Take the recursive decimal digit sum of consecutive binary numbers $2^n$ as $n \to \infty$. You'll see something like this:
n 2^n (recursive) sum of digits ----------------------------------------- 1 2 2 2 4 4 3 8 8 4 16 7 5 32 5 6 64 1 7 128 2 (1+2+8=11), (1+1=2) 8 256 4 9 512 8 10 1024 7 11 2048 5 12 4096 1 13 8192 2 14 16384 4 15 32768 8 16 65536 7 17 131072 5
Notice the cycle $1,2,4,8,7,5$ and exclusion of $3,6,9$? No such cycle for $3$ ($3^1=3$, rest is a $9$) but I see one for $5 $ $(1,5,7,8,4,2) $ and haven't checked any higher.
I'm curious if there's something (theorem/result) in number theory explains such recursive sum of digit cycles, why do they exist and for some (prime?) numbers alone?
* EDIT **: I just ran a quick computation of this recursive-sum-of-digits of $x^n$ for $x = 10,000$ and $n = 25$ and see a general pattern. The sequence repeats for every 9th x:
1, Sequence: 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1 2, Sequence: 1,2,4,8,7,5,1,2,4,8,7,5,1,2,4,8,7,5,1,2,4,8,7,5,1 3, Sequence: 1,3,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 4, Sequence: 1,4,7,1,4,7,1,4,7,1,4,7,1,4,7,1,4,7,1,4,7,1,4,7,1 5, Sequence: 1,5,7,8,4,2,1,5,7,8,4,2,1,5,7,8,4,2,1,5,7,8,4,2,1 6, Sequence: 1,6,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 7, Sequence: 1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1 8, Sequence: 1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1 9, Sequence: 1,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 10, Sequence: 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1 11, Sequence: 1,2,4,8,7,5,1,2,4,8,7,5,1,2,4,8,7,5,1,2,4,8,7,5,1 12, Sequence: 1,3,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 13, Sequence: 1,4,7,1,4,7,1,4,7,1,4,7,1,4,7,1,4,7,1,4,7,1,4,7,1 14, Sequence: 1,5,7,8,4,2,1,5,7,8,4,2,1,5,7,8,4,2,1,5,7,8,4,2,1 15, Sequence: 1,6,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 16, Sequence: 1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1,7,4,1 17, Sequence: 1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1,8,1 18, Sequence: 1,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 19, Sequence: 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1 20, Sequence: 1,2,4,8,7,5,1,2,4,8,7,5,1,2,4,8,7,5,1,2,4,8,7,5,1
Notice that there are vertical patterns as well for every consecutive 9th n. Or rather the matrix below repeats vertically and horizontally(with 9's replacing 3,6 for n > 2, skipping 1 repeat entirely):
1,1,1,1,1,1,1, 1 1,2,4,8,7,5,1, 2 1,3,9,9,9,9,9, 9 (horizontal repeat start, 9 replacement) 1,4,7,1,4,7,1, 4 1,5,7,8,4,2,1, 5 1,6,9,9,9,9,9, 9 1,7,4,1,7,4,1, 7 1,8,1,8,1,8,1, 8 1,9,9,9,9,9,9, 9
The digital sum of $m$ is $m \pmod 9$, except that if $m$ is divisible by $9$ you get $9$, not zero. This is the basis of the classic divisibility rule for $9$. Now once you note the cycle $1, 2, 4, 8, 7, 5, 1$ you will repeat. For other bases you have the same behavior.