OFFSET
1,2
COMMENTS
This sequence is finite.
If 3^n = x^3 + y^3 + z^3, then 3^(n+3) = 3^3 * 3^n = 3^3 * (x^3 + y^3 + z^3) = (3*x)^3 + (3*y)^3 + (3*z)^3. So if we can find the integer n such that 3^n, 3^(n+1) and 3^(n+2) are sums of 3 positive cubes, we will exactly know that this sequence must be finite. Obviously, the minimum value of n such that 3^n, 3^(n+1) and 3^(n+2) are sums of 3 positive cubes determines the last term of this sequence. Since we can solve the 3^n = x^3 + y^3 + z^3 for n = 21, 22, 23 as in example section, all terms of this sequence is computed in limited range that is 0 <= n <= 20.
Corresponding 3^n values are 1, 9, 27, 243, 6561, 177147, 4782969, 129140163, 3486784401.
EXAMPLE
21 is not a term because 3^21 = (3^5)^3 + (6*3^5)^3 + (8*3^5)^3.
22 is not a term because 3^22 = (3^7)^3 + (3^7)^3 + (3^7)^3.
23 is not a term because 3^23 = 1658^3 + 3202^3 + 3843^3.
MATHEMATICA
Select[Range[0, 20], Length[PowersRepresentations[3^#, 3, 3] /. {{0, __} -> Nothing}] == 0 &] (* Michael De Vlieger, Apr 16 2016 *)
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Altug Alkan, Apr 16 2016
STATUS
approved