6
$\begingroup$

Question : Let $n,a,b$ be positive integers. Are there infinitely many triplets $(n,a,b)$ which satisfy the following equality?$$n!=2^a-2^b$$ If Yes, then how can we prove that? If No, then how can we find every such triplet $(n,a,b)$?

The followings are what I've got :

$(1)$ We can get $(n,a,b)=(2,2,1),(3,3,1),(4,5,3),(5,7,3)$ easily.

$(2)$ $a$ can be represented by $n$ :

Let $2^{A_n}$ be the least number in the form of $2^i$ larger than $n!$, namely $$2^{A_n-1}\le n!\lt 2^{A_n}\iff A_n-1\le \log_2{n!}\lt A_n\iff A_n=\lfloor \log_2{n!}\rfloor+1$$ where $\lfloor x\rfloor $ represents the largest integer not greater than $x$.

Supposing $a\gt A_n$ gives us $$n!=2^a-2^b\ge 2^{A_n+1}-2^{A_n}=2^{A_n}\gt n!.$$ This is a contradiction. Hence, we have $a=A_n=\lfloor \log_2{n!}\rfloor+1$.

$(3)$ $b$ can be represented by $n$ :

We have $n!=2^b(2^{a-b}-1).$ Since $2^b$ is even and $2^{a-b}-1$ is odd, we know that $b$ is the number of $2$ as a prime factor of $n!$, namely $$b=\sum_{k=1}^{\lfloor \log_2 n\rfloor}\left\lfloor\frac{n}{2^k}\right\rfloor.$$

$(4)$ From $(2),(3)$, the question can be represented as the following :

Question : Are there infinitely many positive integer $n$ which satisfy the following equality? $$n!=2^{\lfloor \log_2{n!}\rfloor+1}-2^{\sum_{k=1}^{\lfloor \log_2 n\rfloor}\left\lfloor\frac{n}{2^k}\right\rfloor}$$ If Yes, then how can we prove that? If No, then how can we find every such $n$?

$(5)$ It seems that $n\ge6$ don't satisfy the equation in $(4)$ using computer.

Can anyone help?

$\endgroup$
  • $\begingroup$ I would strongly suspect that there are only finitely many solutions, but unless there is some trick it is likely to be really hard to prove. We can always find $n$ so that $2^n-1$ has any particular odd factor we want. The difficulty is that it will also have factors we do not want. $\endgroup$ – almagest Sep 4 '14 at 10:32
  • $\begingroup$ @almagest: I agree with you and that's exaclty what I've thought about:) $\endgroup$ – mathlove Sep 4 '14 at 10:35
  • $\begingroup$ Information on the integer sequence described [here][1] may shed light on this problem. [1]: oeis.org/… $\endgroup$ – MartinG Sep 4 '14 at 11:58
3
$\begingroup$

Begin by noting that, if $n \geq 9$, then $3^k \mid n!$ where $k > n/3$. It follows that, if we have $$ n! = 2^a - 2^b $$ with $n \geq 9$, then $$ 2^{a-b} \equiv 1 \mod{3^k} $$ for some $k > n/3$. An elementary argument shows that necessarily $a-b$ is a multiple of $2 \cdot 3^{k-1}$ and, in particular, that $$ a-b \geq 2 \cdot 3^{k-1} > 2 \cdot 3^{n/3-1}. $$ Since $n! \geq 2^{a-1}$, we thus have that $$ \log (n!) > 2 \cdot 3^{n/3-1} \log (2). $$ On the other hand, from Stirling's formula, $$ \log (n!) < 1-n+(n+1/2) \log n $$ and hence $$ 2 \cdot 3^{n/3-1} \log (2) < 1-n+(n+1/2) \log n. $$ A short computation shows that this is a contradiction for $n \geq 10$. Checking the smaller values gives the solutions noted by the OP.

$\endgroup$
  • $\begingroup$ Could you elaborate the part that "An elementary argument shows that necessarily $a-b$ is a multiple of $2 \cdot 3^{k-1}$"? $\endgroup$ – mathlove Sep 6 '14 at 10:21
  • $\begingroup$ $2$ is a primitive root modulo $9$ and hence modulo $3^k$ for all $k$. $\endgroup$ – Mike Bennett Sep 6 '14 at 14:55
  • $\begingroup$ Thank you so much for this answer! $\endgroup$ – mathlove Sep 6 '14 at 15:39
  • $\begingroup$ "$2$ is a primitive root modulo $9$ and hence modulo $3^k$ for all $k$." This is true, but it doesn't extend that easily to all powers of $3$. One has to check whether $9\mid2^{3-1}-1$, luckily this is not the case. $\endgroup$ – punctured dusk Sep 8 '14 at 7:40
  • 1
    $\begingroup$ Ummm, if $r$ is a primitive root mod $p^2$, then $r$ is a primitive root mod $p^k$ for all $k$. Luck is not involved… $\endgroup$ – Mike Bennett Sep 9 '14 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.