OFFSET
1,1
COMMENTS
a(1) = 2 because the first minimum of sum(mu(n)/n), n>1, is 1-1/2 = 1/(2).
PROG
(PARI) t = 0; t1 = 1; v = []; for( n = 1, 200, t = t + moebius( n) / n; if( ( t / t1)^2 < 1, t1 = t; v = concat( v, denominator( t)), )); v
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Donald S. McDonald, May 18 2002
STATUS
approved