Quote:
Originally Posted by xilman
Another possibly interesting question: the quantity hammingweight(PowerSmooth(N))/N appears to converge to a value close to 0.72 can one prove that it converges and can one find a relatively simple expression for its value?

I assume that half the bits are ones, so that's about equivalent to length of the number x being 1.44N bits, or log(x) = 1.44N*log(2) = N. The prime number theorem says the sum of log(p) for primes p less than L is about L. The PowerSmooth function multiplies the largest power of the prime less than L instead of just the prime itself, but that shouldn't make too much difference.
So the 0.72 value is probably 1/log(2)/2, where the 1/log(2) comes from the size of the product as predicted by the prime number theorem, and an additional /2 for only half the bits being ones.