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a(n) = maximum value among all k where 1<=k<=n of gcd(k,floor(n/k)).
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%I #14 Jun 04 2019 14:44:40

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

%T 4,6,6,6,6,6,6,3,3,3,3,3,3,4,7,7,7,7,7,7,7,5,5,5,5,3,3,3,3,8,8,8,8,8,

%U 8,8,8,6,6,6,6,6,6,6,6,6,9,9,9,9,9,9,9,9,9,4,4,4,4,4,4,4,4,7,7,10,10,10,10

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

%C a(n) is the greatest m such that floor(n/m^2) > n/(m^2+m). - _Robert Israel_, Jun 04 2019

%H Robert Israel, <a href="/A120423/b120423.txt">Table of n, a(n) for n = 1..10000</a>

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

%p a:=n->max(seq(gcd(k,floor(n/k)),k=1..n)): seq(a(n),n=1..112); # _Emeric Deutsch_, Jul 24 2006

%p # Alternative:

%p f:= proc(n) local m,a;

%p for m from floor(sqrt(n)) by -1 do

%p a:= floor(n/m^2);

%p if n < a*(m^2+m) then return m fi

%p od

%p end proc:

%p map(f, [$1..200]); # _Robert Israel_, Jun 04 2019

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

%K nonn,look

%O 1,4

%A _Leroy Quet_, Jul 11 2006

%E More terms from _Stefan Steinerberger_ and _Emeric Deutsch_, Jul 22 2006