A180336 Positive numbers l of the form l = A007913(4*k*m^3 - k^4), where 1 <= k <= 5*l, 1 <= m <= 5*l.

2, 3, 5, 6, 11, 15, 23, 31, 35, 47, 51, 59, 86, 106, 107, 109, 110, 129, 143, 155, 167, 174, 202, 203, 215, 230, 246, 255, 283, 307, 314, 318, 327, 341, 358, 362, 383, 419, 426, 430, 431, 433, 439, 449, 451, 499, 503, 509, 526, 527, 533, 557, 602, 606, 635, 643

Offset: 1

Author

Vladimir Shevelev, Aug 28 2010, Aug 29 2010

Keywords

• nonn

Comments

The equation x^3 + y^3 + z^3 = 0 is solvable in numbers of the form N + M*sqrt(-a(n)), where M and N are integers. Moreover, it is solvable in numbers of the form N + M*sqrt(-l), where l>0 has the form l = A007913(4*k*m^3 - k^4), where k,m >= 1 (without restrictions k,m <= 5*l). But in this more general case there could be unknown numbers l having this form; this circumstance does not allow construction of the full sequence of such l. Therefore we restrict ourselves by the condition k,m <= 5*l. Note that testing l with respect to this condition is rather simple by sorting all values of k,m <= l. One can prove that, at least, in case the Fermat numbers (A000215) are squarefree, the sequence is infinite. Conjecture (necessity of the form of l): If the equation x^3 + y^3 + z^3 = 0 is solvable in numbers of the form N + M*sqrt(-l) with integer N,M, then there exist positive integers k,m such that l = A007913(4*k*m^3 - k^4).

Examples

			We have 2=A007913(4*k*m^3-k^4) for k=2,m=3. Therefore a(1)=2; furthermore, 3=A007913(4*k*m^3-k^4) for k=m=1. Therefore a(2)=3.
		

Crossrefs

Cf. A180323, A007913, A008833, A000215.

Programs

Mathematica

squareFreePart[n_] := Times @@ (#[[1]] ^ Mod[ #[[2]], 2] & /@ FactorInteger@n);
fQ[n_] := If[b = 0; SquareFreeQ@n, Block[{k = 1, m}, While[k < 5 n + 1, m = -5 n; While[m < 5 n + 1, a = 4 k*m^3 - k^4; If[a > 0, a = squareFreePart@ a, a = 0]; If[a == n, b = a; Print[{a, k, m}]; Goto@ fini, 0]; m++ ]; k++ ]]; Label@ fini; b == n, False];
k = 1; lst = {}; While[k < 300, If[ fQ@k, AppendTo[lst, k]]; k++ ]; lst
(* Robert G. Wilson v, Aug 29 2010 *)

Formula

Let a(n)=A007913(4*k*m^3-k^4). Put h=sqrt(A008833(4*k*m^3-k^4)). Then identity A^3+B^3+C^3=0 is valid for A=2*m^6-k^3*m^3-k^6+k*(k^3+5*m^3)*h*sqrt(-a(n)); B=3*m*(k^3-m^3)*(k^2-h*sqrt(-a(n))); C=k^6-8*k^3*m^3-2*m^6-k*(k^3-4*m^3)*h*sqrt(-a(n)).

Extensions

Missing term 35 added and a(14)-a(30) from Robert G. Wilson v, Aug 29 2010

a(31)-a(63) from Robert G. Wilson v, Sep 04 2010