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%I #12 Apr 19 2015 00:59:08
%S 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,
%T 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,
%U 1,1,0,0,2,2,2,2,2,1,1,1,0,3,2,2,1,1
%N A257213 - A003059, where A257213(n) = min{d>0 | floor(n/d) = floor(n/(d+1))}, A003059(n) = ceiling(sqrt(n)).
%C One has a(n) <= a(n-1) except for n = k^2. The positive jumps occur exactly at the squares, cf. formula.
%F 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)).
%F 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.
%F 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).
%t 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 *)
%o (PARI) A257217(n)=A257213(n)-A003059(n)
%Y Cf. A003059, A257213.
%K nonn
%O 0,10
%A _M. F. Hasler_, Apr 18 2015
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