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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.

%I A385714 #51 Jul 21 2025 22:31:08
%S A385714 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,
%T A385714 6,3,2
%N 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.
%F A385714 For 4 < n <= 22, if n == 0 (mod 3) then a(n) = 3, else a(n)=6.
%F A385714 From _Chai Wah Wu_, Jul 15 2025: (Start)
%F A385714 Conjecture: if n >= 24 and n == 0 (mod 3), then a(n) = 2. Verified for n <= 69.
%F A385714 Conjecture: if n >= 23 and n == 2 (mod 3), then a(n) = 3. Verified for n <= 44.
%F A385714 Conjecture: if n >= 7 and n == 1 (mod 3), then a(n) = 6. (End)
%F A385714 From _David A. Corneth_, Jul 15 2025: (Start)
%F A385714 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:
%F A385714 2^(27 + 3*m) = (18 * 2^m)^3 + (366 * 2^m)^3 + (440 * 2^m)^3.
%F A385714 In general 2 <= a(n+3) <= a(n) therefore a(n) <= 6 for n >= 4. (End).
%e A385714 n [x, y, z]         k^n
%e A385714 1 [1, 2, 3]       36 = 36^1;
%e A385714 2 [1, 2, 3]       36 = 6^2;
%e A385714 3 [3, 4, 5]      216 = 6^3;
%e A385714 4 [12, 17, 20] 14641 = 11^4;
%e A385714 5 [6, 12, 18]   7776 = 6^5;
%e A385714 6 [1, 6, 8]      729 = 3^6;
%Y A385714 Cf. A385354, A385565, A384430, A385566.
%K A385714 nonn,more
%O A385714 1,1
%A A385714 _Jean-Marc Rebert_, Jul 07 2025
%E A385714 a(7)-a(12) from _David A. Corneth_, Jul 07 2025
%E A385714 a(16)-a(27) from _Chai Wah Wu_, Jul 15 2025
%E A385714 a(28)-a(34) from _David A. Corneth_, Jul 15 2025, Jul 17 2025