A262824 - OEIS

%I #21 Jul 23 2023 01:45:13

%S 1,2,2,3,4,3,3,3,2,3,3,3,4,2,3,2,2,5,2,4,5,3,2,1,4,5,5,6,8,5,4,5,3,7,

%T 3,4,8,1,4,3,4,7,4,5,4,3,3,3,3,6,5,3,9,3,4,7,3,7,3,5,4,2,6,5,4,6,8,7,

%U 8,5,5,5,1,6,4,3,7,2,5,5,5,8,8,10,9,6,3,7,6,8,9,9,8,5,6,4,3,6,7,4,7

%N Number of ordered ways to write n as w^2 + x^3 + 2*y^3 + 3*z^3, where w, x, y and z are nonnegative integers.

%C Conjecture: (i) For any m = 3, 4, 5, 6 and n >= 0, there are nonnegative integers w, x, y, z such that n = w^2 + x^3 + 2*y^3 + m*z^3.

%C (ii) For P(w,x,y,z) = w^2 + x^3 + 2*y^3 + z^4, w^2 + x^3 + 2*y^3 + 3*z^4, w^2 + x^3 + 2*y^3 + 6*z^4, 2*w^2 + x^3 + 4*y^3 + z^4, we have {P(w,x,y,z): w,x,y,z = 0,1,2,...} ={0,1,2,...}.

%C Conjectures (i) and (ii) verified up to 10^11. - _Mauro Fiorentini_, Jul 22 2023

%C See also A262827 and A262857 for similar conjectures.

%H Zhi-Wei Sun, <a href="/A262824/b262824.txt">Table of n, a(n) for n = 0..10000</a>

%e a(0) = 1 since 0 = 0^2 + 0^3 + 2*0^3 + 3*0^3.

%e a(8) = 2 since 8 = 2^2 + 1^3 + 2*0^3 + 3*1^3 = 0^2 + 2^3 + 2*0^3 + 3*0^3.

%e a(23) = 1 since 23 = 2^2 + 0^3 + 2*2^3 + 3*1^3.

%e a(37) = 1 since 37 = 6^2 + 1^3 + 2*0^3 + 3*0^3.

%e a(72) = 1 since 72 = 8^2 + 2^3 + 2*0^3 + 3*0^3.

%t SQ[n_]:=IntegerQ[Sqrt[n]]

%t Do[r=0;Do[If[SQ[n-x^3-2y^3-3z^3],r=r+1],{x,0,n^(1/3)},{y,0,((n-x^3)/2)^(1/3)},{z,0,((n-x^3-2y^3)/3)^(1/3)}];Print[n," ",r];Continue,{n,1,100}]

%Y Cf. A000290, A000578, A262813, A262815, A262816, A262827, A262857.

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, Oct 03 2015