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%I #2 Dec 05 2013 19:56:17
%S 3,4,9,6,10,10,12,12,16,25,18,18,31,26,26,28,41,34,30,30,36,40,42,42,
%T 42,42,50,48,50,56,54,58,75,64,60,60,66,68,70,72,72,93,78,80,78,86,84,
%U 84,84,90,90,98,96,100,100,102,106,127,108,108,110,116,114,118,122
%N Let b(0) = n, b(2*k+1) = c, where c > b(m) is the smallest number such that gcd(c,b(2*k))=1, b(2*k+2) = e where e < b(m) is the smallest number such that gcd(e,b(2*k+1))=1 for m = 0..2*k, until reaching e = 1. Then a(n) = the last c.
%e n = 6: 7, 5, 8, 3, 10, 1, hence a(6) = 10.
%e n = 7: 8, 5, 9, 4, 11, 6, 13, 3, 10, 1, hence a(7) = 10.
%K easy,nonn
%O 2,1
%A _Amarnath Murthy_, Jun 01 2003
%E Edited by Frank Ellermann, Jun 07 2003
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