I’ve noticed some curious patterns in excellent numbers. This relates to the Perl stuff I wrote about earlier, but this post isn’t about the Perl.

Ignoring the spaces which I use to show the pattern, each of these numbers are excellent. The pattern pivots on a 4 which can have zero or more 3s on the left side and the same number of 6s on the right side. End it all with an 8. The right half is twice the left half. I conjecture that every series of excellent numbers of a particular length has such a number. I’ll figure out the proof later:

4 8 3 4 6 8 33 4 66 8 333 4 666 8 3333 4 6666 8 33333 4 66666 8 333333 4 666666 8 3333333 4 6666666 8 33333333 4 66666666 8 333333333 4 666666666 8 3333333333 4 6666666666 8

Mark Jason Dominus proved that 48, 484848, and all other odd number of repetitions of 48 are also excellent. I think 48 is just a special case with zero 3s and 6s.

But, it turns out so far that any excellent number repeated an odd number of times is excellent. So, 346834683468 (n=12) is also excellent. So is 140400140400140400 (n=18).

This means that I can look at the number of digits and make a guess about how many excellent numbers I find. I take the factorization of number and check how many combinations of an even and odd number I can make:

2 → 2 * 1 4 → 4 * 1 6 → 6 * 1, 2 * 3 8 → 8 * 1 10 → 10 * 1, 2 * 5 12 → 12 * 1, 4 * 3 14 → 14 * 1, 2 * 7 16 → 16 * 1 18 → 18 * 1, 2 * 9, 6 * 3 20 → 20 * 1, 4 * 5 22 → 22 * 1, 2 * 11 24 → 24 * 1, 8 * 3 26 → 26 * 1, 2 * 13 28 → 28 * 1, 4 * 7 30 → 30 * 1, 2 * 15, 6 * 5, 10 * 3

There are many 28 numbers with 18 digits that are excellent. That’s the highest count for a particular number of digits so far (up to n=22). But look at it’s factorization. The number 18 can two factorizations with an even and odd number. The sequence for 18-digit numbers includes all the 6-digit numbers repeated three times. Look at 30 digits. It will repeat the 2-digit, 6-digit, and 10-digit numbers.

There’s probably a proof for that too, but I haven’t worked it out just yet.

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