A257217 A257213 - A003059, where A257213(n) = min{d>0 | floor(n/d) = floor(n/(d+1))}, A003059(n) = ceiling(sqrt(n)).

1, 1, 1, 0, 1, 0, 1, 1, 0, 2, 0, 0, 1, 1, 1, 0, 2, 1, 0, 0, 2, 1, 1, 1, 0, 2, 1, 1, 0, 0, 2, 2, 1, 1, 1, 0, 2, 1, 1, 1, 0, 0, 2, 2, 2, 1, 1, 1, 0, 3, 1, 1, 1, 1, 0, 0, 2, 2, 2, 2, 1, 1, 1, 0, 3, 2, 1, 1, 1, 1, 0, 0, 2, 2, 2, 2, 2, 1, 1, 1, 0, 3, 2, 2, 1, 1

Offset: 0

Author

M. F. Hasler, Apr 18 2015

Keywords

• nonn

Comments

One has a(n) <= a(n-1) except for n = k^2. The positive jumps occur exactly at the squares, cf. formula.

Crossrefs

Cf. A003059, A257213.

Programs

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 *)

PARI

A257217(n)=A257213(n)-A003059(n)

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).