Fun with number representation - the factorial basis.

Consider this series:

0,1,10,11,20,21,100,101,110,111,120,121,200,201,210,211,220,221,300,301,310,311,320,321,1000,1001...

The kth digit can take on values 0...k. The last number before a turnover of digits is k.(k-1).(k-2)...321. Thus 7654321 + 1 = 10000000.

Now, it so happens that

100....00 (k zeros) is the representation of (k+1)! = (k+1)k(k-1)(k-2)..1 and by definition 0! = 1.

Thus a number in this representation can be written

A = sum(0..k) An * n!

where 0 <= An <= n.

What are the primes?

2 = 10
3 = 11
5 = 21
7 = 101
11 = 121
13 = 201
17 = 221
19 = 301
23 = 321
29 = 1021
31 = 1101
37 = 1201
41 = 1221

Here are all the primes up to 719 = 6!-1 ;

0 0 0 1 0
0 0 0 1 1
0 0 0 2 1
0 0 1 0 1
0 0 1 2 1
0 0 2 0 1
0 0 2 2 1
0 0 3 0 1
0 0 3 2 1
0 1 0 2 1
0 1 1 0 1
0 1 2 0 1
0 1 2 2 1
0 1 3 0 1
0 1 3 2 1
0 2 0 2 1
0 2 1 2 1
0 2 2 0 1
0 2 3 0 1
0 2 3 2 1
0 3 0 0 1
0 3 1 0 1
0 3 1 2 1
0 3 2 2 1
0 4 0 0 1
0 4 0 2 1
0 4 1 0 1
0 4 1 2 1
0 4 2 0 1
0 4 2 2 1
1 0 1 0 1
1 0 1 2 1
1 0 2 2 1
1 0 3 0 1
1 1 0 2 1
1 1 1 0 1
1 1 2 0 1
1 1 3 0 1
1 1 3 2 1
1 2 0 2 1
1 2 1 2 1
1 2 2 0 1
1 2 3 2 1
1 3 0 0 1
1 3 0 2 1
1 3 1 0 1
1 3 3 0 1
1 4 1 0 1
1 4 1 2 1
1 4 2 0 1
1 4 2 2 1
1 4 3 2 1
2 0 0 0 1
2 0 1 2 1
2 0 2 2 1
2 0 3 2 1
2 1 0 2 1
2 1 1 0 1
2 1 2 0 1
2 1 2 2 1
2 1 3 0 1
2 2 0 2 1
2 2 3 0 1
2 2 3 2 1
2 3 0 0 1
2 3 0 2 1
2 3 3 0 1
2 4 0 0 1
2 4 1 2 1
2 4 2 0 1
2 4 2 2 1
2 4 3 2 1
3 0 1 0 1
3 0 2 0 1
3 0 3 0 1
3 0 3 2 1
3 1 0 2 1
3 1 2 0 1
3 1 2 2 1
3 2 0 0 1
3 2 1 2 1
3 2 2 0 1
3 2 3 2 1
3 3 0 0 1
3 3 1 0 1
3 3 1 2 1
3 3 2 2 1
3 4 0 0 1
3 4 0 2 1
3 4 1 0 1
3 4 1 2 1
3 4 3 2 1
4 0 1 0 1
4 0 1 2 1
4 0 3 0 1
4 0 3 2 1
4 1 0 2 1
4 1 2 2 1
4 1 3 0 1
4 2 2 0 1
4 2 3 0 1
4 3 0 2 1
4 3 1 2 1
4 3 2 2 1
4 3 3 0 1
4 4 0 0 1
4 4 1 2 1
4 4 2 2 1
4 4 3 2 1
5 0 0 0 1
5 0 1 0 1
5 0 2 0 1
5 0 2 2 1
5 0 3 0 1
5 1 1 0 1
5 1 2 2 1
5 1 3 0 1
5 1 3 2 1
5 2 0 2 1
5 2 1 2 1
5 2 2 0 1
5 3 0 0 1
5 3 0 2 1
5 3 1 2 1
5 3 3 0 1
5 4 0 2 1
5 4 2 0 1
5 4 3 2 1

Can you see a pattern?