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Consider an $n\times n$ matrix $M_n$ where the sequence $1,2,3,\dots,n^2$ forms a clock-wise spiral, in that given order. For example, $$M_4=\begin{bmatrix} 1&2&3&4\\ 12&13&14&5\\ 11&16&15&6 \\ 10&9&8&7 \end{bmatrix} \qquad \text{and} \qquad M_5=\begin{bmatrix} 1&2&3&4&5\\ 16&17&18&19&6 \\ 15&24&25&20&7 \\ 14&23&22&21&8 \\ 13&12&11&10&9 \end{bmatrix}.$$

Question. What are the diagonal entries in the Smith normal form of the matrix $M_n$, over $\mathbb{Z}$?

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  • $\begingroup$ You should specify the PID. If you consider $M_n$ as a matrix with real entries, then the Smith normal form is just $I_n$, because $M_n$ is non-singular. I suspect that you have in mind the PID $\mathbb Z$. $\endgroup$
    – Denis Serre
    Commented May 24, 2017 at 15:17

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not a complete answer, but too long for a comment:

some experimentation$^*$ suggests that the diagonal entries of the diagonal matrix $S_n$ in the integer decomposition $M_n=U_nS_nV_n$, with $U_n,V_n$ unimodular, have the following form for $n> 2$:

$${\rm diag}\,S_n=1,1,2,2,2,\cdots 2,2,2,\{Z_n\},c_n$$

where the $\cdots$ indicate padding to length $n$ with a string of $2$'s and $\{Z_n\}$ is a string of integers given by $$\{Z_n\}=\emptyset\;\;{\rm for}\;\;n\leq 5,$$ $$\{Z_n\}=\{6\},\{30\},\{6\},\{6,30\},\{6,30\}\;\;{\rm for}\;\;n=6,7,8,9,10,$$ $$\{Z_n\}=\{6,30\},\{6,30,210\},\{6,6,210\},\{6,30,210\},\{6,6,30,6930\}\;\;{\rm for}\;\;n=11,12,13,14,15,$$ $$\{Z_n\}=\{6,6,30,630\},\{6,30,30,630\},\{6,6,6,30,6930\},\{6,6,30,210,6930\},\{6,6,30,210,6930\}\;\;{\rm for}\;\;n=16,17,18,19,20,$$ $$\{Z_n\}=\{6,6,6,30,210,90090\},\{6,6,30,30,210,90090\},\{6,6,6,30,210,1531530\},\{6,6,6,30,30,630,90090\},\{6,6,6,30,30,630,90090\}\;\;{\rm for}\;\;n=21,22,23,24,25,$$ and so on. Unfortunately, I have been unable to detect a pattern in this sequence.

The final diagonal entry of $S_n$ is $$c_n=\frac{2^{3-n}|{\rm det}\,M_n|}{\prod_{i}\tfrac{1}{2}(Z_n)_i}$$ following from the formula OEIS A023999 for the determinant of a spiral matrix: $$|{\rm det}\,M_n|={\rm det}\, S_n=(3n-1) \frac{ (2n-3)!}{(n-2)!}$$

$^*$ if you would like to experiment further with spiral matrices, here are a few lines of relevant Mathematica code; I am intrigued by this $Z_n$ pattern, what is the logic behind it?

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  • $\begingroup$ The pattern has to a) accommodate the powers of primes occurring in the determinant and b) account for minors with small determinant. Thus, for example, the powers of 7 dividing the determinant are fewer than n in number, and have to be allocated somewhat evenly among the last few coefficients. If there is a large minor with determinant coprime to 7, that pushes the distribution of 7's closer to the end. Gerhard "Can Justify Instead Of Explain" Paseman, 2017.05.24. $\endgroup$
    – Gerhard Paseman
    Commented May 24, 2017 at 15:28
  • $\begingroup$ In a related problem where the poster asks for determinants mod 4, it would be helpful to show which minors have determinant that are nonzero mod 4. Gerhard "Likes Breaking Things Up Multiplicatively" Paseman, 2017.05.24. $\endgroup$
    – Gerhard Paseman
    Commented May 24, 2017 at 15:38
  • $\begingroup$ Carlo: I appreciate your Mathematica code, although I use Maple. By the way, it could worth a discussion if you propose a new MO question regarding your curiosity about $Z_n$. Just a suggestion. $\endgroup$
    – T. Amdeberhan
    Commented May 25, 2017 at 18:41

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