I accepted this solution after some thought:

The conjecture on your web page about taking differences of digits is true.

Suppose to begin with that all digits are 0 or 1. Then after

t iterations you have, mod 2, a_new(k) = sum {0..t} of (t choose r)

times a(k+r mod n).

1. If n is a power of 2, then take t=n and note that all those

binomial coefficients are even other than (n choose 0)

and (n choose n), and that those two catch the same a(k),

so to speak; so all the a_new(k) are 0. That was all

assuming that the digits are 0 or 1, but notice that the same

argument works with the (digits mod 2), so we've proved

that after iterating n times all your digits are even. So do it

again and get all the digits to be multiples of 4, then again

making them multiples of 8, and so on until you reach a

power of 2 larger than any of the original digits, at which

point all the digits must equal 0. So if n is a power of 2,

you must end up with the all-zero string.

2. If n is odd and > 1, start with any string of 0s and 1s other than

all-0 or all-1. I claim you never end up at all-0. Why? Because

the only possible predecessors of all-0 are all-0 and all-1, and

there is*no*possible predecessor of all-1 (easy exercise for

the reader). So when n is odd and > 1, there must be some possible

fate other than all-0.

3. If n is neither odd nor a power of 2, write it as m*2^k where

m is odd. Take any string S of 0s and 1s of length m, other than

all-0 and all-1, and repeat it 2^k times. Now the t'th iterate of

this string is the t'th iterate of S, repeated 2^k times (easy

exercise for the reader), and*that*is never all-0. So when

n is an odd number > 1 times a power of 2, there must be some

possible fate other than all-0. We're done.

*

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