I said: >The incidence of 1's in the second place from the right (excluding >p=2) is 16/(16+21)=43.2%; This indicates the degree to which Mp for p=1 mod 4 is more probably prime than is p=3 mod 4. Of Mp for p=3 and up, p mod 4 # % 1 21 56.8 3 16 43.2 >the incidence in the remaining interior bits is 172/358=48.0% This last line was wrong. For p=3, the second bit from the right is what I've been calling an exterior bit (most significant or least significant). So the one bit for p=3 should not be subtracted from either the one bit's count or the total bit positions counts for interior bits. So the correct incidence in remaining interior bits is 173/359~=48.2%. Credit for spotting this error goes to, who else, but George Woltman! George asks, what about the bit second from the left? In total, 26 0 bits vs only 12 1 bits. (0 26 of 38 times or ~68.4%; 26/12=2.16666...) Excluding p=2 and p= 3, to look only at interior bits, 25 0 bits vs only 11 1 bits; 0 25 of 36 times, ~69.4%; 25/11=2.2727...) Now why would that be, that a 0 bit is more than twice as likely as a 1 bit? (Note that Mp#38, M6972593, has the less likely 1 bit in the second most significant position; searching biased for these probabilities would have delayed finding it. But the 4 previous Mersenne primes all had a 0 bit in that position.) Ken ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
