login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A091508
Let b(1)=n; b(k+1)=b(k)/gcd(k,b(k)) if gcd(k,b(k))>1; b(k+1)=b(k)+k otherwise, sequence gives least k such that b(k)=1.
3
1, 2, 111, 7, 5, 3, 25, 22, 25, 111, 111, 4, 7, 5, 5, 6, 22, 25, 22, 9, 111, 25, 25, 4, 111, 111, 11, 111, 9, 7, 6, 8, 19, 5, 6, 19, 9, 22, 22, 111, 8, 22, 19, 9, 9, 111, 111, 111, 111, 25, 15, 11, 9, 111, 8, 111, 16, 11, 7, 5, 9, 9, 9, 15, 111, 6, 111, 10, 7, 19, 19, 19, 9, 6, 25
OFFSET
1,2
COMMENTS
I conjecture a(n) always exists. That means sequence (b(k)) becomes ultimately regular for any n. i.e. there is always k0 such that b(k0)=1, so b(k0+1)=b(k0)+k0=k0+1 since gcd(k0,b(k0))=1 and gcd(k0+1,b(k0+1))=k0+1 implies b(k0+2)=b(k0+1)/(k0+1)=1 and from that point k0 sequence (b(k)) continues : 1, k0+1, 1, k0+2, 1, k0+3,1,... and is "regular".
PROG
(PARI) a(n)=if(n<0, 0, s=1; b=n; while(b>1, s++; b=if(gcd(s, b)-1, b/gcd(b, s), b+s)); s)
CROSSREFS
Sequence in context: A063668 A006334 A093425 * A075399 A079840 A231278
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Mar 03 2004
STATUS
approved