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Let $n$ be a positive integer.

In how many ways can one write $n!$ as a product of consecutive integers?

For example: $4!=1\times2\times3\times4=2\times3\times4$. Here, $2$ possibilities exist.

$5!=1\times2\times3\times4\times5=2\times3\times4\times5=4\times5\times6$. Here, $3$ possibilities exist.

153
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    I don't understand what you're asking. Could you try to reformulate your question? – hmakholm left over Monica Sep 11 '15 at 14:30
  • i know its grammar really bad, but i coldn't write. Can somebody change it? – 153 Sep 11 '15 at 14:32
  • example: $4!=1.2.3.4=2.3.4=24$. We can write it 3 ways. – 153 Sep 11 '15 at 14:33
  • I think it would be more proper to say that you can write $4!$ two ways, one as $1\cdot 2\cdot 3\cdot 4$, and the other without the one. I suppose that is trivial and therfore all $n!$ can be written as a product of consecutive integers a minimum of two ways, provided $n$ is a positive Integer greater than 1. – John Molokach Sep 11 '15 at 14:45
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    But I do not know of a general procedure for counting the ways above the two trivial ones. For this reason, I'm up voting your question. – John Molokach Sep 11 '15 at 14:46
  • i changed it, thanks. – 153 Sep 11 '15 at 14:47
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    In other words, the question is: what is the cardinality of the following set?

    $$ A_n = {(\ell,s)\in\mathbb{N}^2,s>\ell | \prod_{k=\ell}^s k = n!} $$

    Clearly $(2,n)\in A_n$. As the user showed, for $n=5$, $(4,6)\in A_5$, so the question is non trivial.

     – bartgol Sep 11 '15 at 14:52
  • The question says "a product of consecutive integers" but the subject line says "a multiple of consecutive integers". $720=6!$ is a multiple of the consecutive integers $15$ and $16$ but it is not their product. (It is also a multiple of $8,9,10$ and it is their product.) ${}\qquad{}$ – Michael Hardy Sep 11 '15 at 14:53
  • @153, it looks like you meant "product" rather than "multiple". Is that right? If so, you should edit the question title. – bartgol Sep 11 '15 at 14:55
  • $5040=7!$ is a multiple of the consecutive integers $1,2,3,4,5,6,7,8,9,10$, but it is not their product. It is also a multiple of the consecutive integers $14,15,16$ but not their product. And a multiple of $20,21$ but not their product. And a multiple of $35,36$ but not their product. ${}\qquad{}$ – Michael Hardy Sep 11 '15 at 14:59
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    So $1\times2\times3\times4\times5\times6 = 8\times9\times10$ is an example of what is sought. ${}\qquad{}$ – Michael Hardy Sep 11 '15 at 15:01
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    @Michael, It is just for fun. Today is Friday. – 153 Sep 11 '15 at 15:04
  • See the reference in Gerry Myerson's answer at the near-duplicate question: http://math.stackexchange.com/a/112709/30402 – Erick Wong Sep 11 '15 at 15:23
  • $1\times2\times3\times4\times5\times6\times7 = 7\times8\times9\times10$. ${}\qquad{}$ – Michael Hardy Sep 11 '15 at 15:30

3 Answers3

5

The products of consecutive integers, $(m+1)\cdot (m+2)\cdot(m+3)\cdots n$ is a factorial from which you have removed the first factors, i.e $\dfrac{n!}{m!}$. To make this equal another factorial, you need to discard the largest factors of $n!$, i.e. $n\cdot(n-1)\cdot(n-2)\cdots$.

It is always possible to discard $n$ itself. This yields $$\frac{(m!)!}{m!}=(m!-1)!$$ For instance $$5\cdot6\cdots24=1\cdot2\cdots23.$$ and there are infinitely many solutions, extremely large. Other solutions are accidental and rare.

3

There are infinitely many $n$ for which $n!$ can be written in 3 ways. All you have to do is set $n_k = 2 \cdot 3 \cdot 4 \ldots \cdot k$. Then whenever $n = n_k - 1$ for some $k$, you can remove the first $k$ numbers in the product and add $n_k$ to the end. It seems much more complicated to prove that you can add, say, two extra consecutive numbers to the end of the product for infinitely many $n$ (and determine for which $n$ when you can, in analytic or fast computational form). If you add two numbers to the end, $n+1$ and $n+2$, then $GCD(n+1,n+2) = 1$ so somehow $n+1$ and $n+2$ must partition the prime factors of $n_k!$. I'm currently running a computer search to find the first $n$ (if any) where this is possible. I'll update if I find anything.

user2566092
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It exists at least 2 trivial combinations (1*2*...*n and 2*3*...*n).

It will also exist a third combination if n+1 is the potencial of other number, although it is not an enough condition.

For example:

while 23! = 1*2*3*...*23 = 2*3*4*5*...*23 = 5*6*...*23*24

To get a fourth combination continuing this method, it would have to be a number that reach next conditions:

  • n+1 = m!
  • n+2 = p!/m!

but as soon as p!/m! is always an even number (p>m+1), the difference between it and m! (also even) cannot be one.

For that reason, it is not possible to have 4 combinations.

Zero point
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    But $6!=8\cdot9\cdot10$. – 153 Sep 11 '15 at 15:16
  • I had consider that you had to do new combinations in order. Then you should always consider one more combination: 6! = 720 – Zero point Sep 11 '15 at 15:33
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    @Zeropoint The point is that your argument is not a proof that it's impossible to have $4$ combinations, because the third/fourth combination might occur in a way that you haven't considered. You said "You can never substitute a new number in the series unless it respects that rule." – Erick Wong Sep 11 '15 at 15:36
  • Yeah, sorry. I understimated the freedom of the exercise. – Zero point Sep 11 '15 at 15:48