A262880 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.

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, 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, 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

Offset: 1

Author

Zhi-Wei Sun, Oct 04 2015

Keywords

• nonn

Comments

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).

(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,...}.

See also A262813, A262824 and A262857 for similar conjectures.

Examples

			a(2) = 1 since 2 = 1*2/2 + 0^3 + 1^3 + 2*0^3.
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.
a(41) = 1 since 41 = 3*4/2 + 2^3 + 3^3 + 2*0^3.
a(51) = 1 since 51 = 6*7/2 + 1^3 + 3^3 + 2*1^3.
a(104) = 1 since 104 = 5*6/2 + 2^3 + 3^3 + 2*3^3.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

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

Programs

Mathematica

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