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Summary: What are the non-trivial solutions to the question: Find two sequences of consecutive integers whose products are the same. There are four known solutions, all of thich consist of small integers, and there is no clear pattern connecting them. Given the small number of known solutions and the lack of any clear pattern among them, I suspect this question contains more number theory than at first appears.

The "Puzzle Corner" of MIT News for March/April 2020 gives a "speed" problem by Sorab Vatcha: "Find seven consecutive integers whose product equals the product of four consecutive integers." The obvious "speed" solution is: 0, 1, 2, 3, 4, 5, and 6; and 0, 1, 2, and 3.

This leads to the general question of whether there are any "nontrivial" solutions to find two distinct sequences of integers whose products are the same. The obvious trivial solutions involve (1) sequences containing 0, (2) replacing all of the integers in a sequence with their negatives, (3) adding or deleting values 1 or -1, and (4) one sequence has length 1 (and hence contains one integer that is the product of the integers in the other sequence). These criteria reduce the problem to finding two distinct sequences of integers $\ge 2$ of length $\ge 2$ whose products are the same.

There is a less obvious criterion of non-triviality: (5) The two sequences must not overlap. If the sequences overlap, then the overlapping part can be deleted from both of them, yielding a soltion with shorter sequences. This interacts with the prohibition (4) against sequences of length 1: Removing the overlapped part may reduce one sequence to length 1, and the shorter solution may be trivial also. And indeed, there is a large family of solutions constructed this way: If the product of $a \cdots b$ is $P$, then $a \cdots (P-1) = (b+1) \cdots P$.

There are non-trivial solutions. The one with the smallest product is $5 \cdot 6 \cdot 7 = 14 \cdot 15 = 210$. Is there an enumeration of all solutions?

The known non-trivial solutions are: $5 \cdots 7 = 14 \cdot 15 = 210$, $2 \cdots 6 = 8 \cdots 10 = 720$, $19 \cdots 22 = 55 \cdots 57 = 175560$, and $8 \cdots 14 = 63 \cdots 66 = 17297280$.

I have run a number of computer searches that have not revealed any further solutions: (a) all sequences with products less than $10^{30}$, (b) sequences with numbers less than 10,000,000 and length less than 10, (c) sequences with numbers less than 1,000,000 and length less than 100, (d) sequences with numbers less than 100,000 and length less than 1,000, and (e) sequences with numbers less than 10,000 and length less than 10,000.

See also https://math.stackexchange.com/questions/991728/equal-products-of-consecutive-integers/ and https://math.stackexchange.com/questions/3346618/non-trivial-solutions-to-equal-products-of-consecutive-integers.

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    $\begingroup$ It appears no further solutions are known: oeis.org/A163263 $\endgroup$ Mar 27, 2020 at 3:40
  • $\begingroup$ Note the literature references in the above OEIS entry. $\endgroup$
    – Dale
    Mar 28, 2020 at 15:51
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    $\begingroup$ A connection linking those numbers in pairs is that $210$ divides $175560$ and $720$ divides $17297280$. $\endgroup$ Dec 28, 2023 at 11:58
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    $\begingroup$ Suppose $m(m+1)\times\dotsb\times(m+j)=n(n+1)\times\dotsb\times(n+k)$, $m<n$. Multiplying by $(n-1)!(m-1)!$ we get $(m+j)!(n-1)!=(n+k)!(m-1)!$ So the problem is equivalent to finding solutions to $a!b!=c!d!$. $\endgroup$ Dec 28, 2023 at 15:26

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The observation is made in many places (see the linked questions and OEIS sequence) that the sequence with larger numbers cannot contain a prime. It should also be noted that sufficiently large solutions cannot contain both three times a prime and two times a different prime. This can be extended using information on maximal gaps to limit how many large primes can appear as factors in a large sequence. An interesting variant is when a product of n consecutive positive integers divides another sequence of n consecutive positive integers. Except for n factorial dividing every such, or a sequence dividing itself, I am not aware of examples beyond the ones listed above and trivial modifications. Listing such pairs of n tuples would go far toward solving this problem.

Gerhard "Unless Erdos Already Did It" Paseman, 2020.03.26.

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  • $\begingroup$ "Unless Erdos Already Did It" - you made my day! 😁 $\endgroup$
    – Wolfgang
    Mar 27, 2020 at 14:12
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At least the first two numbers are related to remarkable approximations to $\pi$.

$210\pi \approx 660$, which simplifies to $$\pi \approx \frac{22}{7}$$ (Archimedes)

$720\pi \approx 2262$, or $$\pi \approx \frac{377}{120}$$ (Ptolemy)

$\frac{377}{120}$ is not a convergent but a semiconvergent, actually closer to $\pi$ than the more usual $\frac{333}{106}$.

Both approximations are given by integrals.

$$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx = \frac{22}{7}-\pi$$ (Dalzell)

$$\frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx = \frac{377}{120}-\pi$$

The relationship between $\pi$ approximations and products of consecutive integers is shown in series related to these integrals.

The next two integrals following this pattern are

$$\frac{1}{4}\int_0^1 \frac{x^6(1-x)^8}{1+x^2}dx = \frac{566053}{180180}-\pi$$

and

$$\frac{1}{8}\int_0^1 \frac{x^7(1-x)^{10}}{1+x^2}dx = \frac{18113671}{5765760}-\pi$$

The ratios of these denominators to the third and fourth numbers are notably simple as well.

$$\frac{180180}{175560} =\frac{39}{38}$$

$$\frac{5765760}{17297280} =\frac{1}{3}$$

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  • $\begingroup$ On an unrelated note, the third number is the order of the sporadic simple group $J_1$ :) $\endgroup$ Dec 28, 2023 at 7:53
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    $\begingroup$ This is because numbers with lots of small factors tend to coincide. The number of seconds in a week is equal to the order of $J_2$, for example. $\endgroup$ Dec 28, 2023 at 8:41
  • $\begingroup$ 19*20*21*22 is also a good denominator for a rational approximation to $\pi$, but not 63*64*65*66. $\endgroup$ Dec 28, 2023 at 8:45
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    $\begingroup$ @DaveBenson Sure, I was not trying to make a serious point here. $\endgroup$ Dec 28, 2023 at 11:12
  • $\begingroup$ The order of $J_2$ does not look like totally unrelated to these numbers. $604800$ is the product of two small factorials, $5!7!$. $720$ is also $3!5!$, besides $6!$. Sequence oeis.org/A175430 lists both. $\endgroup$ Dec 28, 2023 at 11:41

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