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Calculate an n-order determinant:

$\left|\begin{array}{cccccc}1 & 2 & 3 & \cdots & n-1 & n \\ n & 1 & 2 & \cdots & n-2 & n-1 \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 2 & 3 & 4 & \cdots & n & 1\end{array}\right|$

The result should be:

$(-1)^{n-1} \frac{n+1}{2} n^{n-1}$

But I can't get this result by FindSequenceFunction:

Clear["Global`*"];
a[n_Integer?Positive] := Array[Mod[#2 - #1, n] + 1 &, {n, n}]
detmm[n_] := Det[a[n]]
tab = Table[detmm[k], {k, 1, 20}];
FindSequenceFunction[tab, n]

FindSequenceFunction[{1, -3, 18, -160, 1875, -27216, 470596, -9437184, 215233605, -5500000000, 155624547606, -4829554409472, 163086595857367, -5952860799406080, 233543408203125000, -9799832789158199296, 437950726881001816329, -20766159817517617053696, 1041273502979112415328410, -55050240000000000000000000}, n]

Is this due to the limitations of FindSequenceFunction or some problems in my method? Is there any other way?

Improve this question
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    $\begingroup$ This is it, right? $\endgroup$
    – bmf
    Apr 20, 2022 at 3:19
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    $\begingroup$ I didn't know something like this existed. Thanks for that info, @bmf. $\endgroup$
    – E. Chan-López
    Apr 20, 2022 at 3:28
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    $\begingroup$ @E.Chan-López you're welcome. It's useful as a resource. I have found examples where FindSequenceFunction fails, but there's a known sequence. This example seems to fall in this category $\endgroup$
    – bmf
    Apr 20, 2022 at 3:31
  • $\begingroup$ @bmf Thanks for sharing! This website has the same function as FindSequenceFunction. It's great! $\endgroup$
    – lotus2019
    Apr 20, 2022 at 7:13
  • $\begingroup$ @lotus2019 you are very welcome. Actually, OEIS is a database and people can submit known sequences. You can also find Mathematica implementation for those sequences. You might want to send an email to the support team of Mathematica with this case. It's been a while since they updated FindSequenceFunction and might be helpful for them to have a library of sequences that could be included. $\endgroup$
    – bmf
    Apr 20, 2022 at 12:51

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