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A272382
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Primes p == 1 (mod 3) for which A261029(14*p) = 3.
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7
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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MATHEMATICA
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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]];
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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STATUS
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approved
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