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Fedor Petrov
Fedor Petrov
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Note that $-1$ is a root of the polynomial $q\to {n\choose k}_q$ of multiplicity $$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\ 0,&\text{otherwise}.\end{cases}$$ Thus ${n+1\choose 2}_{-1}$ is always non-zero, and ${n+1\choose 1}_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio $$h(n,k):=\frac{{n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q} {{n+1\choose 1}_q{n+1\choose 2}_q}\,\,\,\text{at}\,\,\,q=-1 $$ is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. Moreover, if $n$ and $k$ are both odd, then we get $h(n,k)=0$: the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root. Next, we have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus $$ a_n(-1)=\sum_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}= \sum_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\ =\sum_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}. $$

  • If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a_n(-1)=-a_n(-1)$ and thus $a_n(-1)=0$.
  • If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$. Thus in this case we also get $a_n(-1)=-a_n(-1)=0$.
  • Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t+1\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$ we get $$ a_{4s}(-1)=\frac2{2s}\sum_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s} $$ (because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that $$ \left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}= \frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit) $$ This is similar to Dixon's identity, and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained in this answer, so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to $$ f(x,y,z)=\prod_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i) $$ and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.
  • For $n=4s+1$ we again get $h(n,k)=0$ for odd $k$, and for the sum over even $k=2i$ we apply the identity ${n+1\choose k+1}_q={n+1\choose 1}_q{n\choose k}_q/{k+1\choose 1}_q$. This allows to cancel out the singular term ${n+1\choose 1}_q$, and applying ${k+1\choose 1}_{-1}=0$ for even $k$ we get $$ a_{4s+1}(-1)=\frac1{2s+1}\sum_{i=0}^{2s}(-1)^i{2s+1\choose i}{2s+1\choose i+1}{2s\choose i}. $$ This corresponds to Dyson's conjecture (now a theorem) $${\rm CT} \prod_{i\ne j}(1-x_i/x_j)^{a_i}=\frac{(\sum a_i)!}{\prod a_i!}$$ for $(a_1,a_2,a_3)=(s,s,s+1)$, and we get Roland Bacher's answer $$ a_{4s+1}(-1)=\frac{(-1)^s}{2s+1}{3s+1\choose s,s,s+1}. $$

Note that $-1$ is a root of the polynomial $q\to {n\choose k}_q$ of multiplicity $$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\ 0,&\text{otherwise}.\end{cases}$$ Thus ${n+1\choose 2}_{-1}$ is always non-zero, and ${n+1\choose 1}_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio $$h(n,k):=\frac{{n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q} {{n+1\choose 1}_q{n+1\choose 2}_q}\,\,\,\text{at}\,\,\,q=-1 $$ is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. Moreover, if $n$ and $k$ are both odd, then we get $h(n,k)=0$: the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root. Next, we have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus $$ a_n(-1)=\sum_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}= \sum_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\ =\sum_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}. $$

  • If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a_n(-1)=-a_n(-1)$ and thus $a_n(-1)=0$.
  • If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$. Thus in this case we also get $a_n(-1)=-a_n(-1)=0$.
  • Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t+1\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$ we get $$ a_{4s}(-1)=\frac2{2s}\sum_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s} $$ (because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that $$ \left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}= \frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit) $$ This is similar to Dixon's identity, and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained in this answer, so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to $$ f(x,y,z)=\prod_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i) $$ and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.
  • For $n=4s+1$ we again get $h(n,k)=0$ for odd $k$, and for the sum over even $k=2i$ we apply the identity ${n+1\choose k+1}_q={n+1\choose 1}_q{n\choose k}_q/{k+1\choose 1}_q$. This allows to cancel out the singular term ${n+1\choose 1}_q$, and applying ${k+1\choose 1}_{-1}=0$ for even $k$ we get $$ a_{4s+1}(-1)=\frac1{2s+1}\sum_{i=0}^{2s}(-1)^i{2s+1\choose i}{2s+1\choose i+1}{2s\choose i}. $$

Note that $-1$ is a root of the polynomial $q\to {n\choose k}_q$ of multiplicity $$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\ 0,&\text{otherwise}.\end{cases}$$ Thus ${n+1\choose 2}_{-1}$ is always non-zero, and ${n+1\choose 1}_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio $$h(n,k):=\frac{{n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q} {{n+1\choose 1}_q{n+1\choose 2}_q}\,\,\,\text{at}\,\,\,q=-1 $$ is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. Moreover, if $n$ and $k$ are both odd, then we get $h(n,k)=0$: the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root. Next, we have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus $$ a_n(-1)=\sum_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}= \sum_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\ =\sum_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}. $$

  • If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a_n(-1)=-a_n(-1)$ and thus $a_n(-1)=0$.
  • If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$. Thus in this case we also get $a_n(-1)=-a_n(-1)=0$.
  • Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t+1\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$ we get $$ a_{4s}(-1)=\frac2{2s}\sum_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s} $$ (because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that $$ \left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}= \frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit) $$ This is similar to Dixon's identity, and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained in this answer, so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to $$ f(x,y,z)=\prod_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i) $$ and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.
  • For $n=4s+1$ we again get $h(n,k)=0$ for odd $k$, and for the sum over even $k=2i$ we apply the identity ${n+1\choose k+1}_q={n+1\choose 1}_q{n\choose k}_q/{k+1\choose 1}_q$. This allows to cancel out the singular term ${n+1\choose 1}_q$, and applying ${k+1\choose 1}_{-1}=0$ for even $k$ we get $$ a_{4s+1}(-1)=\frac1{2s+1}\sum_{i=0}^{2s}(-1)^i{2s+1\choose i}{2s+1\choose i+1}{2s\choose i}. $$ This corresponds to Dyson's conjecture (now a theorem) $${\rm CT} \prod_{i\ne j}(1-x_i/x_j)^{a_i}=\frac{(\sum a_i)!}{\prod a_i!}$$ for $(a_1,a_2,a_3)=(s,s,s+1)$, and we get Roland Bacher's answer $$ a_{4s+1}(-1)=\frac{(-1)^s}{2s+1}{3s+1\choose s,s,s+1}. $$
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Fedor Petrov
Fedor Petrov
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Note that $-1$ is a root of the polynomial $q\to {n\choose k}_q$ of multiplicity $$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\ 0,&\text{otherwise}.\end{cases}$$ Thus ${n+1\choose 2}_{-1}$ is always non-zero, and ${n+1\choose 1}_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio $$h(n,k):=\frac{{n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q} {{n+1\choose 1}_q{n+1\choose 2}_q}\,\,\,\text{at}\,\,\,q=-1 $$ is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. Moreover, if $n$ and $k$ are both odd, then we get $h(n,k)=0$: the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root. Next, we have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus $$ a_n(-1)=\sum_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}= \sum_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\ =\sum_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}. $$

  • If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a_n(-1)=-a_n(-1)$ and thus $a_n(-1)=0$.
  • If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$. Thus in this case we also get $a_n(-1)=-a_n(-1)=0$.
  • Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t+1\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$ we get $$ a_{4s}(-1)=\frac2{2s}\sum_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s} $$ (because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that $$ \left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}= \frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit) $$ This is similar to Dixon's identity, and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained in this answer, so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to $$ f(x,y,z)=\prod_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i) $$ and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.
  • For $n=4s+1$ we again get $h(n,k)=0$ for odd $k$, and for the sum over even $k=2i$ we apply the identity ${n+1\choose k+1}_q={n+1\choose 1}_q{n\choose k}_q/{k+1\choose 1}_q$. This allows to cancel out the singular term ${n+1\choose 1}_q$, and applying ${k+1\choose 1}_{-1}=0$ for even $k$ we get $$ a_{4s+1}(-1)=\frac1{2s+1}\sum_{i=0}^{2s}(-1)^i{2s+1\choose i}{2s+1\choose i+1}{2s\choose i}. $$

Most probably the remaining case $n=4s+1$ is quite analogous.

Note that $-1$ is a root of the polynomial $q\to {n\choose k}_q$ of multiplicity $$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\ 0,&\text{otherwise}.\end{cases}$$ Thus ${n+1\choose 2}_{-1}$ is always non-zero, and ${n+1\choose 1}_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio $$h(n,k):=\frac{{n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q} {{n+1\choose 1}_q{n+1\choose 2}_q}\,\,\,\text{at}\,\,\,q=-1 $$ is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. Moreover, if $n$ and $k$ are both odd, then we get $h(n,k)=0$: the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root. Next, we have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus $$ a_n(-1)=\sum_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}= \sum_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\ =\sum_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}. $$

  • If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a_n(-1)=-a_n(-1)$ and thus $a_n(-1)=0$.
  • If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$. Thus in this case we also get $a_n(-1)=-a_n(-1)=0$.
  • Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t+1\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$ we get $$ a_{4s}(-1)=\frac2{2s}\sum_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s} $$ (because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that $$ \left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}= \frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit) $$ This is similar to Dixon's identity, and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained in this answer, so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to $$ f(x,y,z)=\prod_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i) $$ and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.

Most probably the remaining case $n=4s+1$ is quite analogous.

Note that $-1$ is a root of the polynomial $q\to {n\choose k}_q$ of multiplicity $$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\ 0,&\text{otherwise}.\end{cases}$$ Thus ${n+1\choose 2}_{-1}$ is always non-zero, and ${n+1\choose 1}_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio $$h(n,k):=\frac{{n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q} {{n+1\choose 1}_q{n+1\choose 2}_q}\,\,\,\text{at}\,\,\,q=-1 $$ is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. Moreover, if $n$ and $k$ are both odd, then we get $h(n,k)=0$: the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root. Next, we have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus $$ a_n(-1)=\sum_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}= \sum_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\ =\sum_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}. $$

  • If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a_n(-1)=-a_n(-1)$ and thus $a_n(-1)=0$.
  • If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$. Thus in this case we also get $a_n(-1)=-a_n(-1)=0$.
  • Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t+1\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$ we get $$ a_{4s}(-1)=\frac2{2s}\sum_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s} $$ (because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that $$ \left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}= \frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit) $$ This is similar to Dixon's identity, and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained in this answer, so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to $$ f(x,y,z)=\prod_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i) $$ and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.
  • For $n=4s+1$ we again get $h(n,k)=0$ for odd $k$, and for the sum over even $k=2i$ we apply the identity ${n+1\choose k+1}_q={n+1\choose 1}_q{n\choose k}_q/{k+1\choose 1}_q$. This allows to cancel out the singular term ${n+1\choose 1}_q$, and applying ${k+1\choose 1}_{-1}=0$ for even $k$ we get $$ a_{4s+1}(-1)=\frac1{2s+1}\sum_{i=0}^{2s}(-1)^i{2s+1\choose i}{2s+1\choose i+1}{2s\choose i}. $$
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Fedor Petrov
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Note that $-1$ is a root of the polynomial $q\to {n\choose k}_q$ of multiplicity $$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\ 0,&\text{otherwise}.\end{cases}$$ Thus ${n+1\choose 2}_{-1}$ is always non-zero, and ${n+1\choose 1}_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio $$h(n,k):=\frac{{n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q} {{n+1\choose 1}_q{n+1\choose 2}_q}\,\,\,\text{at}\,\,\,q=-1 $$ is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. WeMoreover, if $n$ and $k$ are both odd, then we get $h(n,k)=0$: the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root. Next, we have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus $$ a_n(-1)=\sum_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}= \sum_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\ =\sum_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}. $$

  • If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a_n(-1)=-a_n(-1)$ and thus $a_n(-1)=0$.
  • If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$ (the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root). Thus in this case we also get $a_n(-1)=-a_n(-1)=0$.
  • Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$$${2t+1\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$ we get $$ a_{4s}(-1)=\frac2{2s}\sum_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s} $$ (because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that $$ \left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}= \frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit) $$ This is similar to Dixon's identity, and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained in this answer, so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to $$ f(x,y,z)=\prod_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i) $$ and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.

Most probably the remaining case $n=4s+1$ is quite analogous.

Note that $-1$ is a root of the polynomial $q\to {n\choose k}_q$ of multiplicity $$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\ 0,&\text{otherwise}.\end{cases}$$ Thus ${n+1\choose 2}_{-1}$ is always non-zero, and ${n+1\choose 1}_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio $$h(n,k):=\frac{{n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q} {{n+1\choose 1}_q{n+1\choose 2}_q}\,\,\,\text{at}\,\,\,q=-1 $$ is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. We have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus $$ a_n(-1)=\sum_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}= \sum_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\ =\sum_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}. $$

  • If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a_n(-1)=-a_n(-1)$ and thus $a_n(-1)=0$.
  • If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$ (the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root). Thus in this case we also get $a_n(-1)=-a_n(-1)=0$.
  • Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$ we get $$ a_{4s}(-1)=\frac2{2s}\sum_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s} $$ (because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that $$ \left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}= \frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit) $$ This is similar to Dixon's identity, and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained in this answer, so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to $$ f(x,y,z)=\prod_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i) $$ and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.

Most probably the remaining case $n=4s+1$ is quite analogous.

Note that $-1$ is a root of the polynomial $q\to {n\choose k}_q$ of multiplicity $$\lfloor n/2\rfloor-\lfloor k/2\rfloor-\lfloor (n-k)/2\rfloor=\begin{cases}1,&\text{if}\,\,n\,\,\text{is even and}\,\,k\,\,\text{is odd}\\ 0,&\text{otherwise}.\end{cases}$$ Thus ${n+1\choose 2}_{-1}$ is always non-zero, and ${n+1\choose 1}_{-1}$ is 0 when $n$ is odd. But in the latter case the ratio $$h(n,k):=\frac{{n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q} {{n+1\choose 1}_q{n+1\choose 2}_q}\,\,\,\text{at}\,\,\,q=-1 $$ is well-defined, since at least one of the numbers $k,k+1$, $k+2$ is odd. Moreover, if $n$ and $k$ are both odd, then we get $h(n,k)=0$: the multiplicity of $q=-1$ as a root of ${n+1\choose k}_q {n+1\choose k+1}_q{n+1\choose k+2}_q$ equals 2 in this case, while for ${n+1\choose 1}_q {n+1\choose 2}_q$ $q=-1$ is a simple root. Next, we have $h(n,k)=h(n,n-k-1)$ (this holds not only at $q=-1$, but for corresponding rational functions in $q$). Thus $$ a_n(-1)=\sum_{k=0}^{n-1}h(n,k)(-1)^{k(k+1)/2}= \sum_{k=0}^{n-1}h(n,n-k-1)(-1)^{(n-k-1)(n-k)/2}\\ =\sum_{k=0}^{n-1}h(n,k)(-1)^{n(n-1)/2+k(k+1)/2-nk}. $$

  • If $n=4s+2$, then $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ always have different parity, so we get $a_n(-1)=-a_n(-1)$ and thus $a_n(-1)=0$.
  • If $n=4s+3$, then for even $k$ the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have different parity, while for odd $k$ we have $h(n,k)=0$. Thus in this case we also get $a_n(-1)=-a_n(-1)=0$.
  • Let $n=4s$. Then the exponents $n(n-1)/2+k(k+1)/2-nk$ and $k(k+1)/2$ have the same parity, thus the terms in the sum for $a_n(-1)$ which correspond to $k$ and to $n-k-1$ are equal. Since exactly one of numbers $k,n-k-1$ is even, the sum over $k$ is twice the sum over even $k=2i$. Using the formula $${2t+1\choose m}_{-1}={t\choose \lfloor m/2\rfloor}$$ we get $$ a_{4s}(-1)=\frac2{2s}\sum_{i=0}^{2s-1}{2s\choose i}{2s\choose i}{2s\choose i+1} (-1)^i\\=-\frac1s\left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s} $$ (because, say, if we take the monomials $x^iy^{2s-i}$, $z^{2s-i}y^i$ and $z^{i+1}x^{2s-i-1}$ from the binomials $(x-y)^{2s}$, $(y-z)^{2s}$, $(x-z)^{2s}$ respectively, the coefficient is $(-1)^{i+1}{2s\choose i}{2s\choose i}{2s\choose i+1}$). So, to confirm the guess of Brian Hopkins (with alternating sign) we should prove that $$ \left[x^{2s-1}y^{2s}z^{2s+1}\right](x-y)^{2s}(x-z)^{2s}(y-z)^{2s}= \frac{(-1)^{s+1}s}{2s+1}{3s\choose s,s,s}.\quad (\spadesuit) $$ This is similar to Dixon's identity, and you may adapt your favourite proof of Dixon here. Mine favourite uses Combinatorial Nullstellensatz formula, as explained in this answer, so let me do the same for $(\spadesuit)$. We change the polynomial adding lower degree terms to $$ f(x,y,z)=\prod_{i=-s}^{s-1} (x-y-i)(y-z-i)(x-z-i) $$ and look at values of $f$ when $x\in \{0,1,2,\ldots,2s-1\}$, $y\in \{0,1,\ldots,2s\}$, $z\in \{0,1,\ldots,2s+1\}$. Note that if $f(x,y,z)\ne 0$, then $|x-y|,|x-z|,|y-z|$ are at least $s$, thus either $\{x,y,z\}=\{0,s,2s\}$, or $z=2s+1$. In the first case, since $x\ne 2s$, we get either $x-y+s=0$ or $x-z+s=0$, so $f(x,y,z)=0$, a contradiction. Thus $z=2s+1$. Since $x-z+s\ne 0$, $y-z+s\ne 0$, we get $x,y\leqslant s$. Therefore $x=s$, $y=0$. So, RHS of CN formula has exactly one non-zero term, which routinely simplifies to the above answer $(\spadesuit)$.

Most probably the remaining case $n=4s+1$ is quite analogous.

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Fedor Petrov
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Fedor Petrov
Fedor Petrov
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Fedor Petrov
Fedor Petrov
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