8
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Let $[k] = \{0,\dots,k-1\}$.

Consider the set $F(n,m)$ of functions $f:[n]\times[m]\rightarrow[m]$.

The cardinality of $F(n,m)$ is $|F(n,m)| = m^{nm}$.

Consider the equivalence relation $f \simeq g$ between functions $f,g \in F(n,m)$ iff

there are permutations $\pi:[n]\rightarrow [n]$ and $\tau:[m]\rightarrow [m]$ such that $\tau(f(n,m)) = g(\pi(n),\tau(m))$

(see Harary/Palmer: Enumeration of Finite Automata).

Harary/Palmer give an explicit formula to count the number $a(n,m)$ of orbits (equivalence classes) of $\simeq$. And they show that $a(2,2) = 7$ (compared to $|F(2,2)|=2^4=16$).

But I find it hard to get a number for $a(4,4)$ to be compared to $|F(4,4)| = 4^{16} \sim 4\cdot10^9 $, even given Harary/Palmer's formulas.

Is there a easy way to get this number?

$\endgroup$
  • $\begingroup$ Do you want an answer for the general case or is your interest mainly in $a(4,4)$? $\endgroup$ – Nitin Oct 27 '16 at 19:06
  • $\begingroup$ For the general case there are the formulas of Harary/Palmer which are very hard to evaluate (at least for me). If I knew how to elegantly evaluate them (maybe only approximately), I would be happy. But I would also be happy, if you have a computer program that gives me this one number (and maybe some others I would ask you). $\endgroup$ – Hans-Peter Stricker Oct 28 '16 at 7:44
  • $\begingroup$ I'm working on a program to do this - I'll let you know if runs in some reasonable amount of time. $\endgroup$ – Nitin Oct 29 '16 at 3:53
  • $\begingroup$ @Nitin: It would be great if you could run your program to also calculate $a(3,3)$. $\endgroup$ – Markus Scheuer Oct 29 '16 at 20:50
  • 1
    $\begingroup$ @Markus Scheuer: I've calculated $a(3,3)$ and corrected my code, thanks to your calculation. Please see my post below. I've also calculated $a(4,4)$, I believe correctly now. $\endgroup$ – Scott Burns Oct 29 '16 at 23:47
6
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[Add-on 2016-10-30]: Case n=4 added.

We calculate the number $a(n,n)$ of equivalent functions $$f:[n]\times[n]\rightarrow[n]$$ for $n=2,3$ and $n=4$ according to the paper of F. Harary and E. Palmer and show

\begin{align*} a(2,2)&=7\\ a(3,3)&=638\\ a(4,4)&=7643021 \end{align*}

The formula to be applied is stated as formula (14) in the paper. In fact we can use a simplified version of it, which is given in connection with the calculation of $a(2,2,1)=7$ at the end of page 505. The third parameter is not of interest for us and so we instead write $a(2,2)$.

Here I follow the notation of the authors and use $[p,q]:=\operatorname{lcm}(p,q)$ and $\langle p,q\rangle:=\operatorname{gcd}(p,q)$.

[Harary, Palmer]: The following is valid \begin{align*} a(n,n)=\frac{1}{\left(n!\right)^2}\sum_{(\alpha,\beta)\in S_n^2}\prod_{p=1}^n\prod_{q=1}^n \left(\sum_{s|[p,q]}sj_s(\alpha)\right)^{j_p(\alpha)j_q(\beta)\langle p,q\rangle}\tag{1} \end{align*} where the sum is taken over all pairs of permutations $(\alpha,\beta)$ of degree $n$ and $j_p(\alpha)$ is denoting the number of cycles of $\alpha$ of length $p$.

Hint: Observe the terms in the sum (1) do not make use of $\alpha$ but instead of $j_p(\alpha)$ only. So, it is not necessary to sum over all $\left(n!\right)^2$ pairs of permutations, as we can conveniently use the cycle index of the permutation group $S_n$ and considerably reduce the number of summands.

Preparatory work: Cycle index

We calculate for $n=2,3,4$ the cycle index based upon the recursion formula \begin{align*} Z(S_0)&=1\\ Z(S_n)&=\frac{1}{n}\sum_{j=1}^nz_jZ(S_{n-j})\qquad\qquad n>0 \end{align*} We obtain \begin{align*} Z(S_1)&=z_1Z(S_0)=z_1\\ Z(S_2)&=\frac{1}{2}\left(z_1Z(S_1)+z_2\right)\\ &=\frac{1}{2}\left(z_1^2+z_2\right)\\ Z(S_3)&=\frac{1}{3}\left(z_1\cdot\frac{1}{2}\left(z_1^2+z_2\right)+z_2z_1+z_3\right)\\ &=\frac{1}{6}\left(z_1^3+3z_1z_2+2z_3\right)\\ Z(S_4)&=\frac{1}{4}\left(z_1\cdot\frac{1}{6}\left(z_1^3+3z_1z+2z_3\right)+z_2\cdot\frac{1}{2}\left(z_1^2+z_2\right) +z_3z_1+z_4\right)\\ &=\frac{1}{24}\left(z_1^4+6z_1^2z_2+8z_1z_3+3z_2^2+6z_4\right) \end{align*}

$$ $$

Case: $n=2$:

We show the following is valid \begin{align*} \color{blue}{a(2,2)=7} \end{align*}

In order to calculate $a(2,2)$ we consider according to (1) \begin{align*} a(2,2)=\frac{1}{4}\sum_{(\alpha,\beta)\in S_2^2}\prod_{p=1}^2\prod_{q=1}^2 \left(\sum_{s|[p,q]}sj_s(\alpha)\right)^{j_p(\alpha)j_q(\beta)\langle p,q\rangle}\tag{2} \end{align*}

It it convenient to do some bookkeeping by use of tables. We list the permutations of $S_2=\{\operatorname{id},(12)\}$ in cycle notation and write a table with the number of cycles of each length for each permutation. We also write the corresponding monomial from the cycle index. \begin{array}{l|ccc} \pi&Z(S_2)&j_1(\pi)&j_2(\pi)\\ \hline \operatorname{id}& z_1^2& 2& 0\\ (12)& z_2^1& 0 &1\\ \end{array}

Since it is too cumbersome to write each summand from (2) in one long line we use instead a table description as follows:

\begin{array}{cc|cccc|cc|cc|rr} \alpha&\beta&p&q&[p,q]&<p,q>&s&j_s(\alpha)&j_p(\alpha)&j_q(\beta)&\text{factors}&\text{result}\\ \hline id&id&1&1&1&1&1&2&2&2&16&16\\ &&1&2&2&1&1&2&2&0&1&\\ &&&&&&2&0&&&&\\ &&2&1&2&1&1&2&0&0&1&\\ &&&&&&2&0&&&&\\ &&2&2&2&2&1&2&0&2&1&\\ &&&&&&2&0&&&&\\ \hline id&(12)&1&1&1&1&1&2&2&0&1&4\\ &&1&2&2&1&1&2&2&1&4&\\ &&&&&&2&0&&&&\\ &&2&1&2&1&1&2&0&0&1&\\ &&&&&&2&0&&&&\\ &&2&2&2&2&1&2&0&1&1&\\ &&&&&&2&0&&&&\\ \hline (12)&id&1&1&1&1&1&0&0&2&1&4\\ &&1&2&2&1&1&0&0&0&1&\\ &&&&&&2&1&0&0&&\\ &&2&1&2&1&1&0&1&2&4&\\ &&&&&&2&1&1&2&&\\ &&2&2&2&2&1&0&1&0&1&\\ &&&&&&2&1&1&0&&\\ \hline (12)&(12)&1&1&1&1&1&0&0&0&1&4\\ &&1&2&2&1&1&0&0&1&1&\\ &&&&&&2&1&0&1&&\\ &&2&1&2&1&1&0&1&0&1&\\ &&&&&&2&1&1&0&&\\ &&2&2&2&2&1&0&1&1&4&\\ &&&&&&2&1&1&1&&\\ \hline &\color{blue}{\text{Total}}&&&&&&&&&&\color{blue}{28} \end{array}

Comment:

The table is organised in blocks for pairs of permutation. Although here not eye-catching since we list all $\left(2!\right)^2=4$ pairs, we need in fact only for each cycle type one representative, since we are only interested in the length of cycles of a permutation. The column result gives the summands in (2). Here are the gory details:

  • Columns: $\alpha,\beta$ correspond to a pair of permutations $(\alpha,\beta)$ which is used as index in the outer sum of (2).

  • Columns: $p,q$ are the indices of the products in (2)

  • Column: $s$ gives the divisors of $\operatorname{lcm}(p,q)$

  • Columns: $j_s(\alpha),j_p(\alpha),j_q(\beta)$ list the cycle lengths

  • Column: $\text{factor}$ gives $$\left(\sum_{s|[p,q]}sj_s(\alpha)\right)^{j_p(\alpha)j_q(\beta)\langle p,q\rangle}$$

  • Column: $\text{result}$ calculates finally the product

$$\prod_{p=1}^2\prod_{q=1}^2 \left(\sum_{s|[p,q]}sj_s(\alpha)\right)^{j_p(\alpha)j_q(\beta)\langle p,q\rangle}$$

Since the total of the table is $28$ we finally conclude according to (2) \begin{align*} \color{blue}{a(2,2)=\frac{1}{4}\cdot 28=7} \end{align*} and the claim follows.

$$ $$

Case: $n=3$:

We do the calculation similar to above and show the following is valid \begin{align*} \color{blue}{a(3,3)=638} \end{align*}

In order to calculate $a(3,3)$ we consider according to (1)

\begin{align*} a(3,3)=\frac{1}{\left(3!\right)^2}\sum_{(\alpha,\beta)\in S_3^2}\prod_{p=1}^3\prod_{q=1}^3 \left(\sum_{s|[p,q]}sj_s(\alpha)\right)^{j_p(\alpha)j_q(\beta)\langle p,q\rangle}\tag{3} \end{align*}

We list the permutations of $S_3=\{\operatorname{id},(12),(13),(23),(123),(132)\}$ in cycle notation and write a table with the number of cycles of each length for each permutation. We also write the corresponding monomial from the cycle index. \begin{array}{l|cccc} \pi&Z(S_3)&j_1(\pi)&j_2(\pi)&j_3(\pi)\\ \hline id&z_1^3&3&0&0\\ (12)&3z_1z_2&1&1&0\\ (123)&2z_3&0&0&1\\ \end{array}

Note: The factors $1,3$ and $2$ in the column $Z(S_3)$ indicate the number of different permutations of the corresponding cycle type. We will use this fact to considerably reduce the calculation of the number of summands in (3).

In the following it is sufficient to calculate tables for the nine pairs \begin{align*} \{id,(12),(123)\}\times\{id,(12),(123)\} \end{align*} the cycle index provides the supplementary information we need to calculate the complete sum.

Note that in the main table above there is some redundancy to ease traceability. We now use a somewhat more compact notation to ease readability and keep the space small.

Table: $j_s(\pi), j_p(\pi),j_q(\pi)$ \begin{array}{cc|cc|ccc|ccc|ccc} &&&&&\pi=id&&&\pi=(12)&&&\pi=(123)&\\ p&q&[p,q]&s&j_s&j_p&j_q&j_s&j_p&j_q&j_s&j_p&j_q\\ \hline 1&1&1&1&3&3&3&1&1&1&0&0&0\\ &2&2&1&3&3&0&1&1&1&0&0&0\\ &&&2&0&&&1&&&0&&\\ &3&3&1&3&3&0&1&1&0&0&0&1\\ &&&3&0&&&0&&&1&&\\ 2&1&2&1&3&0&3&1&1&1&0&0&0\\ &&&2&0&&&1&&&0&&\\ &2&2&1&3&0&0&1&1&1&0&0&0\\ &&&2&0&&&1&&&0&&\\ &3&6&1&3&0&0&1&1&0&0&0&1\\ &&&2&0&&&1&&&0&&\\ &&&3&0&&&0&&&1&&\\ 3&1&3&1&3&0&3&1&0&1&0&1&0\\ &&&3&0&&&0&&&1&&\\ &2&6&1&3&0&0&1&0&1&0&1&0\\ &&&2&0&&&1&&&0&&\\ &&&3&0&&&0&&&1&&\\ &3&3&1&3&0&0&1&0&0&0&1&1\\ &&&3&0&&&0&&&1&&\\ \end{array}

The table above provides all information necessary to calculate the summands in (3) for each of the nine pairs of permutations.

An example of a typical block is given here for $((12),(12))$ as it was done for all four blocks in the case $n=2$ and a summary table follows below.

Table: $\{(12)\}\times\{(12)\}$

\begin{array}{cc|cccc|cc|cc|rr} \alpha&\beta&p&q&[p,q]&<p,q>&s&j_s(\alpha)&j_p(\alpha)&j_q(\beta)&\text{factors}&\text{result}\\ \hline (12)&(12)&1&1&1&1&1&1&1&1&1&81\\ &&&2&2&1&1&1&1&1&3&\\ &&&&&&2&1&&&&\\ &&&3&3&1&1&1&1&0&1&\\ &&&&&&3&0&&&&\\ &&2&1&2&1&1&1&1&1&3&\\ &&&&&&2&1&&&&\\ &&&2&2&2&1&1&1&1&9&\\ &&&&&&2&1&&&&\\ &&&3&6&1&1&1&1&0&1&\\ &&&&&&2&1&&&&\\ &&&&&&3&0&&&&\\ &&3&1&3&1&1&1&0&1&1&\\ &&&&&&3&0&&&&\\ &&&2&6&1&1&1&0&1&1&\\ &&&&&&2&1&&&&\\ &&&&&&3&0&&&&\\ &&&3&3&3&1&1&0&0&1&\\ &&&&&&3&0&&&&\\ \end{array}

$$ $$

Summary:

In order to respect all summands of (3) we write the results of the tables above together with the multiplicity of each permutation according to its cycle type. So, e.g. the permutation $(12)$ has cycle type $z_1z_2$ and there are three permutations of this type $\{(12),(13),(23)\}$, we take a factor $3$.

\begin{array}{ll|r|cc|r} \alpha&\beta&\text{res}&\text{m}_{\alpha}&\text{m}_{\beta}&\text{res}\cdot \text{m}_{\alpha}\cdot \text{m}_{\beta}\\ \hline id&id&19683&1&1&19683\\ id&(12)&729&1&3&2187\\ id&(123)&27&1&2&54\\ \hline (12)&id&27&3&1&81\\ (12)&(12)&81&3&3&729\\ (12)&(123)&3&3&2&18\\ \hline (123)&id&27&2&1&54\\ (123)&(12)&9&2&3&54\\ (123)&(123)&27&2&2&108\\ \hline \color{blue}{\text{Total}}&&&&&\color{blue}{22968}\\ \end{array}

Since the total of the table is $22968$ we finally conclude according to (3) \begin{align*} \color{blue}{a(3,3)=\frac{1}{36}\cdot 22968=638} \end{align*} and the claim follows.

$$ $$

Case: $n=4$:

We do the calculation similar to above and show the following is valid \begin{align*} \color{blue}{a(4,4)=7643021} \end{align*}

In order to calculate $a(4,4)$ we consider according to (1)

\begin{align*} a(4,4)=\frac{1}{\left(4!\right)^2}\sum_{(\alpha,\beta)\in S_4^2}\prod_{p=1}^4\prod_{q=1}^4 \left(\sum_{s|[p,q]}sj_s(\alpha)\right)^{j_p(\alpha)j_q(\beta)\langle p,q\rangle}\tag{4} \end{align*}

We list the permutations of $S_4$ in cycle notation and write a table with the number of cycles of each length for each permutation. We also write the corresponding monomial from the cycle index. \begin{array}{l|ccccc} \pi&Z(S_4)&j_1(\pi)&j_2(\pi)&j_3(\pi)&j_3(\pi)\\ \hline id&z_1^4&4&0&0&0\\ (12)&6z_1^2z_2&1&1&0&0\\ (123)&8z_1z_3&0&0&1&0\\ (12)(34)&3z_2^2&0&2&0&0\\ (1234)&6z_4&0&0&0&1\\ \end{array}

Note: The factors $1,6,8,3$ and $6$ in the column $Z(S_4)$ indicate the number of different permutations of the corresponding cycle type. We will use this fact to considerably reduce the calculation of the number of summands in (4).

In the following it is sufficient to calculate tables for the $25$ pairs \begin{align*} \{id,(12),(123),(12)(34),(1234)\}\times\{id,(12),(123),(12)(34),(1234)\} \end{align*} the cycle index provides the supplementary information we need to calculate the complete sum.

Note that in the main table of $n=2$ above there is some redundancy to ease traceability. We now use analogously to $n=3$ above a somewhat more compact notation to ease readability and keep the space small.

Table: $j_s(\pi), j_p(\pi),j_q(\pi)$ \begin{array}{cc|cc|ccc|ccc|ccc} &&&&&\pi=id&&&\pi=(12)&&&\pi=(123)&\\ p&q&[p,q]&s&j_s&j_p&j_q&j_s&j_p&j_q&j_s&j_p&j_q\\ \hline 1&1&1&1&4&4&4&2&2&2&1&1&1\\ 1&2&2&1&4&4&0&2&2&1&1&1&0\\ &&&2&0&&&1&&&0&&\\ 1&3&3&1&4&4&0&2&2&0&1&1&1\\ &&&3&0&&&0&&&1&&\\ 1&4&4&1&4&4&0&2&2&0&1&1&0\\ &&&2&0&&&1&&&0&&\\ &&&4&0&&&0&&&0&&\\ 2&1&2&1&4&0&4&2&1&2&1&0&1\\ &&&2&0&&&1&&&0&&\\ 2&2&2&1&4&0&0&2&1&1&1&0&0\\ &&&2&0&&&1&&&0&&\\ 2&3&6&1&4&0&0&2&1&0&1&0&1\\ &&&2&0&&&1&&&0&&\\ &&&3&0&&&0&&&1&&\\ 2&4&4&1&4&0&0&2&1&0&1&0&0\\ &&&2&0&&&1&&&0&&\\ &&&4&0&&&0&&&0&&\\ 3&1&3&1&4&0&4&2&0&2&1&1&1\\ &&&3&0&&&0&&&1&&\\ 3&2&6&1&4&0&0&2&0&1&1&1&0\\ &&&2&0&&&1&&&0&&\\ &&&3&0&&&0&&&1&&\\ 3&3&3&1&4&0&0&2&0&0&1&1&1\\ &&&3&0&&&0&&&1&&\\ 3&4&12&1&4&0&0&2&0&0&1&1&0\\ &&&2&0&&&1&&&0&&\\ &&&3&0&&&0&&&1&&\\ &&&4&0&&&0&&&0&&\\ 4&1&4&1&4&0&4&2&0&2&1&0&1\\ &&&2&0&&&1&&&0&&\\ &&&4&0&&&0&&&0&&\\ 4&2&4&1&4&0&0&2&0&1&1&0&0\\ &&&2&0&&&1&&&0&&\\ &&&4&0&&&0&&&0&&\\ 4&3&12&1&4&0&0&2&0&0&1&0&1\\ &&&2&0&&&1&&&0&&\\ &&&3&0&&&0&&&1&&\\ &&&4&0&&&0&&&0&&\\ 4&4&4&1&4&0&0&2&0&0&1&0&0\\ &&&2&0&&&1&&&0&&\\ &&&4&0&&&0&&&0&&\\ &&&&&&&&&&&&\\ &&&&&&&&&&&&\\ \end{array}

$$ $$

Table (cont.): $j_s(\pi), j_p(\pi),j_q(\pi)$ \begin{array}{cc|cc|ccc|cccccc} &&&&&\pi=(12)(34)&&&\pi=(1234)&\\ p&q&[p,q]&s&j_s&j_p&j_q&j_s&j_p&j_q\\ \hline 1&1&1&1&0&0&0&0&0&0&&&\\ 1&2&2&1&0&0&2&0&0&0&&&\\ &&&2&2&&&0&&&&&\\ 1&3&3&1&0&0&0&0&0&0&&&\\ &&&3&0&&&0&&&&&\\ 1&4&4&1&0&0&0&0&0&1&&&\\ &&&2&2&&&0&&&&&\\ &&&4&0&&&1&&&&&\\ 2&1&2&1&0&2&0&0&0&0&&&\\ &&&2&2&&&0&&&&&\\ 2&2&2&1&0&2&2&0&0&0&&&\\ &&&2&2&&&0&&&&&\\ 2&3&6&1&0&2&0&0&0&0&&&\\ &&&2&2&&&0&&&&&\\ &&&3&0&&&0&&&&&\\ 2&4&4&1&0&2&0&0&0&1&&&\\ &&&2&2&&&0&&&&&\\ &&&4&0&&&1&&&&&\\ 3&1&3&1&0&0&0&0&0&0&&&\\ &&&3&0&&&0&&&&&\\ 3&2&6&1&0&0&2&0&0&0&&&\\ &&&2&2&&&0&&&&&\\ &&&3&0&&&0&&&&&\\ 3&3&3&1&0&0&0&0&0&0&&&\\ &&&3&0&&&0&&&&&\\ 3&4&12&1&0&0&0&0&0&1&&&\\ &&&2&2&&&0&&&&&\\ &&&3&0&&&0&&&&&\\ &&&4&0&&&1&&&&&\\ 4&1&4&1&0&0&0&0&1&0&&&\\ &&&2&2&&&0&&&&&\\ &&&4&0&&&1&&&&&\\ 4&2&4&1&0&0&2&0&1&0&&&\\ &&&2&2&&&0&&&&&\\ &&&4&0&&&1&&&&&\\ 4&3&12&1&0&0&0&0&1&0&&&\\ &&&2&2&&&0&&&&&\\ &&&3&0&&&0&&&&&\\ &&&4&0&&&1&&&&&\\ 4&4&4&1&0&0&0&0&1&1&&&\\ &&&2&2&&&0&&&&&\\ &&&4&0&&&1&&&&&\\ \end{array}

The table above provides all information necessary to calculate the summands in (4) for each of the $25$ pairs of permutations.

Hint: Observe, that we only need to consider $25$ pairs of permutations instead of $\left(4!\right)^2=576$ pairs which are summed up in (4). We will consider all other permutations by respecting multiplicities given by the cycle-index $Z(S_4)$.

An example of a typical block is given here for $((123),(123))$ as it was done for all four blocks in the case $n=2$ and a summary table follows below.

Table: $\{(123)\}\times\{(123)\}$

\begin{array}{cc|cccc|cc|cc|rr} \alpha&\beta&p&q&[p,q]&<p,q>&s&j_s(\alpha)&j_p(\alpha)&j_q(\beta)&\text{factors}&\text{result}\\ \hline (123)&(123)&1&1&1&1&1&1&1&1&1&1024\\ &&1&2&2&1&1&1&1&0&1&\\ &&&&&&2&0&&&&\\ &&1&3&3&1&1&1&1&1&4&\\ &&&&&&3&1&&&&\\ &&1&4&4&1&1&1&1&0&1&\\ &&&&&&2&0&&&&\\ &&&&&&4&0&&&&\\ &&2&1&2&1&1&1&0&1&1&\\ &&&&&&2&0&&&&\\ &&2&2&2&2&1&1&0&0&1&\\ &&&&&&2&0&&&&\\ &&2&3&6&1&1&1&0&1&1&\\ &&&&&&2&0&&&&\\ &&&&&&3&1&&&&\\ &&2&4&4&2&1&1&0&0&1&\\ &&&&&&2&0&&&&\\ &&&&&&4&0&&&&\\ &&3&1&3&1&1&1&1&1&4&\\ &&&&&&3&1&&&&\\ &&3&2&6&1&1&1&1&0&1&\\ &&&&&&2&0&&&&\\ &&&&&&3&1&&&&\\ &&3&3&3&3&1&1&1&1&64&\\ &&&&&&3&1&&&&\\ &&3&4&12&1&1&1&1&0&1&\\ &&&&&&2&0&&&&\\ &&&&&&3&1&&&&\\ &&&&&&4&0&&&&\\ &&4&1&4&1&1&1&0&1&1&\\ &&&&&&2&0&&&&\\ &&&&&&4&0&&&&\\ &&4&2&4&2&1&1&0&0&1&\\ &&&&&&2&0&&&&\\ &&&&&&4&0&&&&\\ &&4&3&12&1&1&1&0&1&1&\\ &&&&&&2&0&&&&\\ &&&&&&3&1&&&&\\ &&&&&&4&0&&&&\\ &&4&4&4&4&1&1&0&0&1&\\ &&&&&&2&0&&&&\\ &&&&&&4&0&&&&\\ \end{array}

$$ $$

Summary:

In order to respect all summands of (4) we write the results of the tables above together with the multiplicity of each permutation according to its cycle type. So, e.g. the permutation $(12)$ has cycle type $z_1z_2$ and there are three permutations of this type $\{(12),(13),(23)\}$, we take a factor $3$.

\begin{array}{ll|r|cc|r} \alpha&\beta&\text{res}&\text{m}_{\alpha}&\text{m}_{\beta}&\text{res}\cdot \text{m}_{\alpha}\cdot \text{m}_{\beta}\\ \hline id&id&4294967296&1&1&4294967296\\ id&(12)&16777216&1&6&100663296\\ id&(123)&65536&1&8&524288\\ id&(12)(34)&65536&1&3&196608\\ id&(1234)&256&1&6&1536\\ \hline (12)&id&65536&6&1&393216\\ (12)&(12)&65536&6&6&2359296\\ (12)&(123)&256&6&8&12288\\ (12)&(12)(34)&65536&6&3&1179648\\ (12)&(1234)&256&6&6&9216\\ \hline (123)&id&256&8&1&2048\\ (123)&(12)&64&8&6&3072\\ (123)&(123)&1024&8&8&65536\\ (123)&(12)(34)&16&8&3&384\\ (123)&(1234)&4&8&6&192\\ \hline (12)(13)&id&65536&3&1&196608\\ (12)(13)&(12)&65536&3&6&1179648\\ (12)(13)&(123)&256&3&8&6144\\ (12)(13)&(12)(34)&65536&3&3&589824\\ (12)(13)&(1234)&256&3&6&4608\\ \hline (1234)&id&256&6&1&1536\\ (1234)&(12)&256&6&6&9216\\ (1234)&(123)&16&6&8&768\\ (1234)&(12)(34)&256&6&3&4608\\ (1234)&(1234)&256&6&6&9216\\ \hline \color{blue}{\text{Total}}&&&&&\color{blue}{4402380096}\\ \end{array}

Since the total of the table is $4402380096$ we finally conclude according to (4) \begin{align*} \color{blue}{a(4,4)=\frac{1}{576}\cdot 4402380096=7643021} \end{align*} and the claim follows.

Conclusion:

  • In order to calculate $a(n,n)$ we do not need $(n!)^2$ summands but can calculate summands corresponding to the square of the number of summands of the cycle index and then use multiplicities of the cycle index for final calculations.

\begin{array}{c|rr} n&(n!)^2&\left(\text{via }Z(S_n)\right)^2\\ \hline 2&4&4\\ 3&36&9\\ 4&576&25\\ \end{array}

  • It seems feasible to find an efficient implemention based upon this formula. A nice program which coincides with the results of this answer is already given by @ScottBurns.
$\endgroup$
  • $\begingroup$ I congratulate you on having included cycle indices in your repertoire (as you did with the Egorychev method). I always reference Harary and Palmer and it is nice to know that their papers / the book keep gaining attentive readers. Very good work. (+1.) $\endgroup$ – Marko Riedel Oct 30 '16 at 23:55
  • $\begingroup$ @Markus: Thank you very much! To whom should I award the bounty, in your opinion? $\endgroup$ – Hans-Peter Stricker Oct 31 '16 at 8:28
  • $\begingroup$ @HansStricker: You're welcome. The paper from Harary/Palmer is interesting, so it was nice to work on that problem. Thanks for asking regarding the bounty. I think both answers are useful and worth the bounty. But, since Scott is new to the community, he should get the bounty. I think it's a good stimulus for him to provide further useful answers. It would be nice if you could accept my answer. Best regads, $\endgroup$ – Markus Scheuer Oct 31 '16 at 8:50
  • $\begingroup$ @MarkoRiedel: Thanks for your nice comment! In fact I got acquainted with it roughly 30 years ago. This time it's a refresher, but I also would like to become familiar with it and this is a difference! :-) Could you please add a reference to the paper you are referring? (+1) $\endgroup$ – Markus Scheuer Oct 31 '16 at 8:59
  • 1
    $\begingroup$ @Markus: I did as you proposed to do. Thanks again for your great answer! $\endgroup$ – Hans-Peter Stricker Oct 31 '16 at 9:32
7
+200
$\begingroup$

So here's a better approach. Do what Harary/Palmer tell you. To really do this justice you need a crash course in the representation theory of permutation groups and Schur functors, but we'll get by without it.

The formula we need to calculate is $$ a(n,k) = {1\over n!k!} \sum \displaystyle\prod_{p=1}^n \displaystyle\prod_{q=1}^k ( \displaystyle\sum_{s|[p,q]} s j_s(\alpha))^{j_p(\alpha) j_q(\beta) <p,q>} $$ where the outer sum is over pairs of $\alpha \in S_n$ and $\beta \in S_k$

$j_d(\pi)$ the number of d-cycles in a permutation $\pi$

$[a,b]$ represents lcm and $<a,b>$ represents gcd.

It's easier to calculate if, instead of summing over $\alpha$ and $\beta$, we sum over their respective conjugacy classes in $S_n$ and $S_k$, and weight the sum by the size of each class. This is equivalent to enumerating the partitions of $[n]\times[k]$, as the conjugacy classes in $S_n$ are determined by cycle decomposition.

The code is not difficult, and it's a lot quicker than my first approach. Here's the main blob.

sum=0;
for (alpha=0;alpha<CC;alpha++){
  for (beta=0;beta<DD;beta++){

    pqProd=1;
    for (pp=1;pp<=NN;pp++){
      for (qq=1;qq<=KK;qq++){
        gcd=gcdLut[pp][qq];
        lcm=lcmLut[pp][qq];

        sSum=0;
        for (ss=1;ss<=lcm && ss<=NN;ss++){
          if (lcm%ss!=0) continue;
          sSum += ss * cycleCountN[alpha][ss];
        }

        xx=gcd * cycleCountN[alpha][pp] * cycleCountK[beta][qq];
        pqProd *= pow_uint64(sSum,xx);
      }
    }
    sum += pqProd * cSize[alpha] * dSize[beta];
  }
}
printf("sum %llu\n",sum);

Notes: I set NN and KK at compile-time. The partition numbers are CC and DD respectively. I precomputed the gcds and lcms to save rework. The partitions are stored in the cycleCount arrays, and the sizes of the partitions in cSize and dSize. The rest should be self-explanatory. If you want to copy this, you'll also need an integer exponentiation function, but that's pretty easy to source. Make sure it calculates $0^0 = 1$

I have tested this for $N=K=2$ and it returns 28 (note I haven't divided by the order of the groups in the code).

Finally, the answer, which is a lot less interesting than the question!

$$ a(4,4) = 7643021 $$

Bugs notwithstanding...

UPDATE:

I just saw Markus' post - good to see we are on similar lines. I've run my code for $N=K=3$ and get 23076/36 = 641. So we're close, but not perfect!

The only line on which we differ is $\alpha = (123)$, $\beta = (12)$. I get a result of 27, whereas Markus gets 9 (prior to multiplication by the partition sizes). Looking inside I get a factor 9 from $p = 3$, $q = 2$. Which means I have a bug. I'm overflowing the sum up to the lcm, which should by limited by $N$.

I have just fixed this and now Markus and I agree on 638. I've corrected the code and answer for $a(4,4)$ above.

Here is a table of results for $a(N,K)$ for values up to 9.

+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
| N\K |    1 |            2 |              3 |               4 |              5 |           6 |            7 |       8 |        9 |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
|   1 |    1 |            1 |              1 |               1 |              1 |           1 |            1 |       1 |        1 |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
|   2 |    3 |            7 |             13 |              22 |             34 |          50 |           70 |      95 |      125 |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
|   3 |    7 |           74 |            638 |            4663 |          28529 |      151600 |       713176 | 3028727 | 11773093 |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
|   4 |   19 |         1474 |         118949 |         7643021 |      396979499 | 17265522590 | 646203233957 |         |          |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
|   5 |   47 |        41876 |       42483668 |     33179970333 | 20762461502595 |             |              |         |          |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
|   6 |  130 |      1540696 |    23524514635 | 274252613077267 |                |             |              |         |          |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
|   7 |  343 |     68343112 | 18477841853059 |                 |                |             |              |         |          |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
|   8 |  951 |   3540691525 |                |                 |                |             |              |         |          |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+
|   9 | 2615 | 209612916303 |                |                 |                |             |              |         |          |
+-----+------+--------------+----------------+-----------------+----------------+-------------+--------------+---------+----------+

And here's the code. It's slightly complicated due to me initialising variable-width arrays in subroutines, which leads to some interesting pointer constructs! I'm rather pleased with the partition generator. It's your basic backtracker, but it has come out very neat!

Not all compilers are friendly for arrays with widths set at run-time. I compiled this with gcc.

///////// Calculate a(N,K) /////////////
// Call: ank N K
// N and K should be positive integers. This is not checked.
// Results go to stdout
// There is no test for integer overflow, so large values of N and K will not work!

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef unsigned long long uint64;

///////////////////////////
//// Integer exponentiation
///////////////////////////

uint64 pow_uint64(uint64 base, int exp)
{
  uint64 result = 1;

  // 0**0=1
  if (base==0 && exp==0) return result;

  while (exp){
    if (exp & 1)
      result *= base;
    exp >>= 1;
    base *= base;
  }

  return result;
}

//////////////////////
//// gcd & lcm precalc
//////////////////////

void init_gcd_lcm(int nk, int gcd[][nk+1], int lcm[][nk+1])
{
  int ii,jj,kk;

  for (ii=0;ii<=nk;ii++)
    gcd[0][ii]=gcd[ii][0]=lcm[0][ii]=lcm[ii][0]=0;

  // These are small numbers - it's quite efficient to search and easier to code than Euclid's algorithm.
  for (ii=1;ii<=nk;ii++){
    for (jj=1;jj<ii;jj++){
      gcd[ii][jj]=1;
      for (kk=2;kk<=jj;kk++){
        if (ii%kk==0 && jj%kk==0)
          gcd[ii][jj]=kk;
      }
      gcd[jj][ii]=gcd[ii][jj];
      lcm[jj][ii]=lcm[ii][jj]=ii*jj/gcd[ii][jj];
    }
    gcd[ii][ii]=lcm[ii][ii]=ii;
  }

  return;
}

////////////////////////
//// Generate partitions
////////////////////////

int init_partitions(int nk,int (**store)[nk+1])
{
  int lev;                      /* do this by a superloop */
  int rr[nk+1];                 /* remainder at each level */
  int part[nk+1];               /* partition being built */
  int npart=0;
  int ii;

  *store=malloc((nk+1)*sizeof(int));

  part[nk]=2;
  rr[nk]=nk;
  for (lev=nk;lev<=nk;lev++){
    for (part[lev]--;part[lev]>=0;part[lev]--){

      // if we reach rr[lev]==0 we have a partition
      if (rr[lev]==0){
        for (ii=lev;ii>=0;ii--)
          part[ii]=0;
        // Save it
        memcpy((*store)[npart],part,sizeof(part));
        npart++;
        *store=realloc(*store,(npart+1)*sizeof(part));
        continue;
      }

      // if we reach lev==1 we can complete the partition
      if (lev==1){
        part[1]=rr[1];
        part[0]=0;
        // Save it
        memcpy((*store)[npart],part,sizeof(part));
        npart++;
        *store=realloc(*store,(npart+1)*sizeof(part));
        // Set up to ascend
        part[1]=0;
        continue;
      }

      // How much remains to push to a lower level?
      rr[lev-1]=rr[lev]-part[lev]*lev;
      part[lev-1]=1+rr[lev-1]/(lev-1);
      lev--;
    }
  }

  return npart;
}

///////////
//// MAIN
///////////

int main(int argc, char **argv)
{
  // parameters N, K
  int nn=atoi(argv[1]);
  int kk=atoi(argv[2]);
  int nkMax=(nn>kk) ? nn:kk;

  // Precalculate gcds and lcms
  int (*gcdLut)[nkMax+1];
  int (*lcmLut)[nkMax+1];

  // Conjugacy classes = partitions on N and K
  int (*nClass)[nn+1];
  int *nClassSize;
  int nClassCount;
  int (*kClass)[kk+1];
  int *kClassSize;
  int kClassCount;
  uint64 *fact;                 /* factorials */

  int alpha, beta;              /* perms represented by class */
  int pp, qq;
  int ss, gcd, lcm, xx;
  uint64 sum, sSum, pqProd, yy;
  int ii,jj;

  //// Initialise

  gcdLut=malloc((nkMax+1)*(nkMax+1)*sizeof(int));
  lcmLut=malloc((nkMax+1)*(nkMax+1)*sizeof(int));
  init_gcd_lcm(nkMax,gcdLut,lcmLut);

  nClassCount=init_partitions(nn,&nClass);
  kClassCount=init_partitions(kk,&kClass);

  // Factorials
  fact=malloc((1+nkMax)*sizeof(uint64));
  for (fact[0]=1,ii=1;ii<=nkMax;ii++)
    fact[ii]=fact[ii-1]*ii;

  // Class sizes
  nClassSize=malloc(nClassCount*sizeof(int));
  for (ii=0;ii<nClassCount;ii++){
    nClassSize[ii]=fact[nn];
    for (jj=1;jj<=nn;jj++){
      yy=pow_uint64((uint64)jj,nClass[ii][jj]);
      nClassSize[ii] /= fact[nClass[ii][jj]]*yy;
    }
  } 
  kClassSize=malloc(kClassCount*sizeof(int));
  for (ii=0;ii<kClassCount;ii++){
    kClassSize[ii]=fact[kk];
    for (jj=1;jj<=kk;jj++){
      yy=pow_uint64((uint64)jj,kClass[ii][jj]);
      kClassSize[ii] /= fact[kClass[ii][jj]]*yy;
    }
  } 

  //// Principal calculation

  sum=0;
  for (alpha=0;alpha<nClassCount;alpha++){
    for (beta=0;beta<kClassCount;beta++){

      pqProd=1;
      for (pp=1;pp<=nn;pp++){
        for (qq=1;qq<=kk;qq++){
          gcd=gcdLut[pp][qq];
          lcm=lcmLut[pp][qq];
          sSum=0;
          for (ss=1;ss<=lcm && ss<=nn;ss++){
            if (lcm%ss!=0) continue;
            sSum += ss * nClass[alpha][ss];
          }
          xx=gcd * nClass[alpha][pp] * kClass[beta][qq];
          pqProd *= yy = pow_uint64(sSum,xx);

          // Print contributors
          if (yy>1){
            for (ii=1;ii<=nn;ii++)
              printf("%d",nClass[alpha][ii]);
            printf("\t");
            for (ii=1;ii<=kk;ii++)
              printf("%d",kClass[beta][ii]);
            printf("\t%d\t%d",pp,qq);
            printf("\t%llu\n",sSum);
          }

        }
      }
      sum += pqProd * nClassSize[alpha] * kClassSize[beta];
      // Print separator between perm pairs
      printf("%d\t%d\t%llu\n\n",nClassSize[alpha],kClassSize[beta],pqProd);
    }
  }

  sum = sum / fact[nn] / fact[kk];
  printf("sum %llu\n",sum);


  return 0;
}
$\endgroup$
  • $\begingroup$ Good work! (+1) ...and good to see that both answers now coincide! :-) I also do appreciate that coding is now based upon the formula of Harary/Palmer which is clearly superior to the brute force approach. $\endgroup$ – Markus Scheuer Oct 29 '16 at 23:55
  • $\begingroup$ I've also calculated $a(4,4)$ and the results coincide. Since we definitely used a diverse approach, the answer seems to be ok! :-) $\endgroup$ – Markus Scheuer Oct 30 '16 at 20:02
  • $\begingroup$ I checked your table against my results and they are the same. Your work was posted first and hence (+1). $\endgroup$ – Marko Riedel Oct 31 '16 at 0:47
  • $\begingroup$ @Scott: Thank you very much! To whom should I award the bounty, in your opinion? $\endgroup$ – Hans-Peter Stricker Oct 31 '16 at 8:27
  • $\begingroup$ @ScottBurns If you are interested and can find the time you might want to treat the case of more than one output symbol, perhaps in an extra answer. $\endgroup$ – Marko Riedel Oct 31 '16 at 19:01
5
$\begingroup$

This is computationally feasible for F(4,4). I wouldn't attempt it for higher values though.

For F(4,4), represent the functions as 32-bit strings (really as 16-dibit strings but I'm thinking of computation more than algebra here).

The simple approach is to seive. Maintain a Bloom filter of all the functions you've visited (this is simply a $2^{32}$-long array of bits).

Take each $g$ that you've not yet visited, calculate its orbit under the action of S4 x S4, and sieve out all the functions you reach.

I haven't costed this but it's no worse than around $2^{41}$ permutations, so should run inside a day on a single core processor. I expect it to be rather better than that as the sieving gives a lot of cut-down.

Note also that the equivalence classes preserve the partition of [16] into 4 parts as you look across the image of $g$. You can use that to subdivide your search over partition classes, which will save on memory but not on time. This may be useful for tackling larger values where memory becomes an issue.

UPDATE: I've written it and am now running. Very quick and dirty code, so not guaranteed bug-free! Will be useful to compare results with Nitin.

$\endgroup$
  • $\begingroup$ Thanks for posting. I look forward to comparing results. $\endgroup$ – Nitin Oct 29 '16 at 5:26
  • $\begingroup$ @ScottBurns: Maybe you could additionally calculate $a(3,3)$? $\endgroup$ – Markus Scheuer Oct 29 '16 at 20:33
3
$\begingroup$

What we have here is a special and unique variant of Simultaneous Power Group Enumeration (as presented by Harary and Palmer and Fripertinger, in a different publication), with the group acting on the $n\times m$ slots (think of a rectangular array) where the $m$ different values (colors) are placed being the product of the symmetric groups $S_n$ and $S_m$ on $n$ and $m$ elements acting on rows and columns where the group permuting the columns acts on the values themselves at the same time, with column index mapped to colors.

Choosing to present a solution from first principles we can compute the number $q_n$ of functions by Burnside's lemma which says to average the number of assignments fixed by the elements of the group acting on the slots and values. But this number is easy to compute and it only needs the two cycle indices (as opposed to all $n!\times m!$ permutations). (Complexity of the cycle index is given by the partition function.)

Suppose we have as in the problem definition, a permutation $\pi$ from $S_n$ and a permutation $\tau$ from $S_m.$ Let $\alpha$ be the permutation of the cells of the grid that is obtained when $\pi$ and $\tau$ act simultaneously, and factor it into cycles. If we place the appropriate number of complete, directed and consecutive copies of a cycle from $\tau$ on a cycle from $\alpha$ then this assignment is fixed under the power group action for $(\alpha,\tau)$, and this is possible iff the length of the cycle from $\tau$ divides the length of the cycle from $\alpha$. The process yields as many assignments as the length of the cycle from $\tau.$

The cycle index of the symmetric group can be computed from the classical recurrence by Lovasz for the cycle index $Z(S_n)$, which is

$$Z(S_n) = \frac{1}{n} \sum_{l=1}^n a_l Z(S_{n-l}) \quad\text{where}\quad Z(S_0) = 1.$$

It then remains to apply the Power Group Enumeration formula / algorithm as documented at the following MSE link I.

Actually doing the computation we obtain for the number of functions on a square grid and $n$ colors the following sequence:

$$1, 7, 638, 7643021, 20762461502595, 19903050866658120066632, \\ 10114722264843500593900485682759058, \\ 3861169308385212945415179151162048048461447621051,\ldots$$

We get for an $n \times (n+1)$ grids with $n+1$ colors the sequence

$$3, 74, 118949, 33179970333, 2559276179593762172, \\ 85973094952794304259466151418, \\ 1841148232300929744056375072663778725072045, \ldots$$

We get for $n\times (n+2)$ the sequence

$$7, 1474, 42483668, 274252613077267, 626361440405926396941497, \\ 768186632385442429091738459545921683, \\ 716282843344020261990396181005284021715688233692930,\ldots $$

Holding $m$ fixed as opposed to $n$ we get for $(m+1)\times m$

$$1, 13, 4663, 396979499, 10831034126757463, \\ 132673733865643566661223817, \\ 1041247439945746392774732251877428013424, \ldots $$

and for $(m+2)\times m$ we finally have

$$1, 22, 28529, 17265522590, 4844565331763027596, \\ 773869304738817313660236854435, \\ 95280041156689492626670871782790953125692770,\ldots$$

The Maple code for these was as follows.

with(combinat);

pet_cycleind_symm :=
proc(n)
option remember;

    if n=0 then return 1; fi;

    expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;

pet_flatten_term :=
proc(varp)
local terml, d, cf, v;

    terml := [];

    cf := varp;
    for v in indets(varp) do
        d := degree(varp, v);
        terml := [op(terml), seq(v, k=1..d)];
        cf := cf/v^d;
    od;

    [cf, terml];
end;

cycles_sim :=
proc(cyca, cycb)
local ca, cb, lena, lenb, res, vlcm;

    res := 1;

    for ca in cyca do
        lena := op(1, ca);

        for cb in cycb do
            lenb := op(1, cb);

            vlcm := lcm(lena, lenb);
            res := res*a[vlcm]^(lena*lenb/vlcm);
        od;
    od;

    res;
end;


automaton :=
proc(N, M)
option remember;
local idx_slots, idx_cols, res, a, b, sim, flat_sim,
    flat_a, flat_b, cyc_a, cyc_b, len_a, len_b, p, q;

    if N > 1 then
        idx_slots := pet_cycleind_symm(N);
    else
        idx_slots := [a[1]];
    fi;

    if M > 1 then
        idx_cols := pet_cycleind_symm(M);
    else
        idx_cols := [a[1]];
    fi;


    res := 0;

    for a in idx_slots do
        flat_a := pet_flatten_term(a);
        for b in idx_cols do
            flat_b := pet_flatten_term(b);

            sim := cycles_sim(flat_a[2], flat_b[2]);
            flat_sim := pet_flatten_term(flat_a[1]*sim);

            p := 1;
            for cyc_a in flat_sim[2] do
                len_a := op(1, cyc_a);
                q := 0;

                for cyc_b in flat_b[2] do
                    len_b := op(1, cyc_b);

                    if len_a mod len_b = 0 then
                        q := q + len_b;
                    fi;
                od;

                p := p*q;
            od;

            res := res + p*flat_sim[1]*flat_b[1];
        od;
    od;

    res;
end;
$\endgroup$
  • $\begingroup$ This sequence is at the time not in OEIS. Since even the special case of counting the number of finite automata with $n$ states, $n$ input symbols and one output symbol is of interest by itself, it would be nice to add it to OEIS. Your opinion? $\endgroup$ – Markus Scheuer Oct 31 '16 at 9:48
  • $\begingroup$ I agree and I will add these to the OEIS. However I do not quite see where the output symbol is fixed in the functions introduced by the OP. Can you comment please? I guess this would affect the table, with entries becoming pairs of the next state and the symbol being output? Do you want to compute some of these say $n$ by $n$ with two output symbols? Or do we leave it at one symbol for now? $\endgroup$ – Marko Riedel Oct 31 '16 at 16:56
  • $\begingroup$ Hi Marko! I've added this question addressing the general case where you and/or @ScottBurns are heartily invited to add some more information if you like. $\endgroup$ – Markus Scheuer Oct 31 '16 at 21:21
0
$\begingroup$

Below are tables for $a(N,K,M)$ for values $\le 5$ where

$N = $ number of states, $K = $ number of inputs, $M = $ number of outputs

+-----+---+------+------------+-----------------+-----------------+-----------------+
| M=2 |   | K    |            |                 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     |   | 1    | 2          | 3               | 4               | 5               |
+-----+---+------+------------+-----------------+-----------------+-----------------+
| N   | 1 | 1    | 2          | 2               | 3               | 3               |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 2 | 6    | 44         | 226             | 1036            | 4006            |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 3 | 22   | 2038       | 142336          | 7775708         | 341906882       |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 4 | 114  | 176936     | 238882846       | 244698934716    | 200649261017386 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 5 | 538  | 20943790   | 694540531869    |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     |   |      |            |                 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
| M=3 |   | K    |            |                 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     |   | 1    | 2          | 3               | 4               | 5               |
+-----+---+------+------------+-----------------+-----------------+-----------------+
| N   | 1 | 1    | 2          | 3               | 4               | 5               |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 2 | 6    | 75         | 775             | 7124            | 55668           |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 3 | 29   | 7623       | 1804128         | 329641077       | 48317584819     |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 4 | 190  | 1501516    | 10322146155     | 53512221536494  |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 5 | 1289 | 401371270  | 101367856946674 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     |   |      |            |                 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
| M=4 |   | K    |            |                 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     |   | 1    | 2          | 3               | 4               | 5               |
+-----+---+------+------------+-----------------+-----------------+-----------------+
| N   | 1 | 1    | 2          | 3               | 5               | 6               |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 2 | 6    | 81         | 1183            | 17320           | 223743          |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 3 | 29   | 11676      | 6064606         | 2593640209      | 897009602752    |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 4 | 209  | 3831148    | 81573276196     | 248097408553    |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 5 | 1605 | 1790644262 |                 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     |   |      |            |                 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
| M=5 |   | K    |            |                 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     |   | 1    | 2          | 3               | 4               | 5               |
+-----+---+------+------------+-----------------+-----------------+-----------------+
| N   | 1 | 1    | 2          | 3               | 5               | 7               |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 2 | 6    | 81         | 1283            | 23718           | 427097          |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 3 | 29   | 12621      | 9875766         | 7694431189      | 5108729338005   |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 4 | 209  | 5269634    | 242293771832    | 167930305528968 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
|     | 5 | 1652 | 3522483774 |                 |                 |                 |
+-----+---+------+------------+-----------------+-----------------+-----------------+
$\endgroup$

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