Hello, I discovered something about octal prime numbers. I don't know if anyone has dealt with this before, but thanks to the power of the D programming language, it was very easy. So, by defining a range with the empty(), front() and popFront() functions, I produced an output, something like this:
>~ dmd main.d -ofmain.out
~ ./main.out
00000000000000000005: 5, 0
00000000000000000045: 37, 1
00000000000000000445: 293, 2
00000000000000004445: 2341, 3
00000000000044444445: 9586981, 7
00000000004444444445: 613566757, 9
00000000044444444445: 4908534053, 10
44444444444444444445: 658812288346769701, 19
...
4...4444444444444445: ? (hint: greater than 100 bits)
In order, they are as follows "octal: decimal, and number of repetitions (¹⁰¹) and they are all prime 8 numbers! So what's the ninth? I'm not sharing the code for now because it's a challenge. But you should use std.bigint module and simply shift left and add octal 4 or binary 101.
SDB@79