%I #17 Mar 16 2023 08:37:06
%S 0,8,24,48,49,80,120,168,224,242,288,360,440,528,624,675,728,840,960,
%T 1088,1224,1368,1444,1520,1680,1681,1848,2024,2208,2400,2600,2645,
%U 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
%N Numbers k such that A179682(k) <> A033996(k).
%C 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.
%C If k > 0 is a member, then so is A179682(k).
%C Includes A033996.
%C Conjecture: every member of the sequence is a member of A033996 or is A179682(k) for some k in the sequence.
%C A number k in this list indicates that A083481(k) is the same as some A083481(k') at an earlier place k'<k. E.g., 8 appears because 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
%e 24 is a term because A179682(24) = 242: 24 < 242 < 4*24*25 and 24*25*242*243 = 5940^2.
%p A179682:= proc(n) local F, t, p, k0, d, k, a, j;
%p p:= max(map(t -> `if`(t[2]::odd, t[1], NULL), [op(ifactors(n)[2]), op(ifactors(n+1)[2])]));
%p if n mod p = 0 then k0:= n+p-1; d:= 1;
%p else k0:= n+1; d:= p-1;
%p fi;
%p t:= n*(n+1)/4;
%p for a from k0 by p do
%p for k in [a, a+d] do
%p if issqr(k*(k+1)*t) then return k fi
%p od od
%p end proc:
%p f(0):= 1:
%p select(t -> A179682(t) <> 4*t*(t+1), [$0..11000]);
%Y Cf. A033996, A179682.
%K nonn
%O 1,2
%A _Robert Israel_, Feb 15 2019