# Triples of consecutive primes whose product is equal to $3\cdot2^k+1$

Consider triples of consecutive primes whose product is equal to $$3\cdot2^k+1$$, for some positive integer $$k$$. As an example I found:

$$3\cdot2^7+1=5\cdot7\cdot11$$.

This example is wonderful because $$k=7$$ is prime and $$7$$ is the prime just in the middle of the triple of the three consecutive primes.

Any other examples, in particular with k prime and $$k$$ in the middle of the triple? more generally are there other solutions to $$(2^a)\cdot(3^b)+1=p_n\cdot p_{n+1}\cdot p_{n+2}$$ with a and b non negative integers and p primes?

• @Peter any idea?
– user623145
Commented Mar 28, 2019 at 11:40
• you may want to rewrite $3*2^k+1$ as $6*2^{k-1}+1=6m+1$. There are only 2 ways to get a $6m+1$ out of a product of $3$ primes, either the $3$ must be of the form $6j+1$ or we must have $2$ of them of the form $6j-1$. Commented Mar 28, 2019 at 13:24
• There are no more examples among the first million of primes. Commented Mar 28, 2019 at 15:37
• @why? any reason?
– user623145
Commented Mar 28, 2019 at 16:34