I recently wrote about a problem I cam up with while thinking about number theory, which you can find on this post. Long story short, I'm trying to prove there are infinite natural numbers such that their divisors count ($1$ and $n$ included) equals the sum of their digits.
Among the many useful comments and answers, I read about a really clever way of proving the infinite amount of such numbers (if the following statement is true), which is by proving that there are an infinite amount of primes $p_i$ with digit sum $8$.
In that case, $10p$ will have eight divisors and will have digit sum eight.
I thought of some examples of such numbers, and I found a couple:
- made of eight ones and n zeros (starting with 1 and ending with 1) (e.g. $1001110110101$ - I don't know whether this specific one is or not a prime, it's just an example)
- $10...07$
- $70...01$
- $50...03$
I thought about using Dirichlet theorem, but it states that there are infinite primes in the form:
$a + nb$
With $gcd(a, b) = 1$ and $n\in \mathbb{Z^+}$, but I don't think it's feasible because it doesn't include all possible values of n.
Does anybody know something I can work on, or he can help with?