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%I #7 Mar 30 2012 18:51:27
%S 1,1,1,1,5,1,5,1,5,5,1,1,5,5,1,1,17,5,5,5,17,1,17,1,17,5,5,5,1,1,17,1,
%T 17,17,5,5,17,5,1,5,17,17,1,1,17,17,5,1,17,17,5,5,1,5,17,5,1,1,1,1,17,
%U 17,5,1,1,17,17,17,1,5,1,5,17,17,5,5,1,1,1,5,5,17,17,17,1,1,1,1,5,17
%N In repeated iterations of function m->m/2 if m even, m->3m-1 if m odd, a(n) is minimum value achieved if starting from n.
%C At least for n<10000, the only possible cycles reached include 1,2,1,..., 5,14,7,20,10,5,... and 17,50,25,74,37,110,55,164,82,41,122,61,182,91,272,136,68,34,17,... For n<5 only the first occurs, while for n<17 only the first two occur.
%F a(2n) = a(n)
%e a(9)=5 since iteration starts: 9, 26, 13, 38, 19, 56, 28, 14, 7, 20, 10, 5, 14, 7, 20, 10, 5, ... and 5 is the smallest value
%Y Cf. A001281. If n is in A039500 then a(n)=1, if n is in A039501 then a(n)=5, if n is in A039502 then a(n)=17. If n is negative then this becomes the 3x+1 problem and the minimum values become those which are most negative (i.e. maximum absolute values) as in A056959.
%K nonn
%O 1,5
%A _Henry Bottomley_, Jul 18 2000
%E Edited by Bryce Herdt (mathidentity(AT)yahoo.com), Apr 18 2010
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