==> arithmetic/digits/most.significant/powers.s <==
2^N begins with 603245 iff 603246*10^m > 2^N >= 603245*10^m for some
positive integer m ==> m+log(603246) > N*log(2) >= m+log(603245);
so 2^N begins with 603245 iff frac(log(603246)) > frac(N*log(2))
>= frac(log(603245)).  If we are using natural density then N*log(2)
is uniformly distributed mod 1 since log(2) is irrational, hence the
probability is frac(log(603246)) - frac(log(603245)) =
frac(log(603246)-log(603245)) = frac(log(603246/603245)).

A neat observation is that since it is known p_n*c, where p_n is the
nth prime and c is irrational, is uniformly distributed mod 1, we get
the same answer if we replace 2^N with 2^{p_n}.
-- 
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618