -1$

anonymous · Answer

That is where we are using  Euler totient theorem

asnaseer · Answer

Oh I see - so this is the result of some theorem? in this case the Euler Totient Theorem?

anonymous · Answer

Yes, this theorem is a generalizatio of Fermat's Little Theorem since
$$\phi(p)=p-1
$$
if p is a prime.

asnaseer · Answer

ok - I am beginning to understand a bit more now.
Thanks for taking the time to explain this to me - much appreciated! :)

anonymous · Answer

yw
But, you did the most important part of this proble which is to factorize $a_n$. What is your background in math?

asnaseer · Answer

I always loved studying maths so I keep up with it as a hobby.
I am actually a qualified aeronautical engineer but now work as a software engineer :)

anonymous · Answer

Interesting and nice meeting you.

asnaseer · Answer

and you too - thanks again