1,1

Conjecture: All numbers of the form 4*p with p prime and p > 13 belong to the sequence, and no other numbers greater than 80 do. Empirically verified through 100000.

For k = 14, the fractions for 1 <= m <= 14 are 0, 0, 1/3, 0, 3/5, 2/3, 0, 1/2, 1/9, 4/5, 6/11, 1/3, 2/13, 0. Exactly two values are repeated (0 and 1/3), so 14 is not in this sequence.

For k = 68, the fractions (2*k mod m)/m for 1 <= m <= 68 are 0, 0, 1/3, 0, 1/5, 2/3, 3/7, 0, 1/9, 3/5, 4/11, 1/3, 6/13, 5/7, 1/15, 1/2, 0, 5/9, 3/19, 4/5, 10/21, 2/11, 21/23, 2/3, 11/25, 3/13, 1/27, 6/7, 20/29, 8/15, 12/31, 1/4, 4/33, 0, 31/35, 7/9, 25/37, 11/19, 19/39, 2/5, 13/41, 5/21, 7/43, 1/11, 1/45, 22/23, 42/47, 5/6, 38/49, 18/25, 2/3, 8/13, 30/53, 14/27, 26/55, 3/7, 22/57, 10/29, 18/59, 4/15, 14/61, 6/31, 10/63, 1/8, 6/65, 2/33, 2/67, 0. Exactly 4 values are repeated (0, 1/3, 2/3, 3/7), so 68 is in this sequence.

okQ[k_]:=Module[{fr={}}, Do[AppendTo[fr, Mod[2k, m]/m], {m, 1, k}]; Length[Select[Last/@Tally[fr], #>1&]]==4]; Select[Range[1000], okQ] (* James C. McMahon, Nov 01 2025 *)

(Python)

from fractions import Fraction

def ok(k, target=4):

seen, dup = set(), set()

for m in range(1, k + 1):

f = Fraction((2 * k) % m, m)

if f in seen:

dup.add(f)

else:

seen.add(f)

return len(dup) == target

Cf. A387426 (base case with all distinct fractions), A389863, A001749.

Sequence in context: A141638 A033857 A110264 * A034105 A034115 A212600

Adjacent sequences: A389218 A389219 A389220 * A389222 A389223 A389224

Kenneth J Scheller, Oct 24 2025

approved