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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?

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  • $\begingroup$ @Peter any idea? $\endgroup$
    – user623145
    Commented Mar 28, 2019 at 11:40
  • $\begingroup$ 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 $6j-1$. $\endgroup$
    – user25406
    Commented Mar 28, 2019 at 13:24
  • $\begingroup$ There are no more examples among the first million of primes. $\endgroup$ Commented Mar 28, 2019 at 15:37
  • $\begingroup$ @why? any reason? $\endgroup$
    – user623145
    Commented Mar 28, 2019 at 16:34

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Based on the tabulated results below, there don't seem to be any solutions with consecutive prime triplets. However, for non-consecutive primes, the immediate triples are satisfied for n = 14, 19, 22, 26, 31, 34, 35, 39 and 40... enter image description here

... the remaining cases can be found under factorization.

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