cp's OEIS Frontend

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.

%I A091508 #9 Mar 01 2025 16:44:51
%S A091508 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,
%T A091508 111,11,111,9,7,6,8,19,5,6,19,9,22,22,111,8,22,19,9,9,111,111,111,111,
%U A091508 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
%N 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.
%C A091508 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".
%H A091508 Michel Marcus, <a href="/A091508/b091508.txt">Table of n, a(n) for n = 1..10000</a>
%o A091508 (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)
%K A091508 nonn
%O A091508 1,2
%A A091508 _Benoit Cloitre_, Mar 03 2004