A272382 Primes p == 1 (mod 3) for which A261029(14*p) = 3.

13, 19, 31, 37, 43, 61, 67, 97, 157

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

Table of n, a(n) for n=1..9.

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

Cf. A261029, A272381, A272384, A272404, A272406, A272407, A272409.

Sequence in context: A164318 A307533 A092738 * A376830 A115093 A158195

Adjacent sequences: A272379 A272380 A272381 * A272383 A272384 A272385

Keyword

nonn,fini,full

Author

Vladimir Shevelev, Apr 28 2016

Status

approved