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if there exist positive integer sequence $a_{1},a_{2},a_{3},\cdots,a_{n}$,such that $$a_{1}a_{2},a_{2}a_{3},a_{3}a_{4},\cdots,a_{n-1}a_{n},a_{n}a_{1}$$ is arithmetic sequence,and the common difference $d=a_{i+1}a_{i+2}-a_{i}a_{i+1}\neq 0,i=1,\cdots,n-1.a_{n+1}=a_{1}$

find the value $n$

I Think this problem is very nice,Thank you everyone.

my idea: first, we have $n\ge 3$,and

when $n=3$,then $$2a_{2}a_{3}=a_{1}a_{2}+a_{3}a_{1}$$ $$\Longrightarrow \dfrac{2}{a_{1}}=\dfrac{1}{a_{2}}+\dfrac{1}{a_{3}}$$

and $$\dfrac{2}{5}=\dfrac{1}{3}+\dfrac{1}{15}$$

and when $n=4,5,\cdots,$?

\ \

I think $n=2k-1$ is true,but I can't prove it

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$\frac{2}{5}=\frac{1}{3}+\frac{1}{15}$ – André Nicolas Jun 20 '13 at 4:07
oh,Thank you.+1 – math110 Jun 20 '13 at 4:14
I find this confusing, more with the comments: is it given that $\,a_1=5\,,\,a_2=3\,,\,a_3=15\,$ and because of this you wrote $\,\frac25=\frac13+\frac1{15}\,$ ? – DonAntonio Jun 20 '13 at 5:08
yes,That's my meaning – math110 Jun 20 '13 at 5:22
@math110 Please have a look at my most recently edited answer, in which the $n$ even case is shown impossible (same method as before does work for odd $n$). – coffeemath Jun 22 '13 at 20:08

If $d=a_{i+1}a_{i+2}-a_{i}a_{i+1}$, then $$\sum_{i=1}^{n-1} (a_{i+1}a_{i+2}-a_{i}a_{i+1}) =-a_1a_2+a_na_{n+1} =-a_1a_2+a_na_1 =a_1(a_n-a_2) $$ since the sum telescopes.

Also, $$\sum_{i=1}^{n-1} (a_{i+1}a_{i+2}-a_{i}a_{i+1}) =\sum_{i=1}^{n-1} d = (n-1)d $$

Therefore $(n-1)d = a_1(a_n-a_2)$ or $n = 1+\dfrac{a_1(a_n-a_2)}{d}$.

We now need to determine $a_n$.

Since $d = a_{i+1}a_{i+2}-a_{i}a_{i+1}$, $a_{i+2} = a_{i}+\dfrac{d}{a_{i+1}} $.

So, $a_3=a_1+\dfrac{d}{a_2}$, $a_4=a_2+\dfrac{d}{a_3} =a_2+\dfrac{d}{a_1+\dfrac{d}{a_2}} =a_2+\dfrac{d a_2}{d+a_1 a_2} $, $\begin{align} a_5 &=a_3+\dfrac{d}{a_4}\\ &=a_1+\dfrac{d}{a_2}+\dfrac{d}{a_2+\dfrac{d a_2}{d+a_1 a_2}}\\ &=a_1+\dfrac{d}{a_2}+\dfrac{d(d+a_1 a_2)}{a_2(d+a_1 a_2)+d a_2}\\ &=a_1+\dfrac{d}{a_2}+\dfrac{d(d+a_1 a_2)}{a_2(2d+a_1 a_2)}\\ &=a_1+\dfrac{d(2d+a_1 a_2)+d(d+a_1 a_2)}{a_2(2d+a_1 a_2)}\\ &=a_1+\dfrac{d(3d+2a_1 a_2)}{a_2(2d+a_1 a_2)}\\ \end{align} $

It's 11pm here, and I am very tired, so I'll leave it at this and come back tomorrow to try to find a general form for the $a_i$. If I get that, then I'll put the value for $a_n$ in the formula for $n$.

Or maybe someone else will do it.

Good night until tomorrow.

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oh,It's good job.+1,Thank you very much, HaHa, in china, Now It's 14:00 am – math110 Jun 20 '13 at 5:59
I deal with this $d=1$, we have $a_{n}=2\cos{\dfrac{\pi}{2n+1}}$ – math110 Jun 20 '13 at 6:16
can – math110 Jun 20 '13 at 6:24
@math110 With $a_n$ as in your comment, we do not have integers for the $a_n$. Also your $d$ value of $1$ doesn't work since $a_2a_3-a_1a_2=1.29754$. Not only that, it is not an arithmetic progression even with this other $d$, since $a_3a_4-a_2a_3=0.470939.$ – coffeemath Jun 20 '13 at 22:21

This can be done for any odd $n$. Take $d=n!$ and define $a_1=1\cdot 3 \cdot 5 \cdots$ and $a_2=2 \cdot 4 \cdot 6 \cdots$, in both cases progressing through the odd or even integers up to $n$ until they run out. The sequence of products will be the arithmetic sequence $d,2d,3d,...,nd$ where $nd=n\cdot n!$ The values of the $a_k$ are then determined by the requirement that this be the sequence of products of the adjacent $a_k.$

For example if $n=5$ we put $a_1=1\cdot 3 \cdot 5=15$ and $a_2=2 \cdot 4=8$, obtaining for the sequence of five $a$ values $$15,\ 8,\ 30,\ 12,\ 40$$

Note that for $n=6$ if we put $a_1=1 \cdot 3 \cdot 5=15$ and $a_2=2 \cdot 4 \cdot 6=48,$ the $a$ are $$15,\ 48,\ 30,\ 72,\ 40,\ 90,\ 48,$$ but if as required in the OP we use $a_7=a_1=15$ then

$$a_7a_6-a_6a_5=15\cdot 48 - 48 \cdot 90=-3600.$$ It would come out $720$ had we used the value $105=3\cdot 5 \cdot 7$ for $a_7$, which is closely related to $a_1$, and I believe in the even case if one uses $a_{n+1}=(n+1)a_1$ in the $n$ even case we have a solution to a slightly altered problem than asked by math110.

The relations to compute the $a_k$ come from the requirement that the sequence of products be a progression as stated with common difference $d=n!$. We have in general $$\frac{a_k}{a_{k-2}}=\frac{k-1}{k-2}.$$ This gives each $a_k$ from the one two index values before, and also explains why they all remain integers, since e.g. for the odd indices $a_1$ starts out with the full sequence of odd factors it needs in order not to give a noninteger for $a_3,a_5,\cdots.$

Impossibility for even $n$ (with $d \neq 0$ as stated in OP). First assume that $d>0$. Since the $a_k>0$ we have from $a_{k+1}a_k-a_ka_{k-1}=0$ that $a_{k+1}>a_{k-1}$. Therefore $a_1<a_3<\cdots a_{n-1},$ so that $$a_1<a_{n-1}. \tag{1}$$ But then also from $$a_{n+1}a_{n}-a_na_{n-1}=d$$ from which $a_{n+1}>a_{n-1}$ which contradicts $(1)$. and since $a_1=a_{n+1}$ this means $a_1>a_{n-1}.$

If instead $d<0$ we can proceed in the same way, with inequalities reversed, and arrive at a similar contradiction.

Note on why it works for $n$ odd: In this case $a_1=1\cdot 3 \cdots n$ with $(n+1)/2$ odd factors, while $a_2=2 \cdot 4 \cdots (n-1)$ with $(n-1)/2$ even factors. In computing $a_{n+1}$ via the recursion, since $n+1$ is even we begin with $a_2$ and apply the recursion $a_{k+2}=a_k \cdot \frac{k+1}{k}$ a total of $(n-1)/2$ times. This gives $$a_{n+1}=2\cdot 4 \cdots (n-1)\cdot \frac32 \frac54 \cdots \frac{n}{n-1},$$ i.e. $a_{n+1}=3 \cdot 5 \cdots n$, which happens to equal $a_1$ as desired since the first factor of $a_n$ is $1$ and the rest are $3,5,...,n$.

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Note to anyone who read this already: My construction originally given does not work for $n$ even. (I'll try for that.) I have now altered the answer reflecting this, showing for $n=6$ what goes wrong. – coffeemath Jun 22 '13 at 19:10
I have now added a proof that there are no solutions for $n$ even with $d \neq 0$ as OP requires. [allowing $d=0$ lets in constant sequences.] – coffeemath Jun 22 '13 at 20:06

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