**NOTATION**: $O\_x$ -- the product of all odd primes $\le x$.


E.g. $O\_7=3\cdot 5\cdot 7 = 105$.


**QUESTION**: Are the three ordered pairs $\ (d\ p)=(1\ 3)\ \ (2\ 3)\ \ (4\ 5)\ $ the only solutions of the equation:
$$|O\_p-2^d|=1$$
in natural numbers $d$, and odd primes $p$?


(I don't know an answer).


**MOTIVATION**: Let $\ s\ $ be a prime just after $\ p$, so that $\ p<s$. If
$$|O\_p-2^d|\ne 1$$
then
$$|O\_p-2^d|\ge s$$
Furthermore, sometimes $\ s\ $ can be quite a bit larger than $\ p$.


**ACKNOWLEDGEMENT**: Bjørn Kjos-Hanssen has provided one of the above solutions (see his comment below).



Powers of $2$ and the products of initial odd primes
There is no solution for $O\_n=2^d-1$ with $n \geq 7$. 


If 5 divides $2^d-1$ and 7 divides $2^d-1$, then 9 divides $2^d-1$. [Because 4 divides $d$ and 3 divides $d$; 6 divides $d$ and hence $2^{d}-1$ is divisible by 9.]


