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Let $\mathrm{CO}(m)$ be the set of all compositions of the positive integer $m$. By a composition of $m$, I mean a finite sequence $(m_1,\ldots,m_k)$ of positive integers with sum $\sum m_i=m$. There are $2^{m-1}$ compositions of $m$. The $4$ compositions of $3$ are

$$ \mathrm{CO}(3) =\{ (1,1,1), (1,2), (2,1), (3)\} $$

Let also $q$ be a prime power. Put $[k]_q = \frac{q^k-1}{q-1}$. Write $[m]_q! = \prod_{1 \le k \le m} [k]_q$ for the $q$-factorial of $m$ and

$$ \binom{m}{m_1,\ldots,m_k}_q = \frac{[m]_q!}{[m_1]_q! \cdots [m_k]_q!} $$

for the $q$-multinomial coefficient of the composition $(m_1,\ldots, m_k)$ of $m$. Let $S(m,q)$ be the sequence whose $d$th element, $d=0,1,\ldots$, is $$ S(m,q)(d)= \sum_{(m_1,\ldots,m_k) \in \mathrm{CO}(m)} (-1)^{m-k} \binom{m}{m_1,\ldots,m_k}_q q^{\sum \binom{m_i}{2}}\binom{q^{\binom{m}{2}- \sum \binom{m_i}{2}}}{d} $$ For $m=3$ we get the sequence with $d$th element \begin{equation*} S(3,q)(d) =\binom{3}{1,1,1}_q q^0\binom{q^3}{d} -2 \binom{3}{1,2}_q q^1\binom{q^2}{d} + \binom{3}{3}_q q^3 \binom{q^0}{d} \end{equation*} which for $q=2$ evaluates to \begin{equation*} S(3,2)=1, 64, 420, 1064, 1442, 1176, 588, 168, 21, 0, 0, \ldots \end{equation*} Note that $S(3,2)$ is a log-concave sequence. My question is: Is the sequence $S(m,q)$ always log-concave?

$S(m,q)(d)$, for $q$ a power of the prime $p$, is the number of $d$-subsets of $\mathrm{SL}_m(\mathbf{F}_q)$ generating a $p$-subgroup. These subsets form a contractible simplicial complex with reduced Euler characteristic $\sum_{d \ge 0} (-1)^d S(m,q)(d)=0$.

Let $\mathrm{CO}(m)$ be the set of all compositions of the positive integer $m$. By a composition of $m$, I mean a finite sequence $(m_1,\ldots,m_k)$ of positive integers with sum $\sum m_i=m$. There are $2^{m-1}$ compositions of $m$. The $4$ compositions of $3$ are

$$ \mathrm{CO}(3) =\{ (1,1,1), (1,2), (2,1), (3)\} $$

Let also $q$ be a prime power. Put $[t]_q = \frac{q^t-1}{q-1}$. Write $[m]_q! = \prod_{1 \le t \le m} [t]_q$ for the $q$-factorial of $m$ and

$$ \binom{m}{m_1,\ldots,m_k}_q = \frac{[m]_q!}{[m_1]_q! \cdots [m_k]_q!} $$

for the $q$-multinomial coefficient of the composition $(m_1,\ldots, m_k)$ of $m$. Let $S(m,q)$ be the sequence whose $d$th element, $d=0,1,\ldots$, is $$ S(m,q)(d)= \sum_{(m_1,\ldots,m_k) \in \mathrm{CO}(m)} (-1)^{m-k} \binom{m}{m_1,\ldots,m_k}_q q^{\sum \binom{m_i}{2}}\binom{q^{\binom{m}{2}- \sum \binom{m_i}{2}}}{d} $$ For $m=3$ we get the sequence with $d$th element \begin{equation*} S(3,q)(d) =\binom{3}{1,1,1}_q q^0\binom{q^3}{d} -2 \binom{3}{1,2}_q q^1\binom{q^2}{d} + \binom{3}{3}_q q^3 \binom{q^0}{d} \end{equation*} which for $q=2$ evaluates to \begin{equation*} S(3,2)=1, 64, 420, 1064, 1442, 1176, 588, 168, 21, 0, 0, \ldots \end{equation*} Note that $S(3,2)$ is a log-concave sequence. My question is: Is the sequence $S(m,q)$ always log-concave?

$S(m,q)(d)$, for $q$ a power of the prime $p$, is the number of $d$-subsets of $\mathrm{SL}_m(\mathbf{F}_q)$ generating a $p$-subgroup. These subsets form a contractible simplicial complex with reduced Euler characteristic $\sum_{d \ge 0} (-1)^d S(m,q)(d)=0$.

Let $\mathrm{CO}(m)$ be the set of all compositions of the positive integer $m$. By a composition of $m$, I mean a finite sequence $(m_1,\ldots,m_k)$ of positive integers with sum $\sum m_i=m$. There are $2^{m-1}$ compositions of $m$. The $4$ compositions of $3$ are

$$ \mathrm{CO}(3) =\{ (1,1,1), (1,2), (2,1), (3)\} $$

Let also $q$ be a prime power. Put $[k]_q = \frac{q^k-1}{q-1}$. Write $[m]_q! = \prod [k]_q$ (product over $1 \le k \le m$) for the $q$-factorial of $m$ and

$$ \binom{m}{m_1,\ldots,m_k}_q = \frac{[m]_q!}{[m_1]_q! \cdots [m_k]_q!} $$

for the $q$-multinomial coefficient of the composition $(m_1,\ldots, m_k)$ of $m$. Let $S(m,q)$ be the sequence whose $d$th element, $d=0,1,\ldots$, is $$ S(m,q)(d)= \sum_{(m_1,\ldots,m_k) \in \mathrm{CO}(m)} (-1)^{m-k} \binom{m}{m_1,\ldots,m_k}_q q^{\sum \binom{m_i}{2}}\binom{q^{\binom{m}{2}- \sum \binom{m_i}{2}}}{d} $$ For $m=3$ we get the sequence with $d$th element \begin{equation*} S(3,q)(d) =\binom{3}{1,1,1}_q q^0\binom{q^3}{d} -2 \binom{3}{1,2}_q q^1\binom{q^2}{d} + \binom{3}{3}_q q^3 \binom{q^0}{d} \end{equation*} which for $q=2$ evaluates to \begin{equation*} S(3,2)=1, 64, 420, 1064, 1442, 1176, 588, 168, 21, 0, 0, \ldots \end{equation*} Note that $S(3,2)$ is a log-concave sequence. My question is: Is the sequence $S(m,q)$ always log-concave?

$S(m,q)(d)$, for $q$ a power of the prime $p$, is the number of $d$-subsets of $\mathrm{SL}_m(\mathbf{F}_q)$ generating a $p$-subgroup. These subsets form a contractible simplicial complex with reduced Euler characteristic $\sum_{d \ge 0} (-1)^d S(m,q)(d)=0$.

Let $\mathrm{CO}(m)$ be the set of all compositions of the positive integer $m$. By a composition of $m$, I mean a finite sequence $(m_1,\ldots,m_k)$ of positive integers with sum $\sum m_i=m$. There are $2^{m-1}$ compositions of $m$. The $4$ compositions of $3$ are

$$ \mathrm{CO}(3) =\{ (1,1,1), (1,2), (2,1), (3)\} $$

Let also $q$ be a prime power. Put $[k]_q = \frac{q^k-1}{q-1}$. Write $[m]_q! = \prod_{1 \le k \le m} [k]_q$ for the $q$-factorial of $m$ and

$$ \binom{m}{m_1,\ldots,m_k}_q = \frac{[m]_q!}{[m_1]_q! \cdots [m_k]_q!} $$

for the $q$-multinomial coefficient of the composition $(m_1,\ldots, m_k)$ of $m$. Let $S(m,q)$ be the sequence whose $d$th element, $d=0,1,\ldots$, is $$ S(m,q)(d)= \sum_{(m_1,\ldots,m_k) \in \mathrm{CO}(m)} (-1)^{m-k} \binom{m}{m_1,\ldots,m_k}_q q^{\sum \binom{m_i}{2}}\binom{q^{\binom{m}{2}- \sum \binom{m_i}{2}}}{d} $$ For $m=3$ we get the sequence with $d$th element \begin{equation*} S(3,q)(d) =\binom{3}{1,1,1}_q q^0\binom{q^3}{d} -2 \binom{3}{1,2}_q q^1\binom{q^2}{d} + \binom{3}{3}_q q^3 \binom{q^0}{d} \end{equation*} which for $q=2$ evaluates to \begin{equation*} S(3,2)=1, 64, 420, 1064, 1442, 1176, 588, 168, 21, 0, 0, \ldots \end{equation*} Note that $S(3,2)$ is a log-concave sequence. My question is: Is the sequence $S(m,q)$ always log-concave?

$S(m,q)(d)$, for $q$ a power of the prime $p$, is the number of $d$-subsets of $\mathrm{SL}_m(\mathbf{F}_q)$ generating a $p$-subgroup. These subsets form a contractible simplicial complex with reduced Euler characteristic $\sum_{d \ge 0} (-1)^d S(m,q)(d)=0$.

Let $\mathrm{CO}(m)$ be the set set of all compositions of the positive integer $m$. By a composition of $m$, I mean a finite sequence $(m_1,\ldots,m_k)$ of positive integers with sum $\sum m_i=m$. There are $2^{m-1}$ compositions of $m$. The $4$ compositions of $3$ are

$$ \mathrm{CO}(3) =\{ (1,1,1), (1,2), (2,1), (3)\} $$

Let also $q$ be a prime power. Put $[k]_q = \frac{q^k-1}{q-1}$. Write $[m]_q! = \prod [k]_q$ (product over $1 \le k \le m$) for the $q$-factorial of $m$ and

$$ \binom{m}{m_1,\ldots,m_k}_q = \frac{[m]_q!}{[m_1]_q! \cdots [m_k]_q!} $$

for the $q$-multinomial coefficient of the composition $(m_1,\ldots, m_k)$ of $m$. Let $S(m,q)$ be the sequence whose $d$th element, $d=0,1,\ldots$, is $$ S(m,q)(d)= \sum_{(m_1,\ldots,m_k) \in \mathrm{CO}(m)} (-1)^{m-k} \binom{m}{m_1,\ldots,m_k}_q q^{\sum \binom{m_i}{2}}\binom{q^{\binom{m}{2}- \sum \binom{m_i}{2}}}{d} $$ For $m=3$ we get the sequence with $d$th element \begin{equation*} S(3,q)(d) =\binom{3}{1,1,1}_q q^0\binom{q^3}{d} -2 \binom{3}{1,2}_q q^1\binom{q^2}{d} + \binom{3}{3}_q q^3 \binom{q^0}{d} \end{equation*} which for $q=2$ evaluates to \begin{equation*} S(3,2)=1, 64, 420, 1064, 1442, 1176, 588, 168, 21, 0, 0, \ldots \end{equation*} Note that $S(3,2)$ is a log-concave sequence. My question is: Is the sequence $S(m,q)$ always log-concave?

$S(m,q)(d)$, for $q$ a power of the prime $p$, is the number of $d$-subsets of $\mathrm{SL}_m(\mathbf{F}_q)$ generating a $p$-subgroup. These subsets form a contractible simplicial complex with reduced Euler characteristic $\sum_{d \ge 0} (-1)^d S(m,q)(d)=0$.

Let $\mathrm{CO}(m)$ be the set of all compositions of the positive integer $m$. By a composition of $m$, I mean a finite sequence $(m_1,\ldots,m_k)$ of positive integers with sum $\sum m_i=m$. There are $2^{m-1}$ compositions of $m$. The $4$ compositions of $3$ are

$$ \mathrm{CO}(3) =\{ (1,1,1), (1,2), (2,1), (3)\} $$

Let also $q$ be a prime power. Put $[k]_q = \frac{q^k-1}{q-1}$. Write $[m]_q! = \prod [k]_q$ (product over $1 \le k \le m$) for the $q$-factorial of $m$ and

$$ \binom{m}{m_1,\ldots,m_k}_q = \frac{[m]_q!}{[m_1]_q! \cdots [m_k]_q!} $$

for the $q$-multinomial coefficient of the composition $(m_1,\ldots, m_k)$ of $m$. Let $S(m,q)$ be the sequence whose $d$th element, $d=0,1,\ldots$, is $$ S(m,q)(d)= \sum_{(m_1,\ldots,m_k) \in \mathrm{CO}(m)} (-1)^{m-k} \binom{m}{m_1,\ldots,m_k}_q q^{\sum \binom{m_i}{2}}\binom{q^{\binom{m}{2}- \sum \binom{m_i}{2}}}{d} $$ For $m=3$ we get the sequence with $d$th element \begin{equation*} S(3,q)(d) =\binom{3}{1,1,1}_q q^0\binom{q^3}{d} -2 \binom{3}{1,2}_q q^1\binom{q^2}{d} + \binom{3}{3}_q q^3 \binom{q^0}{d} \end{equation*} which for $q=2$ evaluates to \begin{equation*} S(3,2)=1, 64, 420, 1064, 1442, 1176, 588, 168, 21, 0, 0, \ldots \end{equation*} Note that $S(3,2)$ is a log-concave sequence. My question is: Is the sequence $S(m,q)$ always log-concave?

$S(m,q)(d)$, for $q$ a power of the prime $p$, is the number of $d$-subsets of $\mathrm{SL}_m(\mathbf{F}_q)$ generating a $p$-subgroup. These subsets form a contractible simplicial complex with reduced Euler characteristic $\sum_{d \ge 0} (-1)^d S(m,q)(d)=0$.