A385714 a(n) is the smallest positive integer k such that the Diophantine equation x^3 + y^3 + z^3 = k^n, where 0 < x < y < z has an integer solution, or -1 if no such integer exists.
36, 6, 6, 11, 6, 3, 6, 6, 3, 6, 6, 3, 6, 6, 3, 6, 6, 3, 6, 6, 3, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2, 6, 3, 2
Offset: 1
Examples
n [x, y, z] k^n 1 [1, 2, 3] 36 = 36^1; 2 [1, 2, 3] 36 = 6^2; 3 [3, 4, 5] 216 = 6^3; 4 [12, 17, 20] 14641 = 11^4; 5 [6, 12, 18] 7776 = 6^5; 6 [1, 6, 8] 729 = 3^6;
Formula
For 4 < n <= 22, if n == 0 (mod 3) then a(n) = 3, else a(n)=6.
From Chai Wah Wu, Jul 15 2025: (Start)
Conjecture: if n >= 24 and n == 0 (mod 3), then a(n) = 2. Verified for n <= 69.
Conjecture: if n >= 23 and n == 2 (mod 3), then a(n) = 3. Verified for n <= 44.
Conjecture: if n >= 7 and n == 1 (mod 3), then a(n) = 6. (End)
From David A. Corneth, Jul 15 2025: (Start)
For n >= 24 and n == 0 (mod 3) indeed we have a(n) = 2. Proof: holds for n = 24. For multiples of 3 that are >= 27 and m >= 0 we have:
2^(27 + 3*m) = (18 * 2^m)^3 + (366 * 2^m)^3 + (440 * 2^m)^3.
In general 2 <= a(n+3) <= a(n) therefore a(n) <= 6 for n >= 4. (End).
Extensions
a(7)-a(12) from David A. Corneth, Jul 07 2025
a(16)-a(27) from Chai Wah Wu, Jul 15 2025
a(28)-a(34) from David A. Corneth, Jul 15 2025, Jul 17 2025