We define recursively $p_1=2,p_2=3$ and $$p_{n}= \min_{F_{n-1}}|A-B| $$ Where $F_{n-1}=((A,B) |\gcd (A,B)=1,A,B $ are products of powers of $p_i$s $\\ \mathrm{and} \\ ( p_i |A \mathrm{\\ or\}\\ B) \\ 1 \leq i \leq n-1 , |A-B| \not =1 )$

Is always $p_n$ the $n-th$ prime?

We define recursively $p_1=2,p_2=3$ and $$p_{n}= \min_{(A,B)\in F_{n-1}}|A-B| $$ Where $$ \begin{split} F_n=\{(A,B) |&\gcd (A,B)=1,\quad |A-B| \not =1, \\\ &\text{both $A$ and $B$ are products of powers of $p_i$ for $i\le n$}, \\\ &\text{for each $i\le n$, either $p_i |A$ or $p_i |B$}\} \end{split} $$

Is always $p_n$ the $n-th$ prime?

We define recursively $p_1=2,p_2=3$ and $$p_{n}= \min_{F_{n-1}}|A-B| $$ Where $F_{n-1}=((A,B) |\gcd (A,B)=1 \\ \mathrm{and} \\ ( p_i |A \mathrm{\\ or\}\\ B) \\ 1 \leq i \leq n-1 , |A-B| \not =1 )$

Is always $p_n$ the $n-th$ prime?

We define recursively $p_1=2,p_2=3$ and $$p_{n}= \min_{F_{n-1}}|A-B| $$ Where $F_{n-1}=((A,B) |\gcd (A,B)=1,A,B $ are products of powers of $p_i$s $\\ \mathrm{and} \\ ( p_i |A \mathrm{\\ or\}\\ B) \\ 1 \leq i \leq n-1 , |A-B| \not =1 )$

Is always $p_n$ the $n-th$ prime?