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Let $m,n$ be nonnegative integers.

The sequence $\{a_{m,n}\}$ satisfies the following three conditions.

  1. For any $m$, $a_{m,0}=a_{m,1}=1$
  2. For any $n$, $a_{0,n}=1$
  3. For any $m\ge0, n\ge1$, $a_{m+1,n+1}a_{m,n-1}=a_{m+1,n-1}a_{m,n+1}+a_{m+1,n}a_{m,n}$

Prove that $a_{m,n}$ is an integer for any $m\ge0, n\ge0$.

Could you tell me how to prove it?

mathlove
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    Have you evaluated the recurrence for various small values of $m,n$? Can you take an educated guess at what $a_{m,n}$ might be? Once you have a good guess as to the form, you may be able to prove that your guess satisfies the recurrence - and once you do that, you're done; the fact that $a_{r,s}$ is defined strictly in terms of values $a_{t,u}$ with $t+u\lt r+s$ means that the solution for any given set of initial conditions must be unique as long as $a_{m,n}\neq 0$ for all $m,n$ (which is easy to show here). – Steven Stadnicki May 22 '13 at 15:33
  • I think you need another $a_{1,n}=1$ as well. I don't see how to get terms with $n=1$ out of the recurrence in $3.$ – Ross Millikan May 22 '13 at 15:50
  • @Steven Stadnicki: I got the following. $$a_{m,n-3} a_{m,n-2} a_{m-1,n-2}(a_{m+1,n-1} a_{m,n+1}+a_{m+1,n} a_{m,n} )≡0(mod\ a_{m,n-1}) $$ I need to prove the following.
    1. $a_{m,n-3}$ and $a_{m,n-1}$ are coprime.
    2. $a_{m,n-2}$ and $a_{m,n-1}$ are coprime.
    3. $a_{m-1,n-2}$ and $a_{m,n-1}$ are coprime.

    However, these are not always right, so this idea doesn't seem to be good.I think this sequence is similar to 'Somos sequence',but no good idea.

     – mathlove May 23 '13 at 14:49
  • @Ross Millikan: We can get $a_{1,n}$ from the given conditions;$a_{1,0}=a_{1,1}=1$ and to get $a_{1,n}(n\ge2)$ we can also use the third one. – mathlove May 23 '13 at 14:51
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    If I plug $n=0$ into 3, I have $a_{m+1,1}$ on the left, but $a{m+1,-1}$ on the right so that doesn't work. If I plug $n=1$ in, I have $a_{m+1,2}$ on the left, but $a_{m,1}$ on the right and I haven't found it yet. I don't see how to get $a_{2,1)$ – Ross Millikan May 23 '13 at 15:08
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    According to the first condition, you can plug any nonnegative integer into m, so for example $a_{2,1}=1$. – mathlove May 23 '13 at 16:29
  • @mathlove I think you're missing my point - I'm suggesting that you build a table of the values $a_{m,n}$ for, say, all $m,n\lt 10$ and see what you see there. – Steven Stadnicki May 23 '13 at 17:16
  • @Steven Stadnicki: Thank you for your suggestion. I built a table, but I haven't got any result yet. Could you show me your idea if you have? I really hope to solve this problem. – mathlove May 24 '13 at 06:09
  • This is a very difficult problem for me. Does anyone know the name of this double sequence? Is this a famous sequence? Give me something. – mathlove May 27 '13 at 15:10
  • Interesting, strange sequence indeed. I think there's something special behind this. Please share your idea with me. Anything is OK. – mathlove May 28 '13 at 14:47
  • @mathlove Where did you encounter the sequence? Some additional context might be helpful, if it exists. – J. J. May 28 '13 at 20:59
  • @J.J. One of my friends made this. None of us can prove this and no one can get a counterexample even by using a computer. He, who made this, whose major is math, was interested in 'Somos sequence'. That's why the idea I showed on May 23th came from the proof of this sequence. – mathlove May 29 '13 at 04:46
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    @mathlove This problem seems indeed non-trivial. You could consider asking also on MathOverflow. – J. J. May 29 '13 at 06:16
  • @J.J. Thank you. I'll try. – mathlove May 29 '13 at 06:48

1 Answers1

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I'm posting an answer just to inform that the question has received an answer by Timothy Chow on MO.

https://mathoverflow.net/questions/132191/proving-that-every-term-of-the-sequence-is-an-integer

He gives a link to the paper http://arxiv.org/abs/math/0104241 where it is shown that if one replaces the initial conditions $a_{m,0} = 1$, $a_{m,1} = 1$ and $a_{0,n} = 1$ by indeterminates, $a_{m,0} = x_{m,0}$, $a_{m,1} = x_{m,1}$ and $a_{0,n} = x_{0,n}$, then each $a_{m,n}$ is a Laurent polynomial in these indeterminates with coefficients in $\mathbb{Z}$. In particular if one then substitutes $1$ for each of these indeterminates, the result is an integer.

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