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In the $k=4$ case, $$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!} = \frac{(4n+6)!}{n!(n+2)(n^2+5n+5)}.$$

For a lot of (likely infinitely many) values of $n$, $n^2+5n+5$ is a prime larger than $4n+6$, and so the expression is not an integer for those $n$.

A similar situation arises for larger $k$.

(In the $k=3$ case, we get $$ \frac{(3n+3)!}{n!(n+2)^2} $$ and since $2(n+2)<3n+3$, this simplifies to an integer for $n\ge1$. Similarly, with $k=2$, we have $$ \frac{(2n+1)!}{n!(n+2)} $$ which again simplifiees to an integer for $n\ge1$.)

In the $k=4$ case, $$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!} = \frac{(4n+6)!}{n!(n+2)(n^2+5n+5)}.$$

For a lot of (likely infinitely many) values of $n$, $n^2+5n+5$ is a prime larger than $4n+6$, and so the expression is not an integer for those $n$.

(In the $k=3$ case, we get $$ \frac{(3n+3)!}{n!(n+2)^2} $$ and since $2(n+2)<3n+3$, this simplifies to an integer for $n\ge1$. Similarly, with $k=2$, we have $$ \frac{(2n+1)!}{n!(n+2)} $$ which again simplifies to an integer for $n\ge1$.)

In the $k=4$ case, $$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!} = \frac{(4n+6)!}{n!(n+2)(n^2+5n+5)}.$$

For a lot of (likely infinitely many) values of $n$, $n^2+5n+5$ is a prime larger than $4n+6$, and so the expression is not an integer for those $n$.

A similar situation arises for larger $k$.

In the $k=4$ case, $$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!} = \frac{(4n+6)!}{n!(n+2)(n^2+5n+5)}.$$

For a lot of (likely infinitely many) values of $n$, $n^2+5n+5$ is a prime larger than $4n+6$, and so the expression is not an integer for those $n$.

A similar situation arises for larger $k$.

(In the $k=3$ case, we get $$ \frac{(3n+3)!}{n!(n+2)^2} $$ and since $2(n+2)<3n+3$, this simplifies to an integer for $n\ge1$. Similarly, with $k=2$, we have $$ \frac{(2n+1)!}{n!(n+2)} $$ which again simplifiees to an integer for $n\ge1$.)

In the $k=4$ case, $$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!} = \frac{(4n+6)!}{n!(n+2)(n^2+5n+5)}.$$

For a lot of value of $n$, $n^2+5n+5$ is a prime larger than $4n+6$, and so the expression is not an integer for those $n$.

A similar situation arises for larger $k$.

In the $k=4$ case, $$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!} = \frac{(4n+6)!}{n!(n+2)(n^2+5n+5)}.$$

For a lot of (likely infinitely many) values of $n$, $n^2+5n+5$ is a prime larger than $4n+6$, and so the expression is not an integer for those $n$.

A similar situation arises for larger $k$.