Let $s(n)$ have $n$ ones. $\ n\equiv 1\pmod{3}\,\Rightarrow\, 3\mid s(n)$. $n\equiv 2\pmod{3}\,\Rightarrow\, 37\mid s(n)$ because $37\mid 3811$ and $37\mid 111$, so $37\mid 3811\cdot 1000+111$ and so on. $n\equiv 0\pmod{3}\,\Rightarrow\, 2\underbrace{33\cdots 3}_{n/3\text{ threes}}\mid s(n)$, but I don't have a proof.

$n=3k \implies 2\underbrace{3\cdots 3}_{k\text{ threes}}\mid s(n)$ $s(n)=2\underbrace{3\cdots3}_{k\text{ 3's}}\text{ }\times\text{ }16\underbrace{3\cdots3}_{k-1\text{ 3's}}5\underbrace{6\cdots6}_{k-1\text{ 6's}}7$

My solution : Let $a_n=381....1$ have $n$ ones To solve the problem we three cases : Case 1 If $\ n\equiv 1\pmod{3}$ it is easy too see that $3\mid a_n$ Case 2 If $\ n\equiv 2\pmod{3}$ it is easy too see that $37\mid a_n$ Case 3 If $\ n\equiv 0\pmod{3}$ and $n=3x$ $a_n=3.10^{3x+1}+8.10^{3x}+\frac{10^{3x}-1}{9}=p$ where $p$ is prime and $n=3x$ Then $9a_n=9p=343.10^{3x}-1$ $\Longrightarrow$ $9p=(7.10^{x}-1)(49.10^{2x}+7.10^{x}+1)$ then is easy to see that : $p\mid 7.10^{x}-1$ or $p\mid 49.10^{2x}+7.10^{x}+1$.If $p\mid 7.10^{x}-1$ $\Longrightarrow$ is easy too see that $49.10^{2x}+7.10^{x}+1=1,3$ or $9$ which is contradiction similary if $p\mid 49.10^{2x}+7.10^{x}+1$ $\Longrightarrow$ $7.10^{x}-1=1,3$ or $9$ which is easily contradiction .Therefore there is no $a_n$ which is a prime number.

Best solution: $38\underbrace{11\dots1}_{3k\text{ }1\text{s}}=2\underbrace{33\dots3}_{k}\cdot16\underbrace{33\dots3}_{k-1\text{ }3\text{s}}5\underbrace{66\dots6}_{k-1\text{ }3\text{s}}7$ $38\underbrace{11\dots1}_{3k+1\text{ }1\text{s}}=3\cdot127\underbrace{037037\dots037}_{k\text{ }037\text{s}}$ $38\underbrace{11\dots1}_{3k+2\text{ }1\text{s}}=37\cdot103\underbrace{003003\dots003}_{k\text{ }003\text{s}}$

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