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For which $n$ is the sum:

$$\sum_{k=0}^{n}10^k$$

a prime number? Are they finite?

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After adding $1$, the $n$s for which the sum above is known to be prime or probably prime are given in the OEIS sequence A004023. (The $+1$ is because the OEIS lists numbers $n$ such that $(10^n-1)/9$ is prime, but the sum in the question is instead equal to $(10^{n+1}-1)/9$.) Also, see the Wikipedia article and the Prime Pages entry.

These primes are called “repunit primes” since their decimal expansion consists of a repeated series of $1$s. The repunits corresponding to

$$

n=1, 18, 22, 316, \text{and } 1030

$$

$$\text{ (using the $n$ in the questioner’s formula above)}

$$

are known to be prime. The repunits corresponding to

$$

n=49080, 86452, 109296, \text{and } 270342

$$

$$\text{ (again using the $n$ in the questioner’s formula above)}

$$

are as far as I know only known to be probably prime.

The obvious factorization

$$

\frac{10^{km}-1}{9}=\frac{10^k-1}{9}(1+10^k+\cdots+10^{k(m-1)})

$$

means that, for $(10^{n+1}-1)/9$ to be prime, $n+1$ must also be prime.

There are conjectured to be infinitely many repunit primes.

**Partial solution:**

Given a prime $p \neq 2, 5$, let $n = Q(p – 1) + R$. We use Fermat’s Little Theorem modulo $p$:

$$\begin{align}

\sum_{k = 0}^n10^k &= 1 + \sum_{i = 1}^Q10^{(i – 1)(p – 1)}(10^1 + 10^2 + \dotsb + 10^{p – 1}) + 10^{Q(p – 1)}(10^1 + 10^2 + \dotsb + 10^R)\\

&\equiv 1 + \sum_{i = 1}^Q\frac{p(p – 1)}{2} + (10^1 + 10^2 + \dotsb + 10^R)\\

&\equiv \sum_{k = 0}^R10^k \pmod{p}.

\end{align}$$

The $p(p – 1)/2$ came from the fact that the residues of $10^1, 10^2, \dotsc, 10^{p – 1}$ modulo $p$ form a permutation of $1, 2, \dotsc, (p – 1)$.

Numbers having the shape `n = 1111......1`

have less probability to be a prime.

If sum of all digits are divisible by 3 then that number is divisible by 3. Using this rule If number of digits of `n`

is divisible by 3 then n is surely not a prime. $\therefore 111,111111,111111111, ….$ are divisible by 3

If number of digits are divisible by 2 (even number of digits in n), then n can be partitions into multiple 11s. e.g. `1111`

can be partitioned as `11 11`

. Which is divisible by `11`

$\therefore 11,1111,111111, …. $ are divisible by 11.

So if number of digits are not divisible by 2 or 3 (e.g. $5,7,11,13,17,19,23,25,29,31,35,37,….$) we are not obvious that the number is not prime. However for these numbers we get large prime divisors most of the time. Like `11111 = 41*271`

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