Can it be proven that the reduced digit sum of Mersenne primes, except 3 and 7, is always 1 or 4?

Question

For the first 36 Mersenne numbers (M(n)) I have calculated M(n)(mod 9) and the reduced digit sum of M(n). For n > 0, I found repeating sequences of respectively 1,3,7,6,4,0 and 1,3,7,6,4,9. Also, I observed that, within the given range, all Mersenne primes, for n > 3, have a digit sum of 1 or 4. Is this true for every Mersenne prime with n > 3?

Answer 1

The "reduced digit sum" of a number is the remainder when dividing by $9$ (replacing $0$ with $9$).

The powers of $2$ modulo $9$ form the repeating pattern $$2\mapsto 4\mapsto 8\mapsto 7\mapsto 5\mapsto 1 \mapsto 2\mapsto\cdots $$ So the Mersenne numbers modulo $9$ form the repeating pattern $$1 \mapsto 3\mapsto 7\mapsto 6\mapsto 4\mapsto 9\mapsto 1\mapsto\cdots.$$

A Mersenne prime is of the form $M(p)= 2^p-1$. For $p\gt 2$, $p$ is odd. So the remainders must be $1$, $7$, or $4$. You are asking if you can ever get a $7$.

The cycle has length six; so $M(n)\equiv 7\pmod{9}$ exactly when $n\equiv 3\pmod{6}$. If $n\equiv 3\pmod{6}$, and $n\gt 3$, then $n$ is not prime (it is divisible by $3$). In particular, if $M(n)\equiv 7\pmod{9}$, then $n\equiv 3\pmod{6}$, so $n$ is not prime, so $M(n)$ is not a Mersenne prime.