A272341 - OEIS

%I #17 Nov 16 2017 15:52:26

%S 0,1,2,3,5,6,8,9,12,15

%N Numbers n such that 6^n is not of the form (x^3 + y^3 + z^3) / 3 where x > y > z > 0.

%C Numbers n such that 6^n cannot be written as the average of three distinct cubes.

%C This sequence is complete. So 6^n can be written as the average of three distinct positive cubes for all n > 15.

%C Proof: If 6^n = (x^3 + y^3 + z^3)/3, then 6^(n+3) = 6^3 * 6^n = 6^3 * (x^3 + y^3 + z^3)/3 = ((6*x)^3 + (6*y)^3 + (6*z)^3)/3. It is clear that if x, y and z are distinct, then k*x, k*y and k*z are also distinct for all nonzero values of k. So with 6^n also 6^(n+3) is not in the sequence. Therefore, if we can find an integer n such that 6^n, 6^(n+1) and 6^(n+2) are not in the sequence then all 6^m with m >= n will also not be in the sequence. In the example section it is shown that 6^n is not in the sequence for n = 4, 11 and 18. So (4+3*u), (11+3*v) and (18+3*w) cannot be member of this sequence for all nonnegative values of u, v and w. If we choose u=4, v=2 and w=0, we obtain the consecutive values that are 16, 17, 18. Thus this sequence is finite, and all its terms can be found in limited range 0 <= n <= 15.

%e 2 is a term because 3*6^2 = 108 is not a term of A024975.

%e 4 is not a term because 3*6^4 = 3888 = 1^3 + 8^3 + 15^3.

%e 11 is not a term because 3*6^11 = 1088391168 = 48^3 + 124^3 + 1028^3.

%e 18 is not a term because 3*6^18 = 304679870005248 = 38322^3 + 47840^3 + 51790^3.

%t Select[Range@ 12, Length[PowersRepresentations[3*6^#, 3, 3] /. {x_, y_, z_} /; Or[x == 0, x == y == z] -> Nothing] == 0 &] (* _Michael De Vlieger_, Apr 27 2016, Version 10.2 *)

%Y Cf. A000400, A024975.

%K nonn,fini,full

%O 1,3

%A _Altug Alkan_, Apr 26 2016