OFFSET
1,2
COMMENTS
Since sum(n>0,mu(n)/n)=0, this sequence gives the subset of smallest integers that satisfy this sum.
PROG
(PARI) S=0; a=0; w=2; for(n=1, 100, b=a+1; while(abs(S+moebius(b)/b)>=w, b++); S=S+moebius(b)/b; w=abs(S); a=b; print1(b, ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 06 2003
STATUS
approved