TITLE: Does $(10^{6n+5}-54n-46)/162$ ever be a prime for $n>0$?
QUESTION [13 upvotes]: $(10^{6n+5}-54n-46)/162$ is prime for $n=0$, but after that I didn't find anymore prime up to $n=500$, I didn't even find a single semiprime with that form (!!). Is there prime number of such form for $n>0$ ?

REPLY [2 votes]: Incomplete answer with a factorisation (hopefully somebody could extend it) and some preliminary results ...
From:
$$10^{6n+5}-54n-46=10^{6n+5}-9\cdot6n-9\cdot5-1=10^{6n+5}-9(6n+5)-1=\\
10^{6n+5}-10-9(6n+5)+9$$
noting $p=6n+5$:
$$10^{p}-10-9(p-1)=10(10^{p-1}-1)-9(p-1)=\\ 
9\left(10\cdot(10^{p-2}+10^{p-3}+...+10+1)-(p-1)\right)=\\
9\left(10^{p-1}-1+10^{p-2}-1+...+10^2-1+10-1\right)=\\
81\left((10^{p-2}+10^{p-3}+...+10+1)+(10^{p-3}+10^{p-4}+...+10+1)+...+(10+1)+1\right)=...$$
the sum in brackets is the sum of repunits, which appear in other interesting problems like this one
$$...=81\left(10^{p-2}+2\cdot 10^{p-3}+3\cdot 10^{p-4}+...+(p-2)\cdot 10+(p-1)\right)=\\
162\left(2^{6n+2}5^{6n+3}+2\cdot 2^{6n+1}5^{6n+2}+3\cdot 2^{6n}5^{6n+1}+...+(6n+3)\cdot 5+(3n+2)\right)=...$$
Obviously, for odd $n$ the expression in brackets is divisible by $2$ and can not be prime. So the remaining case is $n=2k$
$$...=162\left(2^{12k+2}5^{12k+3}+2\cdot 2^{12k+1}5^{12k+2}+3\cdot 2^{12k}5^{12k+1}+...+(12k+3)\cdot 5+(6k+2)\right)$$
If $k=5t+3$ (or $n=10t+6$) then 
$$(12k+3)\cdot 5+(6k+2)=5(66t+43)$$
and the expression in brackets is divisible by $5$ and can not be prime. 
So far, from $10$ possible cases $n=10t, n=10t+1, ...$ and $n=10t+9$, $6$ are addressed ...