$\prod\limits_{i=1}^n \frac{3a_i+2}{2a_i+1}, a_i\geq 1$

Question

$$\prod\limits_{i=1}^n \frac{3a_i+2}{2a_i+1}, a_i\geq 1$$ Claim: This product is never an integer ($a_i$ integer).

Answer 1

It's false: $$ \frac{29}{19}\frac{38}{25}\frac{44}{29}\frac{83}{55}\frac{125}{83}=8 $$

Answer 2

Some more counterexamples: $$ \frac{20}{13} \frac{ 38}{25} \frac{ 83}{55} \frac{ 125}{83} \frac{143}{95} = 8$$

$$\frac{8}{5}{\frac {65}{43}}{\frac {74}{49}}{\frac {83}{55}}{\frac {98}{65 }}{\frac {125}{83}}{\frac {143}{95}}{\frac {209}{139}}{\frac {215} {143}}{\frac {278}{185}}=64$$

$${\frac {11}{7}}{\frac {38}{25}}{\frac {65}{43}}{\frac {83}{55}}{ \frac {98}{65}}{\frac {125}{83}}{\frac {143}{95}}{\frac {215}{143}} =28$$

$${\frac {17}{11}}{\frac {20}{13}}{\frac {26}{17}}{\frac {74}{49}}{ \frac {83}{55}}{\frac {98}{65}}{\frac {125}{83}}{\frac {143}{95}}{ \frac {209}{139}}{\frac {278}{185}}=64$$

EDIT: In general, given a finite set of fractions $f_j$, consider the set $P$ of primes that divide their denominators. If the $p$-adic order of $f_j$ is $\nu_p(f_j)$, the condition for $\prod_j f_j$ to be an integer is $\sum_j \nu_p(f_j) \ge 0$ for all $p \in P$. Let $x_j$ be binary variables (i.e. possible values $0$ and $1$). Then $\prod_j f_j^{x_j}$ is an integer iff $\sum_j \nu_p(f_j)\; x_j \ge 0$ for all $p \in P$. So what we want to do is find solutions to the system of linear inequalities $$ \sum_j \nu_p(f_j)\; x_j \ge 0, \ p \in P $$ $$ \sum_j x_j \ge 1 $$ $$ x_j \in \{0,1\}$$ Integer linear programming software can be used for this (I used the Optimization package in Maple).