# Does $(10^{6n+5}-54n-46)/162$ ever be a prime for $n>0$?

$(10^{6n+5}-54n-46)/162$ 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 $n>0$ ?

• Better to ask on the Math stack exchange.
– MikeY
Jun 24 '17 at 13:46
• I hope you were only testing even $n$, because $$\frac{10^{6n + 5} - 54n - 46}{162}$$ is even for odd $n$. Something else to take note of, if you haven't already had Mathematica spit out a few values for you: the nine digits 617283950 are repeated almost as many times as $6n + 5$ will allow. However, the three least significant digits make this problem delightfully more difficult.
– Lisa
Jun 24 '17 at 20:18
• They're never multiples of $11$. That's all I can tell. Hoping someone cleverer than I comes along to shed light on this. Jun 26 '17 at 20:24
• Unless somebody finds a factorization, it is very likely that there is a prime in the series. If we use the heuristic that $N$ is prime with probability $\frac 1{\log N}$, your number is prime with probability about $\frac 1{13.8n}$. The sum of this diverges, so there ought to be primes there. Jun 28 '17 at 17:05
• What is the motivation behind this question? How did you come up with this formula?
– Paul
Jun 30 '17 at 20:27

WolframAlpha says that $\frac{10^{6n + 5} - 54n - 46}{162}$ is a semiprime for $n=8$:
$$617283950617283950617283950617283950617283950617281\\ =\\ 3186349113015533 \times 193727657806175483856523861962184357$$
• it is also a semiprime for $n=2$ wolframalpha.com/input/… Jun 28 '17 at 19:36
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 ...