|
|
A079643
|
|
a(n) = floor(n/floor(sqrt(n))).
|
|
3
|
|
|
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
|
|
|
FORMULA
|
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
|
|
MATHEMATICA
|
Table[Floor[n/Floor[Sqrt[n]]], {n, 100}] (* Harvey P. Dale, Sep 22 2011 *)
|
|
PROG
|
(PARI) a(n)=floor(n/sqrtint(n))
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|