OFFSET
1,1
COMMENTS
Peter J. C. Moses did not find any term > 157. The author proved that the sequence is full. Moreover, he proved the following more general result.
Theorem. If p,q == 1 (mod 3) are prime and A261029(2*q*p) > 2, then sqrt(q)/2 < p < 4*q^2.
In this sequence q=7, so a(n) < 196.
Proof of Theorem is similar to proof of the theorem in A272384.
LINKS
Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.
MATHEMATICA
r[n_] := Reduce[0 <= x <= y <= z && z >= x + 1 && n == x^3 + y^3 + z^3 - 3 x y z, {x, y, z}, Integers];
a29[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
Select[Select[Range[1, 1000, 3], PrimeQ], a29[14 #] == 3&] (* Jean-François Alcover, Nov 21 2018 *)
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Vladimir Shevelev, Apr 28 2016
STATUS
approved