OFFSET
0,2
COMMENTS
For n > 1: a(A043548(n)) = n. - Reinhard Zumkeller, Apr 19 2015
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..10000
FORMULA
a(n) >= A003059(n+1) = floor(sqrt(n))+1 >= A003059(n) = ceiling(sqrt(n)) >= A257212(n), with strict inequality (in the middle relation) when n is a square.
a(k^2-1) = k for k > 1. Proof: For n=k^2-1=(k-1)*(k+1), floor(n/k) = k-1 = n/(k+1), but n/(k-1)=k+1 and when denominators decrease further, this keeps increasing.
a(k^2) >= k+d when k > d*(d-1). Proof: This follows from k^2/(k+d) = k-d+d^2/(k+d), which shows that a(k) >= d when k > d*(d-1).
a(n) = A259361(n) + 1 + floor(sqrt((A232091(n+1) - 1 - n) + A079813(n+1)) + A079813(n+1)/2) = floor((sqrt(4*n+1)+1)/2) + floor(sqrt(ceiling((n+1) / ceiling(sqrt(n+1)) + 1) * ceiling(sqrt(n+1)) - round(sqrt(n+1)) - n - 1) + (ceiling(sqrt(n+1)) - round(sqrt(n+1)))/2). - Haofen Liang, Aug 25 2021
a(n) = floor(sqrt(p*q - n) + (p + q)/2), where p = floor(sqrt(n)) and q = floor(sqrt(n+1) + 3/2). - Ridouane Oudra, Jan 24 2023
EXAMPLE
a(0)=1 because 0/1 = 0/2.
a(1)=2 because [1/1] = 1 > [1/2] = 0 = [1/3], where [x] := floor(x).
a(2)=3 because [2/1] = 2 > [2/2] = 1 > [2/3] = 0 = [2/4].
MATHEMATICA
f[n_] := Block[{d, k}, Reap@ For[k = 0, k <= n, k++, d = 1; While[Floor[k/d] != Floor[k/(d + 1)], d++]; Sow@ d] // Flatten // Rest]; f@ 79 (* Michael De Vlieger, Apr 18 2015 *)
PROG
(PARI) A257213(n)=for(d=sqrtint(n)+1, n+1, n\d==n\(d+1)&&return(d))
(Haskell)
a257213 n = head [d | d <- [1..], div n d == div n (d + 1)]
-- Reinhard Zumkeller, Apr 19 2015
CROSSREFS
KEYWORD
AUTHOR
M. F. Hasler, Apr 18 2015
STATUS
approved