🚀 go-pugleaf

In a reply to Bob Silverman I wrote:
> (2^488 + 1)/257 had 146 digits, last I counted.

Which proves once and for all that number theorists can't count. :-)
The factorization is still the record non-networked factorization, of
course, but (2^488 + 1)/257 has only 145 digits. While I'm at it let me
correct my assertion that 2^512 + 1 is the record difficult number; the
record difficult number is (2^512 + 1)/2424833, which has 148 digits.

In case anyone's wondering, I was misled by the UNIX dc program, which
acts like this:

% dc
2 488^ 1+ 257/p
310960166879725296199878945847101390767175534833553546544725362515212\
1265127859917318281565697486165824574686253229943428928722855633467075\
526401
10 145^p
1000000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000000000000000000000000000000000000000000000000\
000000

Notice how the first line of output is chopped off a digit early. This
is what I get for counting digits with the help of a computer instead of
by hand.

---Dan