A257212 Least d>0 such that floor(n/d) - floor(n/(d+1)) <= 1.

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, 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, 7, 7, 6, 6, 7, 7, 7, 7, 7, 6, 8, 8, 7, 7, 7, 7, 8, 8, 8

Offset: 0

Author

M. F. Hasler, Apr 18 2015

Keywords

• nonn

Comments

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

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A257213, A024916.

Programs

Haskell

a257212 n = head [d | d <- [1..], div n d - div n (d+1) <= 1]
-- Reinhard Zumkeller, Apr 19 2015

Mathematica

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 *)
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 *)

PARI

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

Formula

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