How can I prove that $b = \prod_{i=1}^n (3 + \frac{1}{x_i})$ results not in an even, natural number, except for $n=1$ .and $x1 = 1$

Question

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.

Answer 1

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

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

Answer 2

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 $b=\frac{(3x_{1}+1)\cdot(3x_{2}+1)\cdot...\cdot(3x_{n}+1)}{x_{1}\cdot x_{2}\cdot...\cdot x_{n}}$

Answer 3

Interestingly the sequence provided by @Collag3n contains the Engel expansion: 31, 47, 71, 107, 161 which is given by $\frac{3^n(x_1+1)-2^n}{2^n}$