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Define the recurrence \begin{align*} n(2n+x-3)u_n(x) &=2(2n+x-2)(4n^2+4nx-8n-3x+3)u_{n-1}(x) \\ &-4(n+x-2)(2n-3)(2n+2x-3)(2n+x-1)u_{n-2}(x) \end{align*} with initial conditions $u_0(x)=0$ and $u_1(x)=x+1$.

The subject of "Laurent phenomenon" was motivated by Somos sequences. In the same spirit, I ask:

QUESTION. Is it true that each $u_n(x)$ is a polynomial in $x$? In fact, with positive integer coefficients.

EXAMPLES. $u_2(x)=5x^2 + 13x + 6$ and $u_3(x)=22x^3 + 114x^2 + 164x + 60$ and \begin{align*} u_4(x)&=93x^4 + 814x^3 + 2367x^2 + 2606x + 840 \\ u_5(x)&=386x^5 + 5140x^4 + 25030x^3 + 54500x^2 + 51024x + 15120. \end{align*}

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    $\begingroup$ Are you mixing $a$'s and $x$'s? $\endgroup$
    – Sam Hopkins
    Commented Dec 9, 2020 at 17:59
  • $\begingroup$ @SamHopkins: there were typos. Thanks! $\endgroup$
    – T. Amdeberhan
    Commented Dec 9, 2020 at 18:04
  • $\begingroup$ Is it a Somos sequence? $\endgroup$
    – Alexey Ustinov
    Commented Dec 10, 2020 at 8:17
  • $\begingroup$ @AlexeyUstinov: I don't think it is. $\endgroup$
    – T. Amdeberhan
    Commented Dec 10, 2020 at 14:16

1 Answer 1

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In fact, $$ u_n(x) = {2}^{n-1}\prod _{k=0}^{n-1}(2x+2k+1) -{2\,n-1\choose n-1}\prod _{k=0}^{n-1}(x+k) , \tag1$$ which is a polynomial with integer coefficients.

P.S. the proof rests on a routine verification that (1) satisfies the given recurrence relation and initial conditions.

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    $\begingroup$ Interesting. How did you find this? $\endgroup$
    – T. Amdeberhan
    Commented Dec 9, 2020 at 18:47
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    $\begingroup$ @T.Amdeberhan I guess this gave some clues: oeis.org/… $\endgroup$
    – Wolfgang
    Commented Dec 9, 2020 at 19:27
  • $\begingroup$ Is $u_n(x)/(x+n)$ always irreducible ? I have verified it numerically up to $n=170$. $\endgroup$
    – CHUAKS
    Commented Dec 13, 2020 at 2:07
  • $\begingroup$ @CHUAKS: I agree. $\endgroup$
    – T. Amdeberhan
    Commented Dec 13, 2020 at 16:10

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