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When does the sum $\sum_a^b k$ of $b-a+1$ consecutive positive integers divide the product $\prod_a^b k$?

I know that the sum of the first $n$ natural numbers divides the product of the first $n$ natural numbers for all $n$ except when $n=p-1$, $p$ being any odd prime. But what about the more general case?

I tried a number of the particular cases, but from there it seems not clear how the general case would look like.

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    $\begingroup$ You can rewrite it as asking when $\frac{(b+a)(b-a+1)}{2}$ divides $\frac{b!}{(a-1)!}$ and then maybe make some progress with Legendre's formula which will turn it into asking when this inequality is satisfied for each prime divisor $p$: $$\frac{b-a+1+s_p(a-1)-s_p(b)}{p-1} \ge v_p(b+a)+v_p(b-a+1)-v_p(2)$$ $\endgroup$
    – Merosity
    Jun 26 at 17:28
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    $\begingroup$ Do you have any reason to believe that this would have an easy characterization? $\endgroup$
    – jjagmath
    Jun 26 at 19:24
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    $\begingroup$ @ Merosity, would you do that? $\endgroup$
    – Ashiq
    Sep 17 at 3:35
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This is a partial answer. Let $$f(a,b):=\bigg(\prod_{k=a}^bk\bigg)/\bigg(\sum_{k=a}^bk\bigg)=\frac{2\times b(b-1)\cdots a}{(b-a+1)(a+b)}.$$

This answer proves the following claims :

Claim 1 :

  • $f(2,b)$ is an integer for all $b$ except when $b=4,7,p-2$ where $p\geqslant 5$ is an odd prime.

  • $f(3,b)$ is an integer for all $b$ except when $b=7,p-3$ where $p\geqslant 7$ is an odd prime.

  • $f(4,b)$ is an integer for all $b$ except when $b=5,10,p-4$ where $p\geqslant 11$ is an odd prime.

Claim 2 : If $(b-a)$ is even, then $f(a,b)$ is an integer for all $(a,b)$ except when $b-a+1$ is a prime number satisfying $a+b\equiv 0\pmod{b-a+1}$.

Claim 3 : If $(b-a)$ is odd, then it is necessary that $(a+b)\mid ((b-a)!!)^2$, and it is sufficient that $(a+b)\mid \bigg(\dfrac{b-a-1}{2}\bigg)!(b-a)!!$.

Claim 4 :

  • $f(a,a+1)$ is not an integer.

  • $f(a,a+3)$ is an integer if and only if $a=3$.

  • $f(a,a+5)$ is an integer if and only if $a=5,10,35$.

Claim 5 : If $d\ (\geqslant 3)$ is a proper divisor of an odd composite number $a+b$ with $b\geqslant \dfrac{2d+1}{d-1}a+d$, then $f(a,b)$ is an integer.


Claim 1 :

  • $f(2,b)$ is an integer for all $b$ except when $b=4,7,p-2$ where $p\geqslant 5$ is an odd prime.

  • $f(3,b)$ is an integer for all $b$ except when $b=7,p-3$ where $p\geqslant 7$ is an odd prime.

  • $f(4,b)$ is an integer for all $b$ except when $b=5,10,p-4$ where $p\geqslant 11$ is an odd prime.

Proof :

  • One has $f(2,b)=2b\times\dfrac{(b-2)!}{b+2}$ with $f(2,4)=\dfrac 83$ and $f(2,7)=186+\dfrac{2}{3}$. If $b+2=2M\ (\geqslant 8)$ is even, then $b-2=M+(M-4)\geqslant M$. Since the product of $M$ consecutive integers is divisible by $M$, $(b-2)!$ is divisibe by $M$. If $b+2$ is an odd prime, then $f(2,b)$ is not an integer. If $b+2=PQ$ is an odd composite number where $3\leqslant P\leqslant Q$ and $(P,Q)\not=(3,3)$, then $b-2-(P+Q)=(P-1)(Q-1)-5\geqslant 0$, i.e. $b-2\geqslant P+Q$, so $(b-2)!$ is divisible by $PQ$.

  • One has $f(3,b)=b(b-1)\times\dfrac{(b-3)!}{b+3}$ with $f(3,5)=5,f(3,6)=20$ and $f(3,7)=100+\dfrac 45$. If $b+3=2M\ (\geqslant 12)$ is even, then $b-3=M+(M-6)\geqslant M$, so $(b-3)!$ is divisible by $M$. If $b+3$ is an odd prime, then $f(3,b)$ is not an integer. If $b+3=PQ$ is an odd composite number where $3\leqslant P\leqslant Q$ and $(P,Q)\not=(3,3)$, then $b-3-(P+Q)=(P-1)(Q-1)-7\geqslant 0$, i.e. $b-3\geqslant P+Q$, so $(b-3)!$ is divisible by $PQ$.

  • One has $f(4,b)=\dfrac{b(b-1)(b-2)}{3}\times\dfrac{(b-4)!}{b+4}$ with $f(4,5)=\dfrac{20}{9}$, $f(4,6)=8,$ $f(4,8)=224,$ $f(4,10)=12342+\dfrac{6}{7}$ and $f(4,11)=110880$. If $b+4=2M\ (\geqslant 16)$ is even, then $b-4=M+(M-8)\geqslant M$, so $(b-4)!$ is divisible by $M$. If $b+4$ is an odd prime, then $f(4,b)$ is not an integer. If $b+4=PQ$ is an odd composite number where $3\leqslant P\leqslant Q$ and $(P,Q)\not=(3,3),(3,5)$, then $b-4-(P+Q)=(P-1)(Q-1)-9\geqslant 0$, i.e. $b-4\geqslant P+Q$, so $(b-4)!$ is divisible by $PQ$.$\quad\blacksquare$


Claim 2 : If $(b-a)$ is even, then $f(a,b)$ is an integer for all $(a,b)$ except when $b-a+1$ is a prime number satisfying $a+b\equiv 0\pmod{b-a+1}$.

Proof : If $b-a=2m$ where $m$ is a positive integer, then $$f(a,a+2m)=\frac{\overbrace{(a+2m)\cdots (a+m+1)}^{m\ \text{consecutive integers}}\ \overbrace{(a+m-1)\cdots a}^{m\ \text{consecutive integers}}}{2m+1}$$Note here that for every pair $(s,t)$ satisfying $0\leqslant s\lt t\leqslant 2m$, $a+s\not\equiv a+t\pmod{2m+1}$.

  • If $a+m\not\equiv 0\pmod{2m+1}$, then $f(a,a+2m)$ is an integer.

  • If $a+m\equiv 0\pmod{2m+1}$, and $2m+1$ is a prime number, then $f(a,a+2m)$ is not an integer.

  • If $2m+1$ is a composite number, then there are integers $P,Q$ such that $2m+1=PQ$ and $3\leqslant P\leqslant Q$ for which $m-P\geqslant m-Q=\dfrac{Q(P-2)-1}{2}\geqslant 0$ holds, so $f(a,a+2m)$ is an integer.$\quad\blacksquare$


Claim 3 : If $(b-a)$ is odd, then it is necessary that $(a+b)\mid ((b-a)!!)^2$, and it is sufficient that $(a+b)\mid \bigg(\dfrac{b-a-1}{2}\bigg)!(b-a)!!$.

Proof : If $b-a=2m+1$ where $m$ is a non-negative integer, then one has $$f(a,a+2m+1)=\frac{(a+2m+1)(a+2m)\cdots a}{(m+1)(2a+2m+1)}$$

Multiplying the both sides by $2^{2m+2}(m+1)$ gives $$2^{2m+2}(m+1)f(a,a+2m+1)=\frac{(2a+4m+2)(2a+4m)\cdots 2a}{2a+2m+1}$$ Now since $$\begin{align}(\text{the numerator})&\equiv (2m+1)(2m-1)\cdots (-2m-1)\pmod{2a+2m+1} \\\\&\equiv (-1)^{m+1}((2m+1)!!)^2\pmod{2a+2m+1}\end{align}$$ So, it is necessary that $(2a+2m+1)\mid ((2m+1)!!)^2$.

Also, since the product of $N$ consecutive integers is divisible by $N!$, there is an integer $k$ such that $$f(a,a+2m+1)=\frac{k\cdot (2m+2)!}{(m+1)(2a+2m+1)}=\frac{2^{m+1}k\cdot m!(2m+1)!!}{2a+2m+1}$$from which one can say that it is sufficient that $(2a+2m+1)\mid m!(2m+1)!!$.$\quad\blacksquare$


Claim 4 :

  • $f(a,a+1)$ is not an integer.

  • $f(a,a+3)$ is an integer if and only if $a=3$.

  • $f(a,a+5)$ is an integer if and only if $a=5,10,35$.

Proof :

  • From Claim 3, it is necessary that $2a+1\mid 1$ which is impossible.

  • From Claim 3, it is necessary that $2a+3\mid 3^2$ implying $a=3$ which is sufficient.

  • From Claim 3, it is necessary that $2a+5\mid 3^2\cdot 5^2$ implying $a=2,5,10,20,35,110$, and only $a=5,10,35$ are sufficient.$\quad\blacksquare$


Claim 5 : If $d\ (\geqslant 3)$ is a proper divisor of an odd composite number $a+b$ with $b\geqslant \dfrac{2d+1}{d-1}a+d$, then $f(a,b)$ is an integer.

Proof : If $b-2a+1\geqslant d+\dfrac{a+b}{d}$, i.e. $b\geqslant \dfrac{2d+1}{d-1}a+d$, then since $$f(a,b)=\frac{2\times b(b-1)\cdots (b-a+2)\overbrace{(b-a)(b-a-1)\cdots a}^{(b-2a+1)\ \text{consecutive integers}}}{a+b}$$ one can see that $a+b=d\cdot\dfrac{a+b}{d}$ divides $(b-a)(b-a-1)\cdots a$.$\quad\blacksquare$

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    $\begingroup$ That is quite an effort. $\endgroup$
    – Ashiq
    Sep 19 at 15:50
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    $\begingroup$ Although, as you know, your answer is not complete. The effort is appreciable. I am awarding this bounty to you. $\endgroup$
    – Ashiq
    Sep 22 at 4:07

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