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The coefficients of the multinomial formula are integers where the numerator is the factorial of a sum of integers and the denominator is the product of the factorials of those same integers.

Suppose instead of taking a product of those factorials in the denominator we took the sum of those factorials. When would that expression be an integer?

Based on some calculations, I suspect, but I cannot prove in general, that if the integers are consecutive forming the sum, then there exists a non-negative integer $m$ depending on the number of integers $k$ in the sum such that if $n \ge m$ then the expression is always an integer.

In other words, there exists $m$ depending on $k$ such that for all $n \ge m$ the following is always an integer:

$$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!}$$

For $k = 1, m = 0$. For $k = 2, m = 1$. Based on some calculations, I conjecture, but I have no proof, for $k = 3, m = 1$ and for $k = 4, m = 4$.

Is this type of problem familiar to anyone? Perhaps someone has a solution to it or suggestions where I could go for more information.

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    $\begingroup$ For $k=4$, I find your expression equal to $7805560609566720000/41$ when $n=4$, to be an integer with $n=5$, but then not an integer for $n=6,7,8,9$. The sequence of for which $n$ for which it is an integer begins $n=5,10,12,15,16,21,23,27,31,32,34,38,40,41,45,47,50,59,60,61,69,72,73,76,77,78,79,82,86,90,96,100,102,103,106,108,109,114,115,117,121,126,127,131,135,137,140,141,145,148,149,155,163,165,166,170,176,177,185,187,192,195,196,...$. (Not in the OEIS). $\endgroup$ Commented Nov 2, 2016 at 21:13
  • $\begingroup$ For $k=4$ I get the numerator being $(4+5+6+7)! = 22! = 1124000727777607680000$ and the denominator being $4! + 5! + 6! +7! = 24 + 120 + 720 + 5040 = 5904$. I used Python's is_integer() to evaluate whether it was an integer or not and it said it was. $\endgroup$ Commented Nov 5, 2016 at 1:31
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    $\begingroup$ It's not an integer. The denominator has a factor of 41 (since $5904=41\cdot144$), and the numerator does not, so the ratio is not an integer. $\endgroup$ Commented Nov 5, 2016 at 1:50
  • $\begingroup$ Right. I just factored $5904$ as well to check the is_integer() operation. I see now why you put the fraction in your comment with $41$ in the denominator. $\endgroup$ Commented Nov 5, 2016 at 2:20

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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 $n$.

(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 $n\ge1$.)

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  • $\begingroup$ I agree with your arguments for $k \in (2,3,4)$. I wonder whether the statement, "A similar situation arises for larger k", is true. The reason I have my doubts is for $k=3$ you showed there was a factorization of the denominator with factors being small enough to be in the numerator. How do we know that does not happen for some larger $k$? $\endgroup$ Commented Nov 5, 2016 at 1:56
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    $\begingroup$ Since I only specified explicitly the $k = 3$ and $k = 4$ values and not something more general, I marked this as a complete answer. I am still wondering about a more general case. $\endgroup$ Commented Nov 5, 2016 at 2:25
  • $\begingroup$ I removed the claim about $k>4$. Cheers! $\endgroup$ Commented Nov 5, 2016 at 2:39

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