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%I #14 Apr 13 2024 01:58:32
%S 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,
%T 11,13,17,13,13,15,21,17,17,19,18,15,15,17,21,17,12,14,19,14,14,16,24,
%U 18,20,22,22,14,14,16,19,20,14,16,25,16,16,18,20,24,18,20,28,20,20,22
%N Let u(n,k) be the recursion: u(n,1)=1, u(n,2)=n, u(n,k+2) = (1/2) * (u(n,k+1)+u(n,k)) if u(n,k+1)+u(n,k) is even, and u(n,k+2) = 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).
%C It seems that Sum_{i=1..n} a(i) ~ C*n*log(n) asymptotically with C=0.2...
%H Sean A. Irvine, <a href="https://github.com/archmageirvine/joeis/blob/master/src/irvine/oeis/a069/A069211.java">Java program</a> (github)
%F a(2n) = a(n) + 2 + [n>1]. - _Ralf Stephan_, Oct 08 2003
%e 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 we have u(7,k)=1.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Apr 11 2002
%E Name corrected by _Sean A. Irvine_, Apr 12 2024
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