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.

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

Author

Benoit Cloitre, Mar 03 2004

Keywords

• nonn

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".

Links

• Michel Marcus, Table of n, a(n) for n = 1..10000

Programs

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)