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A272384 Primes p == 1 (mod 3) for which A261029(22*p) = 2. 7
7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, 139, 151, 181, 211, 229, 307, 313, 421 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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

LINKS

Table of n, a(n) for n=1..22.

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[22 #] == 2&] (* Jean-Fran├žois Alcover, Nov 21 2018 *)

CROSSREFS

Cf. A261029, A272381, A272382, A272404, A272406, A272407, A272409.

Sequence in context: A144921 A272409 A272406 * A040079 A038160 A106870

Adjacent sequences:  A272381 A272382 A272383 * A272385 A272386 A272387

KEYWORD

nonn,fini,full

AUTHOR

Vladimir Shevelev, Apr 28 2016

EXTENSIONS

All terms (after author's first terms) were calculated by Peter J. C. Moses, Apr 28 2016

STATUS

approved

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Last modified September 24 08:31 EDT 2021. Contains 347623 sequences. (Running on oeis4.)