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Least d>0 such that floor(n/d) - floor(n/(d+1)) <= 1.
2

%I #15 Oct 18 2015 12:57:45

%S 1,1,1,2,2,2,2,2,3,2,3,3,3,3,3,4,3,3,4,4,3,4,4,4,5,4,4,4,5,5,4,4,5,5,

%T 5,4,5,5,5,5,6,6,5,5,5,6,6,6,5,5,6,6,6,6,5,7,6,6,6,6,7,7,7,6,6,6,7,7,

%U 7,7,6,6,7,7,7,7,7,6,8,8,7,7,7,7,8,8,8

%N Least d>0 such that floor(n/d) - floor(n/(d+1)) <= 1.

%C An efficient formula for this sequence could be useful for faster computation of A024916.

%H Reinhard Zumkeller, <a href="/A257212/b257212.txt">Table of n, a(n) for n = 0..10000</a>

%F a(n) <= ceiling(sqrt(n)) <= A257213(n) for all n>0.

%t f[n_] := Block[{d, k}, Reap@ For[k = 0, k <= n, k++, d = 1; While[Floor[k/d] - Floor[k/(d + 1)] > 1, d++]; Sow[d]] // Flatten // Rest]; f@ 86 (* _Michael De Vlieger_, Apr 18 2015 *)

%t ld[n_]:=Module[{d=1},While[Floor[n/d]-Floor[n/(d+1)]>1,d++];d]; Array[ ld,90,0] (* _Harvey P. Dale_, Oct 18 2015 *)

%o (PARI) a(n)=for(d=1,n+1,1>=n\d-n\(d+1)&&return(d))

%o (Haskell)

%o a257212 n = head [d | d <- [1..], div n d - div n (d+1) <= 1]

%o -- _Reinhard Zumkeller_, Apr 19 2015

%Y Cf. A257213, A024916.

%K nonn

%O 0,4

%A _M. F. Hasler_, Apr 18 2015