%I #40 Nov 19 2024 15:03:31
%S 0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,2,1,1,0,0,0,1,1,0,1,1,1,1,0,0,0,0,
%T 0,2,0,1,0,1,2,1,0,1,1,2,0,1,0,1,0,0,1,3,0,1,1,2,0,2,0,0,1,0,0,1,1,0,
%U 1,1,1,3,0,0,2,2,0,1,0,1,2,3,0,3,1,0,4
%N a(n) is the number of solutions to the Diophantine equation i^3 + j^3 + k^3 = n^3, where 0 < i <= j <= k.
%C The first number to have a nonzero number of solutions is 6, which is 3^3 + 4^3 + 5^3 = 6^3. Its cube 216 has been called Plato's number in reference to this.
%C First occurrence of k=0,1,2...: 0, 6, 18, 54, 87, 108, 216, 174, 348, 396, 324, 696, 864, 492, etc. - _Robert G. Wilson v_, Jul 02 2018
%H Arlu Genesis A. Padilla, <a href="/A316359/b316359.txt">Table of n, a(n) for n = 1..10000</a>
%e a(18)=2, because 18^3 = 9^3 + 12^3 + 15^3 = 2^3 + 12^3 + 16^3.
%t Array[Count[PowersRepresentations[#^3, 3, 3], _?(FreeQ[Differences@ #, 0] &)] &, 105] (* _Michael De Vlieger_, Jun 30 2018 *)
%o (PARI) a(n) = sum(i=1, n, sum(j=1, i, sum(k=1, j, i^3 + j^3 + k^3 == n^3))); \\ _Michel Marcus_, Jul 02 2018
%o (PARI) a(n)={sum(i=1, n, sum(j=1, i, my(k); ispower(n^3-j^3-i^3, 3, &k) && k>=1 && k<=j ))} \\ _Andrew Howroyd_, Jul 07 2018
%o (Python)
%o from sympy.solvers.diophantine.diophantine import power_representation
%o def A316359(n): return len(list(power_representation(n**3,3,3))) # _Chai Wah Wu_, Nov 19 2024
%Y Cf. A046080.
%K nonn
%O 1,18
%A _Arlu Genesis A. Padilla_, Jun 30 2018