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A257212
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Least d>0 such that floor(n/d) - floor(n/(d+1)) <= 1.
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2
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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
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OFFSET
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0,4
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COMMENTS
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An efficient formula for this sequence could be useful for faster computation of A024916.
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LINKS
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FORMULA
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a(n) <= ceiling(sqrt(n)) <= A257213(n) for all n>0.
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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n)=for(d=1, n+1, 1>=n\d-n\(d+1)&&return(d))
(Haskell)
a257212 n = head [d | d <- [1..], div n d - div n (d+1) <= 1]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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