On the factorial equations $A! B! =C!$ and $A!B!C! = D!$
$begingroup$
I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient $binom{10}{3}$. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything
$$
frac {8cdot9cdot10}{2cdot3} = 120
$$
And then it hit me that $8cdot9cdot10 = 6!$ and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this
$$
a!b = prod_{n=b}^an = frac{a!}{(b-1)!},quad a, b in mathbb{Z}^+, quad age b
$$
so that $a! = a!1$ (the notation was invented just now, and inspired by the $nCr$-notation for binomial coefficients). Now, apart from the trivial examples $(n!)!(n!) = n!$ and $a!1 = a!2 = a!$, when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, $10!8$ is both.
elementary-number-theory factorial
edited Feb 24 '12 at 2:00
Aryabhata
70.2k6156246
asked Feb 24 '12 at 0:23
ArthurArthur
112k7107190
$endgroup$
add a comment |
$begingroup$
I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient $binom{10}{3}$. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything
$$
frac {8cdot9cdot10}{2cdot3} = 120
$$
And then it hit me that $8cdot9cdot10 = 6!$ and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this
$$
a!b = prod_{n=b}^an = frac{a!}{(b-1)!},quad a, b in mathbb{Z}^+, quad age b
$$
so that $a! = a!1$ (the notation was invented just now, and inspired by the $nCr$-notation for binomial coefficients). Now, apart from the trivial examples $(n!)!(n!) = n!$ and $a!1 = a!2 = a!$, when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, $10!8$ is both.
elementary-number-theory factorial
edited Feb 24 '12 at 2:00
Aryabhata
70.2k6156246
asked Feb 24 '12 at 0:23
ArthurArthur
112k7107190
$endgroup$
2
$begingroup$
You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
$endgroup$
– Henry
Feb 24 '12 at 0:41
$begingroup$
You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
$endgroup$
– Henry
Feb 24 '12 at 1:05
$begingroup$
As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
$endgroup$
– Arthur
Feb 24 '12 at 13:24
4
$begingroup$
The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
$endgroup$
– John Bentin
Sep 15 '13 at 12:56
$begingroup$
Similar question was answered at MathOverflow.
$endgroup$
– Alexey Ustinov
Mar 30 '18 at 11:47
add a comment |
$begingroup$
I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient $binom{10}{3}$. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything
$$
frac {8cdot9cdot10}{2cdot3} = 120
$$
And then it hit me that $8cdot9cdot10 = 6!$ and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this
$$
a!b = prod_{n=b}^an = frac{a!}{(b-1)!},quad a, b in mathbb{Z}^+, quad age b
$$
so that $a! = a!1$ (the notation was invented just now, and inspired by the $nCr$-notation for binomial coefficients). Now, apart from the trivial examples $(n!)!(n!) = n!$ and $a!1 = a!2 = a!$, when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, $10!8$ is both.
elementary-number-theory factorial
edited Feb 24 '12 at 2:00
Aryabhata
70.2k6156246
asked Feb 24 '12 at 0:23
ArthurArthur
112k7107190
$endgroup$
I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient $binom{10}{3}$. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything
$$
frac {8cdot9cdot10}{2cdot3} = 120
$$
And then it hit me that $8cdot9cdot10 = 6!$ and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this
$$
a!b = prod_{n=b}^an = frac{a!}{(b-1)!},quad a, b in mathbb{Z}^+, quad age b
$$
so that $a! = a!1$ (the notation was invented just now, and inspired by the $nCr$-notation for binomial coefficients). Now, apart from the trivial examples $(n!)!(n!) = n!$ and $a!1 = a!2 = a!$, when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, $10!8$ is both.
elementary-number-theory factorial
elementary-number-theory factorial
edited Feb 24 '12 at 2:00
Aryabhata
70.2k6156246
asked Feb 24 '12 at 0:23
ArthurArthur
112k7107190
edited Feb 24 '12 at 2:00
Aryabhata
70.2k6156246
asked Feb 24 '12 at 0:23
ArthurArthur
112k7107190
edited Feb 24 '12 at 2:00
Aryabhata
70.2k6156246
edited Feb 24 '12 at 2:00
Aryabhata
70.2k6156246
edited Feb 24 '12 at 2:00
Aryabhata
70.2k6156246
70.2k6156246
asked Feb 24 '12 at 0:23
ArthurArthur
112k7107190
asked Feb 24 '12 at 0:23
ArthurArthur
112k7107190
asked Feb 24 '12 at 0:23
ArthurArthur
112k7107190
112k7107190
2
$begingroup$
You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
$endgroup$
– Henry
Feb 24 '12 at 0:41
$begingroup$
You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
$endgroup$
– Henry
Feb 24 '12 at 1:05
$begingroup$
As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
$endgroup$
– Arthur
Feb 24 '12 at 13:24
4
$begingroup$
The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
$endgroup$
– John Bentin
Sep 15 '13 at 12:56
$begingroup$
Similar question was answered at MathOverflow.
$endgroup$
– Alexey Ustinov
Mar 30 '18 at 11:47
add a comment |
2
$begingroup$
You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
$endgroup$
– Henry
Feb 24 '12 at 0:41
$begingroup$
You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
$endgroup$
– Henry
Feb 24 '12 at 1:05
$begingroup$
As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
$endgroup$
– Arthur
Feb 24 '12 at 13:24
4
$begingroup$
The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
$endgroup$
– John Bentin
Sep 15 '13 at 12:56
$begingroup$
Similar question was answered at MathOverflow.
$endgroup$
– Alexey Ustinov
Mar 30 '18 at 11:47
2
2
$begingroup$
You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
$endgroup$
– Henry
Feb 24 '12 at 0:41
$begingroup$
You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
$endgroup$
– Henry
Feb 24 '12 at 0:41
$begingroup$
You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
$endgroup$
– Henry
Feb 24 '12 at 1:05
$begingroup$
You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
$endgroup$
– Henry
Feb 24 '12 at 1:05
$begingroup$
As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
$endgroup$
– Arthur
Feb 24 '12 at 13:24
$begingroup$
As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
$endgroup$
– Arthur
Feb 24 '12 at 13:24
4
4
$begingroup$
The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
$endgroup$
– John Bentin
Sep 15 '13 at 12:56
$begingroup$
The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
$endgroup$
– John Bentin
Sep 15 '13 at 12:56
$begingroup$
Similar question was answered at MathOverflow.
$endgroup$
– Alexey Ustinov
Mar 30 '18 at 11:47
$begingroup$
Similar question was answered at MathOverflow.
$endgroup$
– Alexey Ustinov
Mar 30 '18 at 11:47
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).
answered Feb 24 '12 at 1:40
Gerry MyersonGerry Myerson
146k8147298
$endgroup$
$begingroup$
Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
$endgroup$
– Arthur
Feb 24 '12 at 13:25
add a comment |
$begingroup$
Immediately we see that any $n$ that's a power of 2 should work. e.g.
$4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$
Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.
However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by
$n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:
$n(n-1)(n-2) ... (n-k) = P(n) = A!$
Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf
answered Feb 11 '14 at 0:12
Alex MeiburgAlex Meiburg
1,805517
$endgroup$
$begingroup$
Very interesting reference, thank you!
$endgroup$
– Jose Brox
Feb 18 '16 at 20:43
add a comment |
$begingroup$
Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.
Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.
Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
$$
P(n(n+1)) > 4 log n
$$
would imply that the general equation $(1)$ has finitely many non-trivial solutions.
The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.
More references:
- Discussion on ArtofProblemSolving
- Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.
- Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.
answered May 5 '18 at 8:37
Amir HosseinAmir Hossein
2,38111650
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).
answered Feb 24 '12 at 1:40
Gerry MyersonGerry Myerson
146k8147298
$endgroup$
$begingroup$
Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
$endgroup$
– Arthur
Feb 24 '12 at 13:25
add a comment |
$begingroup$
See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).
answered Feb 24 '12 at 1:40
Gerry MyersonGerry Myerson
146k8147298
$endgroup$
$begingroup$
Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
$endgroup$
– Arthur
Feb 24 '12 at 13:25
add a comment |
$begingroup$
See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).
answered Feb 24 '12 at 1:40
Gerry MyersonGerry Myerson
146k8147298
$endgroup$
See Chris Caldwell, The diophantine equation $A!B!=C!$, J. Recreational Math. 26 (1994) 128-133. $9!=7!3!3!2!$, $10!=7!6!=7!5!3!$, and $16!=14!5!2!$ were the only known non-trivial examples of a factorial as a product of factorials as of the 3rd edition of Guy, Unsolved Problems In Number Theory (Problem B23).
answered Feb 24 '12 at 1:40
Gerry MyersonGerry Myerson
146k8147298
answered Feb 24 '12 at 1:40
Gerry MyersonGerry Myerson
146k8147298
answered Feb 24 '12 at 1:40
Gerry MyersonGerry Myerson
146k8147298
answered Feb 24 '12 at 1:40
Gerry MyersonGerry Myerson
146k8147298
146k8147298
$begingroup$
Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
$endgroup$
– Arthur
Feb 24 '12 at 13:25
add a comment |
$begingroup$
Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
$endgroup$
– Arthur
Feb 24 '12 at 13:25
$begingroup$
Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
$endgroup$
– Arthur
Feb 24 '12 at 13:25
$begingroup$
Funny I should stumble upon half of the known solutions to an unsolved problem. Thanks for the references, though. I'll look them up.
$endgroup$
– Arthur
Feb 24 '12 at 13:25
add a comment |
$begingroup$
Immediately we see that any $n$ that's a power of 2 should work. e.g.
$4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$
Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.
However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by
$n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:
$n(n-1)(n-2) ... (n-k) = P(n) = A!$
Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf
answered Feb 11 '14 at 0:12
Alex MeiburgAlex Meiburg
1,805517
$endgroup$
$begingroup$
Very interesting reference, thank you!
$endgroup$
– Jose Brox
Feb 18 '16 at 20:43
add a comment |
$begingroup$
Immediately we see that any $n$ that's a power of 2 should work. e.g.
$4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$
Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.
However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by
$n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:
$n(n-1)(n-2) ... (n-k) = P(n) = A!$
Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf
answered Feb 11 '14 at 0:12
Alex MeiburgAlex Meiburg
1,805517
$endgroup$
$begingroup$
Very interesting reference, thank you!
$endgroup$
– Jose Brox
Feb 18 '16 at 20:43
add a comment |
$begingroup$
Immediately we see that any $n$ that's a power of 2 should work. e.g.
$4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$
Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.
However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by
$n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:
$n(n-1)(n-2) ... (n-k) = P(n) = A!$
Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf
answered Feb 11 '14 at 0:12
Alex MeiburgAlex Meiburg
1,805517
$endgroup$
Immediately we see that any $n$ that's a power of 2 should work. e.g.
$4! = 3!2!2!$, $8! = 7!2!2!2!$, $16! = 15!2!2!2!2!$
Similarly, by combining whatever small factorials we want, we can take $n!$ and $(n-1)!$ to have that ratio. For instance, suppose we wanted to use $3!5!7! = 3628800$ somewhere; we can then make $3628800! = 3628799!7!5!3!$. You can obviously generate an infinite number of solutions this way.
However, the instances of $n!/(n-1)!$ are probably what Caldwell (see other answer) referred to as "trivial". The non-trivial instances of $n!/(n-k)!$ would be given by
$n(n-1)(n-2) ... (n-k) = A_1!A_2!...A_i!$ which -- in general -- I know no way of tackling the general case, but the case where $i=1$ (that is, $n!=(n-k)!A!$) can be attacked:
$n(n-1)(n-2) ... (n-k) = P(n) = A!$
Which is referred to as a polynomial-factorial diophantine equation. This is still largely open, but has some interesting results proven about it -- namely, that the set of solutions is finite for any polynomial P. An extensive treatment of the matter is given in http://www.ams.org/journals/tran/2006-358-04/S0002-9947-05-03780-3/S0002-9947-05-03780-3.pdf
answered Feb 11 '14 at 0:12
Alex MeiburgAlex Meiburg
1,805517
edited May 5 '18 at 23:41
answered Feb 11 '14 at 0:12
Alex MeiburgAlex Meiburg
1,805517
answered Feb 11 '14 at 0:12
Alex MeiburgAlex Meiburg
1,805517
answered Feb 11 '14 at 0:12
Alex MeiburgAlex Meiburg
1,805517
1,805517
$begingroup$
Very interesting reference, thank you!
$endgroup$
– Jose Brox
Feb 18 '16 at 20:43
add a comment |
$begingroup$
Very interesting reference, thank you!
$endgroup$
– Jose Brox
Feb 18 '16 at 20:43
$begingroup$
Very interesting reference, thank you!
$endgroup$
– Jose Brox
Feb 18 '16 at 20:43
$begingroup$
Very interesting reference, thank you!
$endgroup$
– Jose Brox
Feb 18 '16 at 20:43
add a comment |
$begingroup$
Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.
Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.
Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
$$
P(n(n+1)) > 4 log n
$$
would imply that the general equation $(1)$ has finitely many non-trivial solutions.
The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.
More references:
- Discussion on ArtofProblemSolving
- Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.
- Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.
answered May 5 '18 at 8:37
Amir HosseinAmir Hossein
2,38111650
$endgroup$
add a comment |
$begingroup$
Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.
Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.
Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
$$
P(n(n+1)) > 4 log n
$$
would imply that the general equation $(1)$ has finitely many non-trivial solutions.
The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.
More references:
- Discussion on ArtofProblemSolving
- Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.
- Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.
answered May 5 '18 at 8:37
Amir HosseinAmir Hossein
2,38111650
$endgroup$
add a comment |
$begingroup$
Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.
Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.
Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
$$
P(n(n+1)) > 4 log n
$$
would imply that the general equation $(1)$ has finitely many non-trivial solutions.
The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.
More references:
- Discussion on ArtofProblemSolving
- Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.
- Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.
answered May 5 '18 at 8:37
Amir HosseinAmir Hossein
2,38111650
$endgroup$
Consider the general equation $$a_1!a_2!cdots a_n! = b!.$$
People like Paul Erdős worked on this kind of equations. So, they are rather serious problems. Even for the simplest case where $n=2$, namely the equation $A!B!=C!$, we don't know whether there is any other non-trivial solution other than $10!=7!6!$.
Erdős proved that in the equation $A!B!=C!$ if we assume $1< A leq B$, then for large enough values of $C$, the difference of $B$ and $C$ does not exceed $5 log log n$. It is clear that $5 log log n$ grows to infinity as $n$ increases. But in fact, this is a very slowly growing function.
For instance, the first $n$ for which $5 log log n = 50$ has more than $9566$ digits. Caldwell proved (1994) that the only non-trivial solution with $C < 10^6$ is $10!=7!6!$. Florian Luca proved in this paper (2007) that considering the $abc$ hypothesis, we have $C-B=1$ for large enough $n$. Read also a remark (2009) on this paper if you're interested.
Suppose that $P(m)$ denotes the largest prime factor of $m$. Erdős also showed in one of his many papers that the assertion
$$
P(n(n+1)) > 4 log n
$$
would imply that the general equation $(1)$ has finitely many non-trivial solutions.
The last paragraph is taken from a recent work of Hajdua, Pappa, and Szakács, who prove proved in this paper (2018) that writing $k=B-A$ for all non-trivial solutions of the equation $A!B!=C!$ different from $(A, B, C) =(6, 7, 10)$, we have $C<5k$. Further, if $k leq 10^6$, then the only non-trivial solution is given by $(A, B, C) =(6, 7, 10)$. I presented their paper in our MathSciNet seminar at UBC and you can find the slides here.
More references:
- Discussion on ArtofProblemSolving
- Luca, Florian. "The Diophantine equation $P (x)= n!$ and a result of M. Overholt." Glasnik matematički 37.2 (2002): 269-273.
- Berend, Daniel, and Jørgen Harmse. "On polynomial-factorial diophantine equations." Transactions of the American Mathematical Society 358.4 (2006): 1741-1779.
answered May 5 '18 at 8:37
Amir HosseinAmir Hossein
2,38111650
edited May 5 '18 at 9:21
answered May 5 '18 at 8:37
Amir HosseinAmir Hossein
2,38111650
answered May 5 '18 at 8:37
Amir HosseinAmir Hossein
2,38111650
answered May 5 '18 at 8:37
Amir HosseinAmir Hossein
2,38111650
2,38111650
add a comment |
add a comment |
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2
$begingroup$
You need to look at rising factorial and falling factorial and n-permutations for some of the notation used.
$endgroup$
– Henry
Feb 24 '12 at 0:41
$begingroup$
You could restate your question as asking when is a factorial equal to the non-trivial product of two factorials as in $10! = 7! times 6!$ and when is it the non-trivial product of three factorials as in $10! = 7! times 5! times 3!$.
$endgroup$
– Henry
Feb 24 '12 at 1:05
$begingroup$
As for the notation, I was aware of the possibility someone had come up with it before me, it's not like I'm THAT good. And, you are completely right. I could've thought that one more through. It was late at night.
$endgroup$
– Arthur
Feb 24 '12 at 13:24
4
$begingroup$
The notation $a!b$ has a well-established and familiar meaning: namely the product of $a!$ and $b$.
$endgroup$
– John Bentin
Sep 15 '13 at 12:56
$begingroup$
Similar question was answered at MathOverflow.
$endgroup$
– Alexey Ustinov
Mar 30 '18 at 11:47