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Question. Let $\mathcal{M}_3$ be the set of $3\times 3$ matrices with non-negative integer entries and unit determinant. What is the number of $M\in \mathcal{M}_3$ with fixed sum of entries? What is the answer if we allow only matrices with strictly positive entries?

The rate of growth is also interesting. Below $n$ stands for the sum of entries.

Origin. For the similar question in the case of $2\times 2$ matrices we get $\phi(n)$ and $\phi(n) - 2$ respectively, see this recent post: Decomposition of a natural number as sum of positive integers.

Computation of the first elements. If we denote the answer sequence for non-negative case by $a_n$ then the first elements are

$a = [0, 0, 0, 3, 18, 54, 126, 261, 404, 667, 955, 1417, 1709, 2603, 2979, \ldots]$.

And if $b_n$ is the answer for the strictly positive case then $b_n = 0$ if $n \le 10$ and further we have

$b = [0,\ldots ,0, 9, 18, 49, 69, 169, 239, 388, 423, 848, 902, 1231, 1559, 2283, 2002, \ldots]$.

Unfortunately, both sequences do not seem to be in OEIS.

Thoughts on the growth. Evidently, $a_n\ge b_n$, and $a_n\le Cn^8$ because the number of partitions of $n$ into $9$ terms is $O(n^8)$. If we consider only upper-triangular matrices with ones on the diagonal, we will get the bound $a_n\ge Cn^2$.

Thus, the answer is of polynomial order. Note that this is different to the $2\times 2$ case where the number-theoretic properties played the main role.

Question. Let $\mathcal{M}_3$ be the set of $3\times 3$ matrices with non-negative integer entries and unit determinant. What is the number of $M\in \mathcal{M}_3$ with fixed sum of entries? What is the answer if we allow only matrices with strictly positive entries?

The rate of growth is also interesting. Below $n$ stands for the sum of entries.

Origin. For the similar question in the case of $2\times 2$ matrices we get $\phi(n)$ and $\phi(n) - 2$ respectively, see this recent post: Decomposition of a natural number as sum of positive integers.

Computation of the first elements. If we denote the answer sequence for non-negative case by $a_n$ then the first elements are

$a = [0, 0, 0, 3, 18, 54, 126, 261, 432, 783, 1134, 1899, 2286, 3960, 4680, \ldots]$.

And if $b_n$ is the answer for the strictly positive case then $b_n = 0$ if $n \le 10$ and further we have

$b = [0,\ldots ,0, 9, 18, 72, 108, 234, 360, 747, 756, 1818, 1782, 3222, 3672, 6615, \ldots]$.

Unfortunately, both sequences do not seem to be in OEIS.

Thoughts on the growth. Evidently, $a_n\ge b_n$, and $a_n\le Cn^8$ because the number of partitions of $n$ into $9$ terms is $O(n^8)$. If we consider only upper-triangular matrices with ones on the diagonal, we will get the bound $a_n\ge Cn^2$.

Thus, the answer is of polynomial order. Note that this is different to the $2\times 2$ case where the number-theoretic properties played the main role.

Question. Let $\mathcal{M}_3$ be the set of $3\times 3$ matrices with non-negative integer entries and unit determinant. What is the number of $M\in \mathcal{M}_3$ with fixed sum of entries? What is the answer if we allow only matrices with strictly positive entries?

The rate of growth is also interesting.

Origin. For the similar question in the case of $2\times 2$ matrices we get $\phi(n)$ and $\phi(n) - 2$ respectively, see this recent post: Decomposition of a natural number as sum of positive integers.

Computation of the first elements. If we denote the answer sequence for non-negative case by $a_n$ then the first elements are

$a = [0, 0, 0, 3, 18, 54, 126, 261, 404, 667, 955, 1417, 1709, 2603, 2979, \ldots]$.

And if $b_n$ is the answer for the strictly positive case then $b_n = 0$ if $n \le 10$ and further we have

$b = [0,\ldots ,0, 9, 18, 49, 69, 169, 239, 388, 423, 848, 902, 1231, 1559, 2283, 2002, \ldots]$.

Unfortunately, both sequences do not seem to be in OEIS.

Thoughts on the growth. Evidently, $a_n\ge b_n$, and $a_n\le Cn^8$ because the number of partitions of $n$ into $9$ terms is $O(n^8)$. If we consider only upper-triangular matrices with ones on the diagonal, we will get the bound $a_n\ge Cn^2$.

Thus, the answer is of polynomial order. Note that this is different to the $2\times 2$ case where the number-theoretic properties played the main role.

Question. Let $\mathcal{M}_3$ be the set of $3\times 3$ matrices with non-negative integer entries and unit determinant. What is the number of $M\in \mathcal{M}_3$ with fixed sum of entries? What is the answer if we allow only matrices with strictly positive entries?

The rate of growth is also interesting. Below $n$ stands for the sum of entries.

Origin. For the similar question in the case of $2\times 2$ matrices we get $\phi(n)$ and $\phi(n) - 2$ respectively, see this recent post: Decomposition of a natural number as sum of positive integers.

Computation of the first elements. If we denote the answer sequence for non-negative case by $a_n$ then the first elements are

$a = [0, 0, 0, 3, 18, 54, 126, 261, 404, 667, 955, 1417, 1709, 2603, 2979, \ldots]$.

And if $b_n$ is the answer for the strictly positive case then $b_n = 0$ if $n \le 10$ and further we have

$b = [0,\ldots ,0, 9, 18, 49, 69, 169, 239, 388, 423, 848, 902, 1231, 1559, 2283, 2002, \ldots]$.

Unfortunately, both sequences do not seem to be in OEIS.

Thoughts on the growth. Evidently, $a_n\ge b_n$, and $a_n\le Cn^8$ because the number of partitions of $n$ into $9$ terms is $O(n^8)$. If we consider only upper-triangular matrices with ones on the diagonal, we will get the bound $a_n\ge Cn^2$.

Thus, the answer is of polynomial order. Note that this is different to the $2\times 2$ case where the number-theoretic properties played the main role.

Question. Let $\mathcal{M}_3$ be the set of $3\times 3$ matrices with non-negative integer entries and unit determinant. What is the number of $M\in \mathcal{M}_3$ with fixed sum of entries? What is the answer if we allow only matrices with strictly positive entries?

The rate of growth is also interesting.

Origin. For the similar question in the case of $2\times 2$ matrices we get $\phi(n)$ and $\phi(n) - 2$ respectively, see this recent post: Decomposition of a natural number as sum of positive integers.

Computation of the first elements. If we denote the answer sequence for non-negative case by $a_n$ then the first elements are

$a = [0, 0, 0, 3, 18, 54, 126, 261, 404, 667, 955, 1417, 1709, 2603, 2979, \ldots]$.

And if $b_n$ is the answer for the strictly positive case then $b_n = 0$ if $n \le 10$ and further we have

$b = [0,\ldots ,0, 9, 18, 49, 69, 169, 239, 388, 423, 848, 902, 1231, 1559, 2283, 2002, \ldots]$.

Unfortunately, both sequences do not seem to be in OEIS.

Thoughts on the growth. Evidently, $a_n\ge b_n$, and $a_n\le Cn^9$ because the number of partitions of $n$ into $9$ terms is $O(n^9)$. If we consider only upper-triangular matrices with ones on the diagonal, we will get the bound $a_n\ge Cn^3$.

Thus, the answer is of polynomial order. Note that this is different to the $2\times 2$ case where the number-theoretic properties played the main role.

Question. Let $\mathcal{M}_3$ be the set of $3\times 3$ matrices with non-negative integer entries and unit determinant. What is the number of $M\in \mathcal{M}_3$ with fixed sum of entries? What is the answer if we allow only matrices with strictly positive entries?

The rate of growth is also interesting.

Origin. For the similar question in the case of $2\times 2$ matrices we get $\phi(n)$ and $\phi(n) - 2$ respectively, see this recent post: Decomposition of a natural number as sum of positive integers.

Computation of the first elements. If we denote the answer sequence for non-negative case by $a_n$ then the first elements are

$a = [0, 0, 0, 3, 18, 54, 126, 261, 404, 667, 955, 1417, 1709, 2603, 2979, \ldots]$.

And if $b_n$ is the answer for the strictly positive case then $b_n = 0$ if $n \le 10$ and further we have

$b = [0,\ldots ,0, 9, 18, 49, 69, 169, 239, 388, 423, 848, 902, 1231, 1559, 2283, 2002, \ldots]$.

Unfortunately, both sequences do not seem to be in OEIS.

Thoughts on the growth. Evidently, $a_n\ge b_n$, and $a_n\le Cn^8$ because the number of partitions of $n$ into $9$ terms is $O(n^8)$. If we consider only upper-triangular matrices with ones on the diagonal, we will get the bound $a_n\ge Cn^2$.

Thus, the answer is of polynomial order. Note that this is different to the $2\times 2$ case where the number-theoretic properties played the main role.