

A069211


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).


0



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
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OFFSET

1,2


COMMENTS

It seems that Sum_{i=1..n} a(i) ~ C*n*log(n) asymptotically with C=0.2...


LINKS

Table of n, a(n) for n=1..72.


FORMULA

a(2n) = a(n) + 2 + [n>1].  Ralf Stephan, Oct 08 2003


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.


CROSSREFS

Sequence in context: A225794 A206769 A141219 * A242822 A295996 A240675
Adjacent sequences: A069208 A069209 A069210 * A069212 A069213 A069214


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Apr 11 2002


STATUS

approved



