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a(n) = maximum value among all k where 1<=k<=n of GCD(k,ceiling(n/k)).
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%I #8 Sep 16 2015 06:17:25

%S 1,1,2,2,2,2,3,3,3,2,2,2,4,4,4,4,3,3,3,3,5,5,5,5,5,4,4,4,4,4,6,6,6,6,

%T 6,6,4,4,4,4,5,5,7,7,7,7,7,7,7,5,4,4,4,4,4,4,8,8,8,8,8,8,8,8,6,6,6,6,

%U 6,6,6,6,9,9,9,9,9,9,9,9,9,5,5,5,7,7,7,7,7,7,10,10,10,10,10,10,10,10,10,10

%N a(n) = maximum value among all k where 1<=k<=n of GCD(k,ceiling(n/k)).

%e For n = 10, we have the pairs {k,ceiling(n/k)} of {1,10},{2,5},{3,4},{4,3},{5,2},{6,2},{7,2},{8,2},{9,2},{10,1}. The GCD's of these 10 pairs are 1,1,1,1,1,2,1,2,1,1. Of these, 2 is the largest. So a(10) = 2.

%t Table[Max[Table[GCD[k, Ceiling[n/k]], {k, 1, n}]], {n, 1, 100}] (* _Stefan Steinerberger_, Jul 22 2006 *)

%K nonn

%O 1,3

%A _Leroy Quet_, Jul 12 2006

%E More terms from _Stefan Steinerberger_, Jul 22 2006