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A257213 Least d>0 such that floor(n/d) = floor(n/(d+1)). 4
1, 2, 3, 2, 3, 3, 4, 4, 3, 5, 4, 4, 5, 5, 5, 4, 6, 6, 5, 5, 7, 6, 6, 6, 5, 7, 7, 7, 6, 6, 8, 8, 7, 7, 7, 6, 8, 8, 8, 8, 7, 7, 9, 9, 9, 8, 8, 8, 7, 10, 9, 9, 9, 9, 8, 8, 10, 10, 10, 10, 9, 9, 9, 8, 11, 11, 10, 10, 10, 10, 9, 9, 11, 11, 11, 11, 11, 10, 10, 10 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

For n > 1: a(A043548(n)) = n. - Reinhard Zumkeller, Apr 19 2015         ~

LINKS

Michael De Vlieger, Table of n, a(n) for n = 0..10000

FORMULA

a(n) >= A003059(n+1) = floor(sqrt(n))+1 >= A003059(n) = ceiling(sqrt(n)) >= A257212(n), with strict inequality (in the middle relation) when n is a square.

a(k^2-1) = k 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 and when denominators decrease further, this keeps increasing.

a(k^2) >= k+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).

EXAMPLE

a(0)=1 because 0/1 = 0/2.

a(1)=2 because [1/1] = 1 > [1/2] = 0 = [1/3], where [x] := floor(x).

a(2)=3 because [2/1] = 2 > [2/2] = 1 > [2/3] = 0 = [2/4].

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] // Flatten // Rest]; f@ 79 (* Michael De Vlieger, Apr 18 2015 *)

PROG

(PARI) A257213(n)=for(d=sqrtint(n)+1, n+1, n\d==n\(d+1)&&return(d))

(Haskell)

a257213 n = head [d | d <- [1..], div n d == div n (d + 1)]

-- Reinhard Zumkeller, Apr 19 2015

CROSSREFS

Cf. A003059, A257212.

Cf. A043548.

Sequence in context: A026256 A079715 A030397 * A205780 A204905 A082597

Adjacent sequences:  A257210 A257211 A257212 * A257214 A257215 A257216

KEYWORD

nonn,nice,hear

AUTHOR

M. F. Hasler, Apr 18 2015

STATUS

approved

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Last modified July 23 21:10 EDT 2021. Contains 346265 sequences. (Running on oeis4.)