Freaky Factorials
n ! ( n − 1 ) ! = m !
Solve the equation above for natural numbers m , n .
Let S = { ( m 1 , n 1 ) , ( m 2 , n 2 ) , … , ( m j , n j ) } denote all the possible solutions.
Evaluate k = 1 ∑ j ( m k + n k ) .
The answer is 23.
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1 solution
I disagree with the posted answer. I found the ordered pairs listed. But I also included (0,1). If you using the def of 0!=1, which is common. Then it also works, making the sume requested one higher, 24.
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( 0 , 1 ) does satisfy the equation, but since we are asked to work within the set of natural numbers this pair is not an element of S . (There is some ambiguity as to whether 0 is considered a natural number, but generally the default is that the set of natural numbers is equivalent to the set of positive integers.)
But now I am curious whether there are any more. I couldn't find any, but I couldn't prove that there weren't any.
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Yes, I'm surprised that there aren't at least a few more. As m gets larger it becomes harder and harder for the prime factors to "double up" so as to have m ! = n ! ( n − 1 ) ! = n ∗ ( ( n − 1 ) ! ) 2 , but a proof that there are no more pairs remains elusive.
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I found it gave me something to work with the expand the factorials, and for examples in the range of the solutions know about, it is perfectly workable.
For example: consider n=5
5!(4!)= 4 3 2(5!) ... The trick now is transform 4 3 2 into the integers we need to complete m!. 2*3 = 6. 7 will never be attainable being in this context a prime number we must derive via a product.
I ran into this situation over and over working examples. I wonder if there is some way to make the distance to the next prime a concrete expression, Something that can be understand in terms of (n-1)!.
Probably not, that's just the most promising route I found to take towards a proof.
Any thoughts?
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@Carl White – Yes, that was similar to what I was finding too. We can't have any primes p between m and n , for that would imply that p would be a factor of m ! but not n ! ( n − 1 ) ! , i.e., there would be no chance they could be equal. We can't have m being prime either, although n can be prime. So we have to work in the "gaps" between primes such that
m ( m − 1 ) ( m − 2 ) . . . . ( n + 1 ) = ( n − 1 ) ! .
Evidently this is a rare occurrence, but the issue with establishing a proof is that the sizes of these "gaps" vary greatly. If the gaps were of a predictable size then a proof would (probably) be straightforward. There is, however, an upper bound on the prime gap , (by way of Bertrand's postulate), so we might yet be able to get somewhere with a proof. A tough problem; definitely fun to think about. :)
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@Brian Charlesworth – An upper bound you say. I would think we should be able to prove that there are no examples above a certain point, since there would be far too many terms between the expanded factorials, to fit in the maximum gap.
I'll have to put some htought into it, and read your link when I have time.
@Brian Charlesworth – I could use some input from the outside. I've been thinking my around in circles for several hours now on how to prove there are or are not more cases of products of consecutive factorials producing a third factorial.
The basic outline of my work... leaning toward the negative on the possibility question is as follows:
I assume Bertrand's Postulate as found in the link http://en.wikipedia.org/wiki/Prime_gap . There is always a prime between c and 2c.
Prove that there cannot be a prime between (k+1)! and m!, because then said prime, I've come to calling it "the offending prime", must be a factor of m!, but cannot be a factor of k! or (k+1)!, and therefore not there product.
Using Bertrand's postulate there will be an offending prime if m!>2(k+1)!.
Assuming that k!(k+1)! = m!
k(k-1)...3 2 (k+1)! = m!
obviously m!, under this assumption, is bigger than 2(k+1)! for any k ≥ 3.
Now if this were correct it would discount the possibility of 6!*7!=10!... so obviously something is off. But I'm not sure what it is.
Any comments?
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@Carl White – Have a look at Mathh Mathh's MSE link, where Calvin Lin applied a stronger version of Bertrand's postulate to solve this problem.
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@Brian Charlesworth – Yeah, I'll probably do that. I was really hoping to figure this out for myself lol... you know what I mean? For pride's sake?
Oh well, maybe it will help with the next difficult problem.
Where did you see it mentioned that this is an open problem? Just because it is open whether there exist other solutions to A ! B ! = C ! doesn't mean it is open whether the special case n ! ( n − 1 ) ! = m ! can be proved to have no others. Unless you prove A ! B ! = C ! implies A , B are consecutive (for non-trivial solutions).
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Good point. The note in the link doesn't say the consecutive integer scenario is an open question, so there may be a relatively simple proof for this question. I just can't see how to do that yet.
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I've found this problem on Math StackExchange .
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@Mathh Mathh – Great find! So it's not an open question after all. Funny to see Calvin Lin showing up on MSE. :)
This appears to be an open question, as mentioned in this link . The only known instances are where ( m , n ) = ( 1 , 1 ) , ( 2 , 2 ) , ( 1 0 , 7 ) , giving us a total of 2 + 4 + 1 7 = 2 3 .
For three consecutive factorials, it would appear that there would just be 0 ! 1 ! 2 ! = 2 ! .