A262880 - OEIS

%I #10 Oct 05 2015 00:14:17

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

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

%U 6,5,9,6,9,3,7,5,7,3,5,9,3,11,6,9,5,3,7,4,4,7,9,8,5,8,7,7,2,6,7,4

%N Number of ordered ways to write n as w*(w+1)/2 + x^3 + y^3 + 2*z^3 with w > 0, 0 <= x <= y and z >= 0.

%C Conjecture: (i) Any positive integer can be written as w*(w+1)/2 + x^3 + b*y^3 + c*z^3 with w > 0 and x,y,z >= 0, provided that (b,c) is among the following ordered pairs: (1,2),(1,3),(1,4),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6),(2,7),(2,20),(2,21),(2,34),(3,3),(3,4),(3,5),(3,6),(4,10).

%C (ii) For (b,c) = (3,4),(3,6),(4,8), we have {w*(w+1)/2 + 2*x^3 + b*y^3 + c*z^3: w,x,y,z = 0,1,2,...} = {0,1,2,...}.

%C See also A262813, A262824 and A262857 for similar conjectures.

%H Zhi-Wei Sun, <a href="/A262880/b262880.txt">Table of n, a(n) for n = 1..10000</a>

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

%e a(34) = 2 since 34 = 4*5/2 + 0^3 + 2^3 + 2*2^3 = 3*4/2 + 1^3 + 3^3 + 2*0^3.

%e a(41) = 1 since 41 = 3*4/2 + 2^3 + 3^3 + 2*0^3.

%e a(51) = 1 since 51 = 6*7/2 + 1^3 + 3^3 + 2*1^3.

%e a(104) = 1 since 104 = 5*6/2 + 2^3 + 3^3 + 2*3^3.

%t TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]

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

%Y Cf. A000217, A000578, A262813, A262824, A262857.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Oct 04 2015