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I have the hypothesis that the term $b = \prod_{i=1}^n (3 + \frac{1}{x_i})$ results not in an even, natural number, except for $n=1$ and $x_1 = 1$. Actually, I believe that it is not even possible that the equation leads to a natural number at all (except the case mentioned). How can I prove that?

Condition: $x_i$ is an odd, natural number.

It is fairly easy to prove my hypothesis for $n=1$ and $n=2$. I find it very difficult to formulate a proof for $n>2$.

I would appreciate it if you could help me.

amWhy
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c4ristian
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3 Answers3

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$(3+\frac{1}{31})(3+\frac{1}{47})(3+\frac{1}{71})(3+\frac{1}{107})(3+\frac{1}{161})(3+\frac{1}{121})(3+\frac{1}{91})(3+\frac{1}{137})(3+\frac{1}{103})=20480$

or this

$(3+\frac{1}{83})(3+\frac{1}{125})(3+\frac{1}{47})(3+\frac{1}{71})(3+\frac{1}{107})(3+\frac{1}{161})(3+\frac{1}{121})(3+\frac{1}{91})(3+\frac{1}{137})(3+\frac{1}{103})(3+\frac{1}{155})(3+\frac{1}{233})(3+\frac{1}{175})(3+\frac{1}{263})(3+\frac{1}{395})(3+\frac{1}{593})(3+\frac{1}{445})(3+\frac{1}{167})(3+\frac{1}{251})(3+\frac{1}{377})(3+\frac{1}{283})(3+\frac{1}{425})(3+\frac{1}{319})(3+\frac{1}{479})(3+\frac{1}{719})=893353197568$

Some have the same result, like this

$(3+\frac{1}{293})(3+\frac{1}{55})(3+\frac{1}{83})(3+\frac{1}{125})(3+\frac{1}{47})(3+\frac{1}{71})(3+\frac{1}{107})(3+\frac{1}{161})(3+\frac{1}{121})(3+\frac{1}{91})(3+\frac{1}{137})(3+\frac{1}{103})(3+\frac{1}{155})(3+\frac{1}{233})(3+\frac{1}{175})(3+\frac{1}{263})(3+\frac{1}{395})(3+\frac{1}{593})(3+\frac{1}{445})(3+\frac{1}{167})(3+\frac{1}{251})(3+\frac{1}{377})(3+\frac{1}{283})(3+\frac{1}{425})(3+\frac{1}{319})(3+\frac{1}{479})(3+\frac{1}{719})(3+\frac{1}{1079})(3+\frac{1}{1619})(3+\frac{1}{2429})(3+\frac{1}{911})(3+\frac{1}{1367})=1970324836974592$

and this

$(3+\frac{1}{347})(3+\frac{1}{521})(3+\frac{1}{391})(3+\frac{1}{587})(3+\frac{1}{881})(3+\frac{1}{661})(3+\frac{1}{31})(3+\frac{1}{47})(3+\frac{1}{71})(3+\frac{1}{107})(3+\frac{1}{161})(3+\frac{1}{121})(3+\frac{1}{91})(3+\frac{1}{137})(3+\frac{1}{103})(3+\frac{1}{155})(3+\frac{1}{233})(3+\frac{1}{175})(3+\frac{1}{263})(3+\frac{1}{395})(3+\frac{1}{593})(3+\frac{1}{445})(3+\frac{1}{167})(3+\frac{1}{251})(3+\frac{1}{377})(3+\frac{1}{283})(3+\frac{1}{425})(3+\frac{1}{319})(3+\frac{1}{479})(3+\frac{1}{719})(3+\frac{1}{1079})(3+\frac{1}{1619})=1970324836974592$

In the Collatz context pick a candidate such as $e_0|e_{n+1}$:

$(3+\frac{1}{e_0})(3+\frac{1}{e_1})...(3+\frac{1}{e_n})=\frac{e_{n+1}}{e_0}\prod_{k=0}^n2^{\nu_2(3e_k+1)}$

which you can find here A005184

Collag3n
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  • Thank you very much. Your provided examples are very interesting and help me a lot, though they falsify my hypothesis. I am indeed working on the Collatz problem. We try to find out under which conditions cycles occur within a sequence. I trained a Machine Learning model to predict the next Collatz number. This model lead me to the above formula. Afterwards we discovered that the formula was already discussed in existing papers – c4ristian Mar 19 '20 at 20:04
  • Interesting question: which even number can be represented in the said form? – Batominovski Mar 19 '20 at 20:20
  • Indeed. The even numbers mentioned above are limited to odd $x_i$ from the same Collatz sequence. – Collag3n Mar 19 '20 at 20:24
  • Very good! I will crunch the sequences tomorrow with Python and try to find out more. That's a great new inspiration :-) – c4ristian Mar 19 '20 at 20:29
  • Some more interesting things: the even number obtained with the OEIS A005184 sequence above are of the form $p\cdot2^k$ with p prime....at least for the ones in the list – Collag3n Mar 19 '20 at 20:33
  • I had a closer look to the self-containedness condition. I studied the examples $e_0 = {31, 83, 293, 347, 671}$. All products lead to natural numbers that have the form $p * 2^\alpha$ where $p$ is prime and $\alpha$ is the number of divisions by 2 that is performed to get from $e_0$ to $e_{n+1}$. For the mentioned examples we obtain $p = {5, 13, 7, 7, 11}$ and $\alpha = {12, 36, 48, 48, 33}$ – c4ristian Mar 21 '20 at 19:02
  • Furthermore, I explored the self-containedness condition for known cycles for $3n+1$ (there is only one) and the generalized form $kn+1$. Obviously, their product results in natural numbers of the form $1 * 2^\alpha$. The list of cycles can be found in our working paper. In our study we empirically discovered that cycles only occur, if the condition $\alpha = \lfloor n * log_23\rfloor + 1$ is met. That is very similar to the discussion @Collag3n has linked above. Exciting :-) – c4ristian Mar 21 '20 at 19:09
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Interestingly the sequence provided by @Collag3n contains the Engel expansion: 31, 47, 71, 107, 161 which is given by $\frac{3^n(x_1+1)-2^n}{2^n}$

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    Any odd integer $x$ can be written in the form $a\cdot2^i-1$ and reaches $a\cdot3^i-1$ by applying $i$ times the collatz function $f(x)=\frac{3x+1}{2}$. A shortcut to this is to add 1 to $x_1$ and multiply by $(\frac{3}{2})^n$ with $n<=i$, than substract 1 (which is equivalent to your equation). Gottfried Helms knows collatz continued fractions better than me. Perhaps he might comment on this....if he comes accross this post – Collag3n Mar 19 '20 at 19:49
  • Great - thank you. We found something interesting in this context: The condition for cycles in a Collatz sequence $v_1,v_2,\ldots,v_{n+1}$ can be more restricted when proving that the product $\prod_{i=1}^{n+1}(1+\frac{1}{v_i})$ is limited. I posted a related question some minutes ago. –  Mar 21 '20 at 15:02
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Far from being a full answer, but worth stating. If $b$ is natural then it is divisible by $2^n$. Moreover, if one of the $x_i's$ is divisible by $3$, then $b$ is not natural. Both of these are obvious when writing $b$ as $b=\frac{(3x_{1}+1)\cdot(3x_{2}+1)\cdot...\cdot(3x_{n}+1)}{x_{1}\cdot x_{2}\cdot...\cdot x_{n}}$

Al-Hasan Ibn Al-Hasan
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