Suppose we have a finite set of numbers $K_i \in \mathbb N$.

Can we prove that the product $$\prod_{i=1}^n \frac{3K_i+2}{2K_i+1}$$

can never be equal a power of $2$?

As you can see, each term is between $\frac{3}{2}$ and $\frac{5}{3}$. The greater $K_i$, the closer it is to $\frac{3}{2}$. Ratio between exact power or $\frac{3}{2}$ and the next greater power of $2$ places the constraint on $K_i$ values: the lowest $K_i$ cannot exceed certain value. For example, for products up to $2000$ terms, the biggest lowest value of $K_i$ was found to be for $1636$ terms, and it is $291643$.

The value would be a power of $2$ if all factors of the numerator and the denominator, except of $2$s in the numerator, completely canceled each other. Note that the denominator is always odd. It appears that there's no set of $K_i$ to make such a product equal to a power of $2$.

But can it be proven?

  • $\begingroup$ We must stipulate $n\ge1$, since an empty product is $2^0$. $\endgroup$ – J.G. Jan 2 at 10:19

Its not true. For instance, define $\mathcal K:=\{5,6,8,24,27,32,41,47,69,92\}$. Then we have, $$\prod_{i\in\mathcal K}\frac{3i+2}{2i+1}=2^6.$$

  • 2
    $\begingroup$ Thank you, that's a nice counter-example. $\endgroup$ – Alexander Grigoriev Jan 2 at 15:13
  • $\begingroup$ How did you obtain this counter example? $\endgroup$ – N.S.JOHN Jan 24 at 4:21

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.