Although this is a follow up to an earlier post, it can be read independently.

Repeatedly squaring the number 5, we get the following.

5^1 = 5 5^2 = 25 5^4 = 625 5^8 = 390625 5^16 = 152587890625 5^32 = 23283064365386962890625

There is a pattern here, which becomes more evident if the table is truncated to a triangle:

5^1 = 5 5^2 = 25 5^4 = 625 5^8 = ...0625 5^16 = ...90625 5^32 = ...890625 5^64 = ...2890625 5^128 = ...12890625 5^256 = ...212890625

What other digits, besides 5, have this property? None in the decimal system (ignoring the trivial 0 and 1). But the same pattern is shown by 3 in base 6 (below, the exponent is still in base 10, for clarity).

3^1 = 3 3^2 = 13 3^4 = 213 3^8 = ...0213 3^16 = ...50213 3^32 = ...350213 3^64 = ...1350213

and so on.

Here is a proof that the pattern appears when the base is 2 mod 4 (except 2), and the starting digit is half the base.

Let be the starting digit, and the base. The claim is that for all .

Base of induction: is divisible by and also by , since is even. Hence it is divisible by .

Step of induction: Suppose is divisible by . Then

The second factor is divisible by and also by , being the sum of two odd numbers. So, it is divisible by . Since the first factor is divisible by , the claim follows: is divisible by .

So that was easy. The part I can’t prove is the converse. Numerical evidence strongly suggests that ,

is also the necessary condition for the triangular pattern to appear.