N = Ak*k! + Ak-1*(k-1)! + ... + A1*1!
with Am <= m. So for example
100 = 4*4! + 0*3! + 2*2! + 0*1! = 4020f
The last number before a rollover of digits is [n][n-1][n-2]...[1]f and represents (n+1)!-1.
In this basis you are 24 + 6 + 2 = 1*4! + 1*3! + 1*2! + 0*1! = 1110f.
Here are all the primes up to 719 = 6! - 1 = 54321f
\n0 0 0 1 0\n0 0 0 1 1\n0 0 0 2 1\n0 0 1 0 1\n0 0 1 2 1\n0 0 2 0 1\n0 0 2 2 1\n0 0 3 0 1\n0 0 3 2 1\n0 1 0 2 1\n0 1 1 0 1\n0 1 2 0 1\n0 1 2 2 1\n0 1 3 0 1\n0 1 3 2 1\n0 2 0 2 1\n0 2 1 2 1\n0 2 2 0 1\n0 2 3 0 1\n0 2 3 2 1\n0 3 0 0 1\n0 3 1 0 1\n0 3 1 2 1\n0 3 2 2 1\n0 4 0 0 1\n0 4 0 2 1\n0 4 1 0 1\n0 4 1 2 1\n0 4 2 0 1\n0 4 2 2 1\n1 0 1 0 1\n1 0 1 2 1\n1 0 2 2 1\n1 0 3 0 1\n1 1 0 2 1\n1 1 1 0 1\n1 1 2 0 1\n1 1 3 0 1\n1 1 3 2 1\n1 2 0 2 1\n1 2 1 2 1\n1 2 2 0 1\n1 2 3 2 1\n1 3 0 0 1\n1 3 0 2 1\n1 3 1 0 1\n1 3 3 0 1\n1 4 1 0 1\n1 4 1 2 1\n1 4 2 0 1\n1 4 2 2 1\n1 4 3 2 1\n2 0 0 0 1\n2 0 1 2 1\n2 0 2 2 1\n2 0 3 2 1\n2 1 0 2 1\n2 1 1 0 1\n2 1 2 0 1\n2 1 2 2 1\n2 1 3 0 1\n2 2 0 2 1\n2 2 3 0 1\n2 2 3 2 1\n2 3 0 0 1\n2 3 0 2 1\n2 3 3 0 1\n2 4 0 0 1\n2 4 1 2 1\n2 4 2 0 1\n2 4 2 2 1\n2 4 3 2 1\n3 0 1 0 1\n3 0 2 0 1\n3 0 3 0 1\n3 0 3 2 1\n3 1 0 2 1\n3 1 2 0 1\n3 1 2 2 1\n3 2 0 0 1\n3 2 1 2 1\n3 2 2 0 1\n3 2 3 2 1\n3 3 0 0 1\n3 3 1 0 1\n3 3 1 2 1\n3 3 2 2 1\n3 4 0 0 1\n3 4 0 2 1\n3 4 1 0 1\n3 4 1 2 1\n3 4 3 2 1\n4 0 1 0 1\n4 0 1 2 1\n4 0 3 0 1\n4 0 3 2 1\n4 1 0 2 1\n4 1 2 2 1\n4 1 3 0 1\n4 2 2 0 1\n4 2 3 0 1\n4 3 0 2 1\n4 3 1 2 1\n4 3 2 2 1\n4 3 3 0 1\n4 4 0 0 1\n4 4 1 2 1\n4 4 2 2 1\n4 4 3 2 1\n5 0 0 0 1\n5 0 1 0 1\n5 0 2 0 1\n5 0 2 2 1\n5 0 3 0 1\n5 1 1 0 1\n5 1 2 2 1\n5 1 3 0 1\n5 1 3 2 1\n5 2 0 2 1\n5 2 1 2 1\n5 2 2 0 1\n5 3 0 0 1\n5 3 0 2 1\n5 3 1 2 1\n5 3 3 0 1\n5 4 0 2 1\n5 4 2 0 1\n5 4 3 2 1\n
I suspect this basis could be useful in studying primes.