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A069211
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Let u(n,k) be the recursion: u(n,1)=1 u(n,2)=n u(n,k+3)= (1/2) *(u(n,k+1)+u(n,k)) if u(n,k+1)+u(n,k) is even u(n,k+3)=abs(u(n,k+1)-u(n,k)) otherwise. Sequence gives integer values a(n) such that u(n,k)=1 for any k>=a(n).
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1, 3, 4, 6, 8, 7, 7, 9, 13, 11, 8, 10, 16, 10, 10, 12, 14, 16, 12, 14, 14, 11, 11, 13, 15, 19, 11, 13, 17, 13, 13, 15, 21, 17, 17, 19, 18, 15, 15, 17, 21, 17, 12, 14, 19, 14, 14, 16, 24, 18, 20, 22, 22, 14, 14, 16, 19, 20, 14, 16, 25, 16, 16, 18, 20, 24, 18, 20, 28, 20, 20, 22
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| It seems that sum(i=1,n,a(i)) ~ C*n*ln(n) asymptotically with C=0.2...
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FORMULA
| a(2n) = a(n) + 2 + [n>1]. - Ralf Stephan (ralf(AT)ark.in-berlin.de), Oct 08 2003
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EXAMPLE
| Let n=7, for k=1,2,3,4,5,6,7,8 u(7,k)=1,7,4,3,1,2,1,1 hence a(7)=7 since for all k>=7 u(7,k)=1.
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CROSSREFS
| Sequence in context: A161001 A139450 A168170 * A072152 A199015 A196098
Adjacent sequences: A069208 A069209 A069210 * A069212 A069213 A069214
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KEYWORD
| nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 11 2002
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