Charlie, note that I am talking about the first digit of A(n), not the last digit. The parity of A(n) does not matter much. The reason powers of 10 are special is that the A(n) is a power of 10, and must start with a 1.
I personally believe the assumption I am making about S, but if you do not, you can replace it with some other sequence that you believe is pseudorandom.
This afterthought confused me. I spent fifteen minutes trying to figure out why you claim that Ackermann numbers are all powers of 10 and start with 1. I guess you wanted to write something like: »The reason […] is that if n is a power of 10, A(n) must be a power of 10, and start with a 1« Right?
Oh, whoops! Managed to confuse myself. I’m not totally sure about Benford’s law, but am happy to assume for the sake of argument now that I’m not actually deluded.
Double Edit: Hm, the converse of the equidistribution theorem doesn’t hold even close to as strictly as I thought it did. Nevermind me.
Charlie, note that I am talking about the first digit of A(n), not the last digit. The parity of A(n) does not matter much. The reason powers of 10 are special is that the A(n) is a power of 10, and must start with a 1.
I personally believe the assumption I am making about S, but if you do not, you can replace it with some other sequence that you believe is pseudorandom.
This afterthought confused me. I spent fifteen minutes trying to figure out why you claim that Ackermann numbers are all powers of 10 and start with 1. I guess you wanted to write something like: »The reason […] is that if n is a power of 10, A(n) must be a power of 10, and start with a 1« Right?
Oh, whoops! Managed to confuse myself. I’m not totally sure about Benford’s law, but am happy to assume for the sake of argument now that I’m not actually deluded.
Double Edit: Hm, the converse of the equidistribution theorem doesn’t hold even close to as strictly as I thought it did. Nevermind me.