Are Mersenne numbers $M_p$ deficient? A positive integer $n$ is called deficient if $\sigma(n)<2n$, i.e., the sum of divisors is less than $2n$. What is known about Mersenne numbers $n=2^p-1$ with $p$ prime in this respect ? Is there a counterexample to $$ \sigma(2^p-1)<2^{p+1}-2 \; $$ for a prime $p$ ?

For $p\ge 3$, let $q$ be a prime dividing $n=2^p-1$, and let $\omega = \operatorname{ord}_q 2$. Then $\omega\
e 1, \omega|q-1$ and $p = k(q-1)+\omega$ for some $k$. But since $p$ is prime we must have $$ k=0, \quad p=\omega, \quad p|q-1, \quad q>p $$ Let $$ n = \prod q_i^{\alpha_i} $$ be the prime factorization of $n$, with each $q_i>p$. Then $$ \begin{align} \frac{\sigma(n)}{n} & = \prod \frac{\sigma(q_i^{\alpha_i})}{q_i^{\alpha_i}} \\\ &\le \prod \frac{\sigma(q_i)^{\alpha_i}}{q_i^{\alpha_i}}\\\ &= \prod \left(1+\frac{1}{q_i}\right)^{\alpha_i} \\\ &< \left(1+\frac{1}{p}\right)^A \end{align} $$ where $A=\sum \alpha_i$. From $n>p^A$ we get $A < \log_p n < p\log_p 2$ and hence $$ \begin{align} \frac{\sigma(n)}{n} &< \left[\left(1+\frac{1}{p}\right)^p\right]^{\log 2/\log p} \\\ & < e^{\log 2/\log p} \\\ &= 2^{1/\log p} < 2 \end{align} $$ So $n$ is deficient when $p\ge 3$, and we can also confirm by inspection for $p=2$.