2

The "ec"-numbers (named after Enzo Creti) are defined as $$(2^{n+1}-1)\cdot 10^m+2^n-1$$ where $m$ is the number of digits in the decimal expansion of $2^n-1$. Or shorter, we concatenate the Mersenne numbers $M(n+1)$ and $M(n)$ (where $M(n)=2^n-1$)

Which primes $p>3$ do never divide an "ec"-number ?

I determined the orders of $2$ and $10$ modulo $p$ and just calculated all the possible values, but this method is quite time-consuming. I found out that $$1321,3191,3541$$ are the primes $p$ in the range $[5,10^4]$ with the desired property. Can I efficiently find more such primes (or even all) ?

Peter
  • 86,576
  • 1
    Note that the decimal expansion of $2^n-1$ has the same number of digits as $2^n$, and thus $$m=\left\lfloor n\cdot\frac{\log(2)}{\log(10)}\right\rfloor$$ –  Apr 09 '18 at 20:59
  • https://math.stackexchange.com/questions/2635516/a-conjecture-about-numbers-of-the-form-10m2k%e2%88%9212k-1%e2%88%921-where-m-is/2636195#2636195 – Peter Apr 09 '18 at 21:02

0 Answers0