Query on $3^x \pm 2^{x-a}$ with relation to prime and semiprime

Question

While doing some research(more closer to some playing) with the formula $3^x\pm2^{x-a}$ for $x\in \mathbb{N}$ and $ \{a\mid a \in \mathbb{Z}_{\geq 0},\hspace{1mm} a\le (x-1)\}$ I've become to observe that there are many primes and semiprimes for each $x$ of the formula.

For example when $x=22$, we can find that $3^{22}-2^{x-a}$ being semiprime for $(x-a)\in\{21,19,9\}$, and $3^{22}+2^{x-a}$ being semiprime for $(x-a)\in\{22,20,18,17,15,14,13,8,6,5\}$.

Also we can find that $3^{22}-2^{15}$, $3^{22}-2^{7}$, $3^{22}-2^{3}$,$3^{22}-2^{1}$, $3^{22}+2^{2}$, $3^{22}+2^{19}$ being prime when $x=22$.

So I've come to expect the following two statements.

1. For $x\in \mathbb{N}$ and $ \{a\mid a \in \mathbb{Z}_{\geq 0},\hspace{1mm} a\le (x-1)\}$, there are at least one semiprime for every $x$ among each set of $\{semiprime\in 3^x-2^{x-a}\}$ and $\{semiprime\in 3^x+2^{x-a}\}$.

2. For $x\in \mathbb{N}$ and $ \{a\mid a \in \mathbb{Z}_{\geq 0},\hspace{1mm} a\le (x-1)\}$, there are at least one prime for every $x$ among the set of $\{\mathbb{P} \in 3^x\pm2^{x-a}\}$.

I've seen a lot of people who have astonishing mathematical abilities in this site so I've become to ask for some opinions with the statements above, of which I wouldn't be able to handle them since they are out of my reach otherwise.

Answer 1

The semiprime-version is promising. I arrived at the following conjecture

Suppose , $\ n\ge 4\ $ is an integer. Then

• there is an integer $\ m\ $ with $\ 0<m<n\ $ such that $\ 3^n\cdot 2^m-1\ $ is a semiprime.
• there is an integer $\ m\ $ with $\ 0<m<n\ $ such that $\ 3^n\cdot 2^m+1\ $ is a semiprime. If we allow also $\ m=0\ $ , the only counterexample for positive integer $\ n\ $ is the minus-case for $\ n=1\ $

The conjecture is true upto $\ n=100\ $

The smallest solutions are listed below.

gp > for(n=4,100,m=1;while(bigomega(3^n*2^m-1)<>2,m=m+1);print(n,"  ",m))

4 1 5 1 6 1 7 3 8 3 9 1 10 1 11 5 12 5 13 1 14 1 15 4 16 3 17 2 18 1 19 1 20 6 21 1 22 3 23 2 24 1 25 3 26 7 27 2 28 1 29 2 30 1 31 5 32 1 33 2 34 15 35 3 36 1 37 7 38 1 39 1 40 9 41 5 42 19 43 1 44 9 45 1 46 11 47 3 48 5 49 2 50 13 51 4 52 25 53 2 54 1 55 1 56 7 57 2 58 23 59 2 60 7 61 2 62 27 63 1 64 3 65 3 66 13 67 2 68 7 69 2 70 29 71 1 72 5 73 3 74 3 75 14 76 11 77 2 78 1 79 4 80 1 81 20 82 23 83 9 84 1 85 4 86 7 87 1 88 5 89 1 90 1 91 1 92 9 93 7 94 11 95 8 96 55 97 2 98 3 99 14 100 29 gp >

gp > for(n=4,100,m=1;while(bigomega(3^n*2^m+1)<>2,m=m+1);print(n,"  ",m))

4 3 5 2 6 3 7 2 8 1 9 2 10 1 11 3 12 1 13 8 14 1 15 1 16 3 17 2 18 1 19 2 20 3 21 4 22 1 23 2 24 5 25 2 26 1 27 2 28 3 29 4 30 7 31 10 32 1 33 7 34 1 35 3 36 23 37 6 38 5 39 7 40 11 41 1 42 2 43 5 44 7 45 1 46 1 47 6 48 1 49 24 50 1 51 4 52 8 53 3 54 10 55 18 56 1 57 5 58 10 59 1 60 4 61 5 62 1 63 2 64 1 65 4 66 13 67 3 68 1 69 1 70 1 71 1 72 4 73 5 74 7 75 2 76 1 77 13 78 2 79 5 80 8 81 2 82 1 83 7 84 1 85 8 86 6 87 2 88 5 89 1 90 17 91 31 92 1 93 20 94 7 95 3 96 15 97 12 98 25 99 1 100 3 gp >

To check larger $\ n\ $ , I have to program a more efficient version to find a solution. I will update the post , if I have arrived at a new milestone.