A306415 Numbers k such that A179682(k) <> A033996(k).

0, 8, 24, 48, 49, 80, 120, 168, 224, 242, 288, 360, 440, 528, 624, 675, 728, 840, 960, 1088, 1224, 1368, 1444, 1520, 1680, 1681, 1848, 2024, 2208, 2400, 2600, 2645, 2808, 3024, 3248, 3480, 3720, 3968, 4224, 4374, 4488, 4760, 5040, 5328, 5624, 5928, 6240, 6560, 6727, 6888, 7224, 7568, 7920, 8280, 8648, 9024, 9408, 9800, 10200, 10608

Offset: 1

Author

Robert Israel, Feb 15 2019

Keywords

• nonn

Comments

0 and numbers k such that for some j with k < j < 4*k*(k+1), k*(k+1)*j*(j+1) is a square.

If k > 0 is a member, then so is A179682(k).

Includes A033996.

Conjecture: every member of the sequence is a member of A033996 or is A179682(k) for some k in the sequence.

A number k in this list indicates that A083481(k) is the same as some A083481(k') at an earlier place k'A083481(8) = A083481(1). 24 appears because A083481(24) = A083481(2). 242 appears because A083481(242) = A083481(24) = A083481(2). - R. J. Mathar, Mar 16 2023

Examples

			24 is a term because A179682(24) = 242: 24 < 242 < 4*24*25 and 24*25*242*243 = 5940^2.
		

Crossrefs

Cf. A033996, A179682.

Programs

Maple

A179682:= proc(n) local F, t, p, k0, d, k, a, j;
  p:= max(map(t -> `if`(t[2]::odd, t[1], NULL), [op(ifactors(n)[2]), op(ifactors(n+1)[2])]));
  if n mod p = 0 then k0:= n+p-1; d:= 1;
    else  k0:= n+1; d:= p-1;
  fi;
  t:= n*(n+1)/4;
  for a from k0 by p do
    for k in [a, a+d] do
       if issqr(k*(k+1)*t) then return k fi
  od od
end proc:
f(0):= 1:
select(t -> A179682(t) <> 4*t*(t+1), [$0..11000]);