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Primes p == 1 (mod 3) for which A261029(14*p) = 3.
7

%I #18 Nov 21 2018 09:23:12

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

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

%C _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.

%C Theorem. If p,q == 1 (mod 3) are prime and A261029(2*q*p) > 2, then sqrt(q)/2 < p < 4*q^2.

%C In this sequence q=7, so a(n) < 196.

%C Proof of Theorem is similar to proof of the theorem in A272384.

%H Vladimir Shevelev, <a href="http://arxiv.org/abs/1508.05748">Representation of positive integers by the form x^3+y^3+z^3-3xyz</a>, arXiv:1508.05748 [math.NT], 2015.

%t 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];

%t a29[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];

%t Select[Select[Range[1, 1000, 3], PrimeQ], a29[14 #] == 3&] (* _Jean-François Alcover_, Nov 21 2018 *)

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

%K nonn,fini,full

%O 1,1

%A _Vladimir Shevelev_, Apr 28 2016