A273615 Numbers k such that k^4 is the average of two positive cubes while k is not.

329, 518, 566, 662, 732, 741, 777, 804, 806, 876, 921, 998, 1029, 1092, 1236, 1238, 1317, 1497, 1526, 1596, 1812, 1862, 1929, 1988, 2181, 2316, 2604, 2632, 2757, 4204, 4396, 4446, 4684, 5068, 5548, 5782, 5838, 5856, 5928, 5982, 6124, 6126, 6216

Offset: 1

Author

Altug Alkan, May 26 2016

Keywords

• nonn

Comments

If k is the average of two positive cubes, then k^4 is also the average of two positive cubes. So this sequence focuses on the solutions that are not trivial.

Examples

			329 is a term because 329 is not the average of two positive cubes while 329^4 = (1833^3 + 2585^3)/2.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..502

Crossrefs

Cf. A003325, A051387.

Programs

Maple

Q:=  proc(x) local t;
  for t in select(t -> t^3<=x and 4*t^3 > x and x/t - t^2 mod 3 = 0,
        numtheory:-divisors(x)) do
    if issqr((x/t - t^2)/3)  then return true fi
  od:
  false
end proc:
select(x -> not(Q(x)) and Q(x^4), [$1..10000]); # Robert Israel, May 26 2016

Mathematica

Q[x_] := Module[{s, t}, s = Select[Divisors[x], #^3 <= x && 4*#^3 > x && Mod[x/# - #^2, 3] == 0 &]; For[t = 1, t <= Length[s], t++, If[IntegerQ@Sqrt[(x/s[[t]] - s[[t]]^2)/3],  Return[True]]]; False];
Reap[For[x = 1, x <= 10000, x++, If[!Q[x] && Q[x^4], Print[x]; Sow[x]]]][[2, 1]] (* Jean-François Alcover, May 18 2023, after Robert Israel *)

PARI

isA003325(n) = for(k=1, sqrtnint(n\2, 3), ispower(n-k^3, 3) && return(1));
lista(nn) = for(n=1, nn, if(isA003325(2*n^4) && !isA003325(2*n), print1(n, ", ")));

PARI

T=thueinit('z^3+1);
isA003325(n)=#select(v->min(v[1], v[2])>0, thue(T, n))>0
is(n)=isA003325(2*n^4) && !isA003325(2*n) \\ Charles R Greathouse IV, May 27 2016