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%I #36 Dec 25 2020 10:53:34
%S 1,2,3,2,2,3,3,4,3,3,3,4,4,4,5,4,4,4,4,5,5,5,5,6,5,5,5,5,5,6,6,6,6,6,
%T 7,6,6,6,6,6,6,7,7,7,7,7,7,8,7,7,7,7,7,7,7,8,8,8,8,8,8,8,9,8,8,8,8,8,
%U 8,8,8,9,9,9,9,9,9,9,9,10,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,11
%N a(n) = floor(n/floor(sqrt(n))).
%C a(n) > a(n+1) iff n = m^2 - 1 with m >= 2; that is the answer to the 4th problem of the 32nd British Mathematical Olympiad (1996) [See link BMO]. - _Bernard Schott_, Oct 28 2019
%D A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 4 pp. 54 and 92-93 (1996).
%H Giovanni Resta, <a href="/A079643/b079643.txt">Table of n, a(n) for n = 1..10000</a>
%H British Mathematical Olympiad, <a href="https://bmos.ukmt.org.uk/home/bmo1-1996.pdf">1996 - Problem 4</a>
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F a(A064801(n)) = sqrtint(A064801(n)); a(A005563(n)) = 2+sqrtint(sqrt(A005563(n))).
%F For m = positive integer, terms a(m^2) through a(m^2+m-1) each equal m; terms a(m^2+m) through a(m^2+2m-1) each equal m+1; term a(m^2+2m) equals m+2. - _Leroy Quet_, Apr 02 2007
%F a(n) = floor(2*sqrt(n+1)) - floor(sqrt(n)). - _Wesley Ivan Hurt_, Dec 25 2020
%t Table[Floor[n/Floor[Sqrt[n]]],{n,100}] (* _Harvey P. Dale_, Sep 22 2011 *)
%o (PARI) a(n)=floor(n/sqrtint(n))
%Y Cf. A000196, A005563, A064801.
%K nonn,easy
%O 1,2
%A _Benoit Cloitre_, Jan 31 2003
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