Sum of Digits is Prime

Sum of Digits is Prime

August 19, 2014

For an integer $q\geqslant2$ let $s_q(n)$ denote the $q$-ary sum-of-digits function of a non-negative integer $n$, that is, if $n$ is given by its $q$-ary digits expansion $n=\sum\limits_{k=0}^{r} a_k q^{k}$ with digits $a_k\in\{0,1,\ldots,q-1\}$ and $a_r\neq0$, then
\[s_q(n)=\sum_{k=0}^{r} a_k.\]

As usual we write $\phi(n)$ for Euler's totient function, and $\pi(x)$ for the number of primes up to $x$. We recall the prime number theorem in the form

\begin{equation}\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{(\log x)^2}\right).\end{equation}

Theorem (Drmota/Mauduit/Rivat) We have uniformly for all integers $k\geqslant 0$ with $(k,q-1)=1$,

\begin{equation}\label{eq:2}\#\{p\leqslant x:s_q(p)=k\}=\frac{q-1}{\phi(q)}\frac{\pi(x)}{\sqrt{22\pi\sigma_{q}^{2}\log_q x}}\left(\exp\left(-\frac{(k-\mu_q\log_q x)^2}{2\sigma_{q}^{2}\log_q x}\right)+O\big((\log x)^{-1/2+\varepsilon}\big)\right),\end{equation}

where $\varepsilon>0$ is arbitrary but fixed, and $\mu_q:=\frac{q-1}{2}$, $\sigma_{q}^{2}:=\frac{q^2-1}{12}$.

From the above result we can deduce some very interesting corollaries. Clearly, \eqref{eq:2} shows that every sufficiently large integer is the sum of digits of a prime number. So, in particular, there are infinitely many primes whose sum of digits is also prime. Also, every sufficiently large prime is the sum of digits of another prime which in turn is the sum of digits of another prime, and so on.

References

C. Mauduit and J. Rivat, Sur un problème de Gelfond: la somme des chiffres des nombres premiers, Ann. of Math. 171 (2010), 1591–1646.

Glyn Harman, Counting Primes whose Sum of Digits is Prime, Journal of Integer Sequences, Vol. 15 (2012), Article 12.2.2.