1,1

Conjecture: The sequence appears to be finite and consists exactly of the 11 terms 42, 49, 55, 60, 63, 65, 66, 75, 81, 85, and 102. Empirically verified through 10^5; no further terms found. Open question: For the family "numbers k with exactly n coincident values among ((2*k mod m)/m) for 1 <= m <= k," are there other n for which the corresponding sequence is finite or empty?

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 = 42, the fractions (2*k mod m)/m for 1 <= m <= 42 are 0, 0, 0, 0, 4/5, 0, 0, 1/2, 1/3, 2/5, 7/11, 0, 6/13, 0, 3/5, 1/4, 16/17, 2/3, 8/19, 1/5, 0, 9/11, 15/23, 1/2, 9/25, 3/13, 1/9, 0, 26/29, 4/5, 22/31, 5/8, 6/11, 8/17, 2/5, 1/3, 10/37, 4/19, 2/13, 1/10, 2/41, 0. Exactly 5 values are repeated (0, 1/2, 4/5, 2/5, 1/3), so 42 is in this sequence.

okQ[k_]:=Module[{m=Range[k]}, Length[Select[Counts[Mod[2k, m]/m], #>1&]]==5]; Select[Range[120], okQ] (* James C. McMahon, Nov 09 2025 *)

(Python)

from fractions import Fraction

def has_k_alignments(k, target=5):

seen, dup = set(), set()

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

r = k % m

f = Fraction(r, m)

if f >= Fraction(1, 2):

f -= Fraction(1, 2)

if f in seen:

dup.add(f)

else:

seen.add(f)

return len(dup) == target

for k in range(1, 1000):

if has_k_alignments(k, target=5):

print(k)

Cf. A387426 (base case with all distinct fractions), A389863, A390100, A389221.

Sequence in context: A124189 A249043 A063998 * A255513 A186456 A181647

Adjacent sequences: A390316 A390317 A390318 * A390320 A390321 A390322

nonn,more

Kenneth J Scheller, Nov 01 2025

approved