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A272384
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Primes p == 1 (mod 3) for which A261029(22*p) = 2.
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7
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7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, 139, 151, 181, 211, 229, 307, 313, 421
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OFFSET
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1,1
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COMMENTS
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Theorem. Let q==2 (mod 3) be a fixed prime and for a prime p==1 (mod 3), A261029(2*q*p) > 1. Then p < (2*q)^2.
In this sequence q=11, a(n) < 484.
Proof of Theorem. Recall that in A261029 we consider the number of representations of nonnegative numbers by the form: F(x,y,z) = x^3 + y^3 + z^3 - 3*x*y*z with the conditions 0 <= x <= y <= z, z >= x+1.
Note that F(x,y,z) = (x+y+z)*G(x,y,z), where G(x,y,z) = x^2 + y^2 + z^2 - x*y - x*z - y*z. It is easy to see that G==0 or 1(mod 3). Therefore, G differs from 2,2*p and q*p. In the case when G=1, x+y+z = 2*q*p, we have the representation [Shevelev] 2*p*q = F(k-1,k-1,k), where k=2(p*q + 1)/3.
Thus one can obtain more representations only in the case G=p, x+y+z = 2*q. Let a = z-x >= 1, then y = 2*q - 2*x - a. Then G=p yields 9*x^2 + (9a - 12q)*x + (3*a^2 - 6*a*q + 4q^2 - p) = 0. Solving this equation gives x = (4*q - 3*a +- sqrt(4*p - 3*a^2))/6. Since x >= 0, the choice of minus is possible only if p <= ((4*q - 3*a)^2 - 3*a^2)/4 or, since 1 <= a < 2*q, p < 4q^2. Now choose plus. Let b = sqrt(4*p - 3a^2). The condition y >= x yields b <= a. So p = (3*a^2 + b^2)/4 <= a^2 = (z-x)^2 < 4q^2.
It remains to show that the case G=2q, x+y+z = p is impossible. Indeed, in this case x = (2*p - 3*a +- sqrt(8*q - 3*a^2))/6. Let c = sqrt(8*q - 3*a^2), i.e., q = (3*a^2 + c*2)/8. Since q is prime, then 3*a^2 + c^2 should be divisible by 2^3. But it is easy to show that this is impossible for any integers a,c. QED
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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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EXTENSIONS
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All terms (after author's first terms) were calculated by Peter J. C. Moses, Apr 28 2016
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STATUS
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approved
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