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question 26336.

question 26336.

question 26336.

question 26336.

So we now turn to the case $ m> k-1\geq i\geq m+1-k\geq 0$. Then $$\text{frac1}(m,k,i)/(6m(m-1)C(i-1))$$ equals $$\frac{i! (2 k-2)! m! (-2 i+2 m-2)! (-2 k+2 m+1)!}{(2 i)! (k-1)! (-i+k-1)! (m-i)! (-2 i+2 m-1)! (-k+m+1)! (2 m-2 k)! (i+k-m-1)!} .$$

If $\text{frac2}(m,k,i)$ is nonzero then $m> k-1\geq i\geq m+1-k>0$ and
$\text{frac2}(m,k,i)/(6m(m-1))$ equals $$\frac{-(2 k-2)! (m-2)!}{i! (k-1)! (-i+k-1)! (m-i)! (m-k)! (i+k-m-1)!}.$$ This can be treated like the previous case. We eliminate $k$, $m$, $i$ in that order and take $n\geq6$ as bound where all $8\times 1278$ Floors are stable.

So we now turn to the case $ m> k-1\geq i\geq m+1-k\geq 0$. Then $$\text{frac1}(m,k,i)/(6m(m-1)C(i-1))$$ equals $$\small\frac{i! (2 k-2)! m! (-2 i+2 m-2)! (-2 k+2 m+1)!}{(2 i)! (k-1)! (-i+k-1)! (m-i)! (-2 i+2 m-1)! (-k+m+1)! (2 m-2 k)! (i+k-m-1)!} $$

If $\text{frac2}(m,k,i)$ is nonzero then $m> k-1\geq i\geq m+1-k>0$ and
$\text{frac2}(m,k,i)/(6m(m-1))$ equals

$$\frac{-(2 k-2)! (m-2)!}{i! (k-1)! (-i+k-1)! (m-i)! (m-k)! (i+k-m-1)!}$$

This can be treated like the previous case. We eliminate $k$, $m$, $i$ in that order and take $n\geq6$ as bound where all $8\times 1278$ Floors are stable.

$$ \frac{-(2 i-2)! (2 m)! (-2 i+2 m-2)!}{2 (i!)^2 (2 i-1)! ((m-i)!)^2 (-2 i+2 m-1)!}$$ and we must show it takes values in $\mathbb{Z}[1/2]$.

So we now turn to the case $ m> k-1\geq i\geq m+1-k\geq 0$. Then \begin{align*}&\text{frac1}(m,k,i)/(6m(m-1)C(i-1))=\\&\frac{i! (2 k-2)! m! (-2 i+2 m-2)! (-2 k+2 m+1)!}{(2 i)! (k-1)! (-i+k-1)! (m-i)! (-2 i+2 m-1)! (-k+m+1)! (2 m-2 k)! (i+k-m-1)!} \end{align*}

$$\small \frac{-(2 i-2)! (2 m)! (-2 i+2 m-2)!}{2 (i!)^2 (2 i-1)! ((m-i)!)^2 (-2 i+2 m-1)!}$$ and we must show it takes values in $\mathbb{Z}[1/2]$.

So we now turn to the case $ m> k-1\geq i\geq m+1-k\geq 0$. Then $$\text{frac1}(m,k,i)/(6m(m-1)C(i-1))$$ equals $$\frac{i! (2 k-2)! m! (-2 i+2 m-2)! (-2 k+2 m+1)!}{(2 i)! (k-1)! (-i+k-1)! (m-i)! (-2 i+2 m-1)! (-k+m+1)! (2 m-2 k)! (i+k-m-1)!} .$$