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A120423 a(n) = maximum value among all k where 1<=k<=n of gcd(k,floor(n/k)). 1
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, 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, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

a(n) is the greatest m such that floor(n/m^2) > n/(m^2+m). - Robert Israel, Jun 04 2019

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

EXAMPLE

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.

MAPLE

a:=n->max(seq(gcd(k, floor(n/k)), k=1..n)): seq(a(n), n=1..112); # Emeric Deutsch, Jul 24 2006

# Alternative:

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

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

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

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

od

end proc:

map(f, [$1..200]); # Robert Israel, Jun 04 2019

MATHEMATICA

Table[Max[Table[GCD[k, Floor[n/k]], {k, 1, n}]], {n, 1, 100}] (* Stefan Steinerberger, Jul 22 2006 *)

CROSSREFS

Sequence in context: A203776 A343559 A242357 * A113137 A220603 A238404

Adjacent sequences: A120420 A120421 A120422 * A120424 A120425 A120426

KEYWORD

nonn,look

AUTHOR

Leroy Quet, Jul 11 2006

EXTENSIONS

More terms from Stefan Steinerberger and Emeric Deutsch, Jul 22 2006

STATUS

approved

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Last modified December 1 14:52 EST 2022. Contains 358468 sequences. (Running on oeis4.)