Prove that the largest power of $2$ dividing $(n+1)(n+2) \cdots (an)$ is greater than$2^{(a-1)n}$

Question

Let $a,n$ be positive integers. Find all $a$ such that for some $n$ the largest power of $2$ dividing $(n+1)(n+2) \cdots (an)$ is greater than $2^{(a-1)n}$.

Since I thought there were no such $a$, I thought about proving this by contradiction. That is, assume that $2^{(a-1)n+1}$ divides $(n+1)(n+2) \cdots (an)$. How can we get to a contradiction from here?

Answer 1

There are $(a-1)n$ factors in the original expression.

$\lfloor \frac {(a-1)n + 1}{2}\rfloor$ are divisible by $2.$ $\lfloor\frac {(a-1)n + 1}{4}\rfloor$ are divisible by $4$. $\lfloor\frac {(a-1)n + 1}{2^i}\rfloor$ are divisible by $2^i$

The largest power of $2$ that divides $(n+1)\cdots(an) = 2^{\left(\lfloor \frac {(a-1)n + 1}{2}\rfloor + \lfloor \frac {(a-1)n + 1}{2^2}\rfloor \cdots +\lfloor \frac {(a-1)n + 1}{2^i}\rfloor\cdots+\lfloor \frac {(a-1)n + 1}{2^{k}}\rfloor\right)}$

Where k is the largest integer $\le \log_2 ((a-1)n + 1) $

$\sum_\limits{i=1}^{k} \lfloor \frac {(a-1)n + 1}{2^i} \rfloor < ((a-1)n + 1)\sum_\limits {i=1}^{\infty} \frac 1{2^i}\\\sum_\limits {i=1}^{\infty} \frac 1{2^i} = 1\\ \sum_\limits{i=1}^{k } \lfloor \frac {(a-1)n + 1}{2^i} \rfloor < (a-1)n + 1\\ \sum_\limits{i=1}^{k } \lfloor \frac {(a-1)n + 1}{2^i} \rfloor \le (a-1)n$

Answer 2

Since $$(n+1)(n+2)...(an)=\frac{(an)!}{n!}$$ According to Legendre's Theorem $n!=2^{n-r_1}\cdot q_1$ where $r_1$-the number of $1$'s in binary expansion of $n$ and similarly $(an)!=2^{an-r_2}\cdot q_2$ and $$\frac{(an)!}{n!}=\frac{q_2}{q_1}\cdot 2^{n(a-1)-r_2+r_1}$$

It is known (and not difficult to show) that multiplication by $2$ "shifts" the bits to the left, e.g. $(3)_{10}=(11)_2 \Rightarrow (2\cdot 3)_{10}=(110)_2$. Multiplying by $2^r$ shifts the bits $r$ times so that if $a=2^r$ then $an$ and $n$ will have the same number of $1$'s in binary representation, or $r_1=r_2$.

Re the contradiction, we don't want to have $r_1 > r_2$. But, let's assume $n=(11)_{10}=(1011)_2$ and $a=3$ then $an=(33)_{10}=(100001)_2$. $$11!=2^{11-3}\cdot 155925$$ $$33!=2^{33-2}\cdot 4043484860477916195764296875$$ or $$2^{11(3-1)+1} | 217535414131691079834009600000=\frac{33!}{11!}$$ This is an example of $r_1 > r_2$.

Update

Few more examples:

• $an=129=2^7+1=3\cdot 43$, where $n=43$ has $4$ of $1$'s in its binary representation and $129$ only 2.
• $an=2^9+1=513=27⋅19$ , where $19$ has $3$ of $1$'s and $27$ has $4$ of $1$'s.

Basically, looking at $an=2^{2k+1}+1$ (which always has 2 of $1$'s in its binary representation and is always divisible by $3$) will always yield infinite examples. This is easy to check, basically $2 \equiv -1 \pmod{3} \Rightarrow 2^{2k+1} \equiv (-1)^{2k+1} \equiv -1 \pmod{3}$, so $$an=2^{2k+1}+1=3\cdot n$$ $n$ will always (except a few exceptions) have more than $2$ of $1$'s in its binary representation, because:

• if $n$ has $1$ of $1$'s in its binary representation, then $n=2^t$, which will lead to $2 | 1$ contradiction, except the $t=0$ case or $2^1+1=3$.
• if $n$ has $2$ of $1$'s in its binary representation, then $n=2^t+2^z, t>z\geq0$. If $z>0$ then $2 | 1$ - contradiction. If $z=0$ then $2^{2k+1}+1=3\cdot (2^t+1)\Rightarrow 2^{2k}-3\cdot 2^{t-1}=1$. It $t>1$ then we have the same $2|1$ contradiction, thus $t=1$ and $2^{2k}=4$ or $k=1$ leading to $2^3+1=3\cdot 3$.

To conclude, apart for the two examples outlined above, $n$ will always have $\geq 3$ of $1$'s in its binary representation, while $an$ only $2$.