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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}$.

the ec-primes are introduced. They emerge by concatenating the Mersenne numbers $M_{n+1}$ and $M_n$ for $n\ge 1$ (where $M_n=2^n-1$ is the $n$ th Mersenne number)

If we define $$ec(n):=(2^n-1)\cdot 10^m+2^{n-1}-1$$ where $m$ is the number of digits in the decimal expansion of $2^{n-1}-1$ , hence then $n$-th ec-number and $$M(n):=2^n-1$$ the $n$-th Mersenne number, we can compare the number of ec-primes for $n\le L$ and the number of Mersenne primes for $n\le L$ , where $L$ is a very large limit.

Who will be the winner in the long run ? Which kind of numbers will produce more primes in the long run ?

I want to assume that a random number $n$ is prime with probability $\frac{1}{\ln(n)}$, but my problem is how to consider the structure of the Mersenne-primes. For the ec-primes, we should have about $5.48$ primes in every exponent-range $[10^k,10^{k+1}]$ , $k\ge 3$. But how many Mersenne-primes can we expect in the range $[10^k,10^{k+1}]$ considering that $M_n$ can only be prime for prime $n$ and that then, every factor is of the form $2kn+1$ ?

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  • $\begingroup$ We don't even know that there are infinitely many Mersenne primes, so it's unlikely this question can be answered without solving some open problem. $\endgroup$ – user334732 Dec 17 '18 at 2:24

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