Voozh

A079643

a(n) = floor(n/floor(sqrt(n))).

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, 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, 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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

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

REFERENCES

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).

LINKS

Giovanni Resta, Table of n, a(n) for n = 1..10000

British Mathematical Olympiad, 1996 - Problem 4

Index to sequences related to Olympiads.

FORMULA

a(A064801(n)) = sqrtint(A064801(n)); a(A005563(n)) = 2+sqrtint(sqrt(A005563(n))).

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

a(n) = floor(2*sqrt(n+1)) - floor(sqrt(n)). - Wesley Ivan Hurt, Dec 25 2020

From Natalia L. Skirrow, May 13 2025: (Start)

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.

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.

(End)

MATHEMATICA

Table[Floor[n/Floor[Sqrt[n]]], {n, 100}] (* Harvey P. Dale, Sep 22 2011 *)

PROG

(PARI) a(n)=floor(n/sqrtint(n))

(Python)

from math import isqrt

def a(n): return n//isqrt(n)