This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A079643 #59 Jun 02 2025 16:49:53
%S A079643 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 A079643 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 A079643 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 A079643 a(n) = floor(n/floor(sqrt(n))).
%C A079643 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 A079643 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 A079643 Giovanni Resta, <a href="/A079643/b079643.txt">Table of n, a(n) for n = 1..10000</a>
%H A079643 British Mathematical Olympiad, <a href="https://bmos.ukmt.org.uk/home/bmo1-1996.pdf">1996 - Problem 4</a>
%H A079643 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F A079643 a(A064801(n)) = sqrtint(A064801(n)); a(A005563(n)) = 2+sqrtint(sqrt(A005563(n))).
%F A079643 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 A079643 a(n) = floor(2*sqrt(n+1)) - floor(sqrt(n)). - _Wesley Ivan Hurt_, Dec 25 2020
%F A079643 From _Natalia L. Skirrow_, May 13 2025: (Start)
%F A079643 G.f.: (t_3(x)-1)/(2*x) + psi(x^2)/(1-x) - 2, where t_3(x) (A000122) is Jacobi's third theta function and psi(x) is Ramanujan's psi function.
%F A079643 a(n) = A010052(n+1) + A000194(n+1) = [m is square] + floor((sqrt(4*m)+1)/2) where m=n+1 and [] is the Iverson bracket.
%F A079643 (End)
%t A079643 Table[Floor[n/Floor[Sqrt[n]]],{n,100}] (* _Harvey P. Dale_, Sep 22 2011 *)
%o A079643 (PARI) a(n)=floor(n/sqrtint(n))
%o A079643 (Python)
%o A079643 from math import isqrt
%o A079643 def a(n): return n//isqrt(n)
%o A079643 print([a(n) for n in range(1, 100)]) # _Michael S. Branicky_, May 25 2025
%Y A079643 Cf. A000196, A005563, A064801.
%K A079643 nonn,easy
%O A079643 1,2
%A A079643 _Benoit Cloitre_, Jan 31 2003