I think my question is rather simple, but I can't wrap my head around it. In "The (new) Turing Omnibus" on page 266, the author writes:
[...], and let A be a [Turing-]machine that converts a blank tape to one with the number n written in binary on it.
and then, further down:
[...], ceil(log(n)) states are required by A [...]
I looked over the Internet, but I found no answer. Why this number of states? Wouldn't it be sufficient to have one state, which writes a 0 when given a 0, and 1 when given 1 and in any case shift left?
thanks in advance!