OFFSET
0,10
COMMENTS
One has a(n) <= a(n-1) except for n = k^2. The positive jumps occur exactly at the squares, cf. formula.
FORMULA
a(k^2-1) = 0 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, thus A257213(n) = k = ceiling(sqrt(n)).
A257213(n) >= floor(sqrt(n))+1 = A257213(n+1) >= A257213(n) = ceiling(sqrt(n)), with strict inequality (in the second relation) when n is a square. Therefore a(n) >= 1 for all n = k^2.
a(k^2) >= 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).
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 - Ceiling[Sqrt@ k]]] // Flatten // Rest]; f@ 85 (* Michael De Vlieger, Apr 18 2015 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Apr 18 2015
STATUS
approved