A conjecture about numbers of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$.

Question

Numbers $n$ of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. For example:

• $k=1$ then $n=10$.
• $k=2$ then $n=31$.
• $k=3$ then $n=73$.
• $k=4$ then $n=157.$

Conjecture:

the number $(2^k-1)\cdot 10^m+2^{k-1}-1$ where $m$ is the number of decimal digits of $2^{k-1}$ is never prime when it is of the form $7s+6$, that is when it is congruent to $6$ $\pmod 7$. Examples: $n=1023511$ ($k=10$)$\equiv 6 \pmod 7$ and thus it is composite $(1023511=19\times103\times523)$, $n=20471023$ ($k=11$) $\equiv 6 \pmod 7$ and thus it is composite ($20471023=479\times42737)$. With PFGW we arrived to $k=565000$ and all the $n's$ congruent to $6 \pmod 7$ are composite. According to Giovanni Resta's calculations in a post which has been canceled, there should be no probable prime congruent to 6 $\pmod 7$ upto k=800.000. The residue $6$ $\pmod 7$ occurs when either $m=6t+3$ and $k=3l+1$ or $m=6t+4$ and $k=3l+2$ with $k$ and $l$ some non-negative integers, but amazingly when it occurs the number is not prime. Can you find a counter-example or give a proof for the conjecture? Here a link to other interesting questions: Is there a number of the form $f(n)=7k+6=5p$ with prime p? and Why do all residues occur in this similar sequence? For primes of this form see: The on-line Encyclopedia of integer sequences The following vector contains all the exponents k<=366800 leading to a prime

$[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$

Exponent $541456$ leads to another probable prime with residue 5 mod 7 and 325990 digits, but it need not be the next in increasing order.

Remark: we found five-in-a-row probable primes with res 5 mod 7. Probable primes with residue 5 are now twice frequent than expected. Exponents of these primes seem to be NOT random at all. Another thing I noticed, i don't know if it has some importance: the exponents leading to a probable prime $215, 69660, 92020, 541456$ are multiples of $43$. I noticed that $\frac{215}{41}, \frac{69660}{41}, \frac{92020}{41}, \frac{541456}{41}$ all have a periodic decimal expansion equal to $\overline{24390}=29^3+1$. This is equivalent to say that when k is a multiple of 43 and the number $10^{m}(2^{k}−1)+2^{k-1}−1$ is prime, then k is of the form $41s+r$ where r is a number in the set (1,10,16,18,37). Is there some mathematical reason for that?

Solution 1:

According to your list, a counter-example, if it exists, must have more than $60,000$ digits. So, a counterexample would be a quite gigantic prime.

Unfortunately, a proof of the conjecture will almost certainly be out of reach.

The search for a counter-example can be painful as well, it is well possible that the smallest is already too big for current algorithms for primality testing.

Solution 2:

@peter told me about this and i found it very interesting.
I made this tool, so anybody can help computing. Simply download and run to participate. Just refresh the page to update stats.

With 30-40 person, getting to $n=10^7$ should not be to long.