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A091508
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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.
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0
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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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".
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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)
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CROSSREFS
| Sequence in context: A077804 A006334 A093425 * A075399 A079840 A117698
Adjacent sequences: A091505 A091506 A091507 * A091509 A091510 A091511
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KEYWORD
| nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 03 2004
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