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I was inspired in the section B48 of Guy's book ([1]) for ask myself a question that arises when one consider the sequence for integers $m\geq 1$

$$\mathcal{S}_m:=\prod_{n=1}^m\frac{s_n+1}{s_n-1}\tag{1}$$ where $s_n$ denotes the sequence of semiprimes, thus $s_1=4$, $s_2=6$, $s_3=9\ldots$, that is A001358 in the OEIS (see the Wikipedia Semiprime).

Seems that it is obvious that exists an integer $m_0$ for which $\forall m\geq m_0$ the denominator of $S_m$ (written as irreducible fraction) is an even integer, while that its numerator is odd, thus seems that the following conjecture is "obvious".

Conjecture. There exists an integer $m_0$ such that $\forall m\geq m_0$ $$\prod_{n=1}^m\frac{s_n+1}{s_n-1}$$ is never an integer.

What I am saying is that semiprimes as $9,33$ or $129$ add in the denominator of the corresponding $S_m$ powers of two with a great exponent. But I don't know how to get a rigorous proof from this heuristic. I've tested the conjecture for the first few dozens of semiprimes.

Question. What is/Is there a proof for the existence of an integer $m_0$ such that $\forall m\geq m_0$ $$\prod_{n=1}^m\frac{s_n+1}{s_n-1}$$ is never an integer? Many thanks.

I hope that this question isn't obvious, and good for this MathOverflow. Thanks all.

References:

[1] Richard K. Guy, Unsolved Problems in Number Theory, Unsolved problems in Intuitive Mathematics Volume I, Second Edition, Springer-Verlag (1994).

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    $\begingroup$ It is not clear to me why the sequence $\{s_n\}$ is particularly interesting in formulating a conjecture like this, rather than some other infinite sequence of positive integers. $\endgroup$
    – Stanley Yao Xiao
    Aug 3, 2019 at 13:10
  • $\begingroup$ (1/2) Yes, @StanleyYaoXiao I am asking just about a sequence, because if with few computational evidence I ask about more sequences then the post doesn't seem serious. I thought, but I did not do the calculations for the sequence of square-free integers (the same product, specialization for squarefree integers, taken for $n\geq 2$), and I tried with variants of some of the best known sequences of figurate numbers, see the table in the encyclopedia MathWorld Figurate Number. Invoking a formula due to Sitaramachandrarao (see the Wikipedia Euler's totient function), then combining with the $\endgroup$
    – user142929
    Aug 4, 2019 at 8:08
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    $\begingroup$ (2/2) convergence criteria for infinite products I know that $\prod_{n=2}^\infty\frac{n\varphi(n)+1}{n\varphi(n)-1}$ will be convergent, but I think that has no a nice closed-form. Thus this is an invitation if you, or some of your colleagues in MO want to study more variants (the problems that have a good mathematical content are those showed in Guy's book). With my computational evidence and mathematical experience, I could not propose a more interesting product. Many thanks again, as was said this is an invitation if some user want try to explore these products or variants of problems $\endgroup$
    – user142929
    Aug 4, 2019 at 8:08

1 Answer 1

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I made wrote a program in python to test it for more numbers. Here are the results:

numerator gets more factors of two again

longer time behaviour

You can see that, after a while, the numerator becomes even again. Then it goes back to being odd. I can't really see much of a pattern here, it looks indistinguishable from a random walk to me. Though it's true that numbers like 9 and 33 contribute many factors of 2 to the denominator, numbers like 15 and 511 contribute many factors of 2 to the numerator. There may be an argument that shows that one type wins out, but if so it isn't obvious to me.

The may still be some way of proving the overall conjecture, but it will probably require thinking about things besides just the number of factors of 2 involved.

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  • $\begingroup$ Many thanks you was generous with your contribution. $\endgroup$
    – user142929
    Aug 3, 2019 at 11:39

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