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$\bullet\ \textbf{Question}$

One can find equivalent products of consecutive integers such as $$8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14=63\cdot64\cdot65\cdot66.$$ Other solutions of this have been posted already (Equal products of consecutive integers). But the implied meaning of 'trivial' solutions in other posts seems unsatisfactory -- I think I can convince you. There appear to be exactly $4$ non-trivial solutions after adjusting our vocabulary. One of which is the previous equation. I'm curious; are their any other non-trivial solutions?

$\bullet\ \textbf{Triviality}$

Firstly, let's use the notation $a^{(b)}=\prod_{k=0}^{b-1}(a+k)$ to denote rising factorial. The previous equation then becomes $8^{(7)}=63^{(4)}$.

All solutions come in pairs -- one overlapping and one disjoint. For example, if we take the solution $$2^{(5)}=2\cdot3\cdot4\cdot5\cdot6=8\cdot9\cdot10=8^{(3)},$$ which we will call disjoint because no integer appears on both sides, we can create an overlapping solution multiplying both sides by $7$: $$2^{(6)}=2\cdot3\cdot4\cdot5\cdot6\cdot\textbf{7}=\textbf{7}\cdot8\cdot9\cdot10=7^{(4)}.$$ This duality means that boring solutions like $2^{(2)}=2\cdot3=6^{(1)}$ can be disguised as interesting by transforming them from disjoint to overlapping: $$2^{(2)}=6^{(1)}\quad\rightarrow\quad 2^{(4)}=4^{(3)}.$$ For example, two answers on a previous post gave $3^{(9)}=5^{(8)}$ as an "interesting" solution. But this is simply the overlapping dual of $3\cdot 4 = 12$.

In general, if $a^{(b)}=c^{(d)}$ is a solution, then so is its dual: $$a^{(c-a)}=(a+b)^{(c+d-a-b)}.$$

So here we define a trivial solution either 1) having overlap or 2) having only one term on one side (previous inquiries seemed to only have considered #2).

$\bullet\ \textbf{Non-trivial Solutions}$

There are $4$ known non-trivial solutions under this definition: $$2^{(5)}=8^{(3)},\quad 5^{(3)}=14^{(2)},\quad 19^{(4)}=55^{(3)},\quad\text{and}\quad 8^{(7)}=63^{(4)}.$$ I ran a computer search on equivalent $a^{(b)}$ for $a<100,000$ and $b<50$ and found nothing new (https://repl.it/@onnomc/EqualProductsOfConsecutiveIntegers).

$\bullet\ \textbf{Proof Attempts}$

One attempt is to note $a^{(b)}=\frac{(a+b-1)!}{(a-1)!}$ and that therefore solutions to $a^{(b)}=c^{(d)}$ are also solutions to $A!B!=C!D!$. But apparently even $A!B!=C!$ has not yet been solved (On the factorial equations $A! B! =C!$ and $A!B!C! = D!$).

I thought also to use Stirling's approximation since $$(a+1)^{(b)}=\frac{(a+b)!}{a!}\approx \Big(\frac{a+b}{a}\Big)^{a+1/2}\Big(\frac{a+b}{e}\Big)^b$$ and ... well, hope for the best. I made no progress here.

I tried also bounding the number of distinct primes dividing $a^{(b)}$. We can bound $c$ from below since $$a^b<a^{(b)}=c^{(d)}<(c+d-1)^d\quad\Rightarrow\quad c>a^{b/d}-d+1.$$ And of course, the non-overlapping criteria gives $c\ge a+b$. Thus if the prime bound rises fast enough, we can show that we are trying to fit too many prime divisors into $c(c+1)...(c+d-1)$. The furthest I got down this trail was finding $r$ consecutive integers with a product divisible by at most $r$ primes. For example, the largest solution for $r=5$ is $c=24$ since $$24\cdot25\cdot26\cdot27\cdot28=2^6\cdot3^4\cdot5^2\cdot7\cdot13.$$ The largest solution for $r=42$ seems to be $c=2175$. But I cannot prove these are the largest such integers -- only that they appear so empirically (https://repl.it/@onnomc/ConsecutiveIntegersOfFewPrimes). Interestingly, the case $r=2$ yields $c=+\infty$ if there are infinitely many Mersenne primes.

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  • 2
    $\begingroup$ One of the first issues that strikes me is that the larger string of consecutive numbers can contain no primes. This also seems relevant to the $A!B!=C!D!$ formulation, where the largest prime on the LHS must be present in the RHS as well. So the search is (primarily?) for strings of consecutive numbers that can all be formed from small prime factors, and as the absolute magnitude of the numbers being examined increases, I would expect such strings to become increasingly rare, or very short. $\endgroup$ – Keith Backman Sep 6 '19 at 19:46
  • $\begingroup$ See also mathoverflow.net/questions/355829/… $\endgroup$ – Dale Mar 27 at 2:36

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