OFFSET
1,7
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 4, 5, 6, 10, 12, 20, 27, 56, 101.
(ii) Each nonnegative integer can be written as x^3 + P(y,z) with x >= 0 and y and z integers, provided that P(y,z) is among y^2+z(3z+1), y(2y+1)+z(3z+1), y(2y+1)+z(3z+2), y(2y+1)+z(5z+2), y(2y+1)+z(5z+3), y(2y+1)+2z(3z+1), y(2y+1)+ 2z(3z+2), y(2y+1)+z(6z+5), y(2y+1)+z(7z+2), y(2y+1)+z(7z+6), y(3y+1)+z(4z+1), y(3y+1)+z(7z+2), y(3y+2)+z(4z+1).
(iii) Every n = 0,1,2,... can be expressed as f(x,y,z) with x >= 0 and y and z integers, provided that f(x,y,z) is among 2x^3+y^2+z(3z+1), 2x^3+y(y+1)+z(3z+2), 3x^3+y(y+1)+z(3z+2), 2x^3+y(2y+1)+z(3z+1), 2x^3+y(2y+1)+z(3z+2), 2x^3+y(2y+1)+z(5z+4), 2x^3+y(3y+1)+z(3z+2), 2x^3+y(3y+2)+z(4z+3).
Note that those y(2y+1) with y integral are just triangular numbers.
See also A262813 for a similar conjecture.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(10) = 1 since 10 = 0^3 + 1*2 + (-2)(3*(-2)+2).
a(12) = 1 since 12 = 1^3 + 2*3 + 1*(3*1+2).
a(20) = 1 since 20 = 0^3 + 3*4 + (-2)*(3*(-2)+2).
a(27) = 1 since 27 = 0^3 + 2*3 + (-3)*(3*(-3)+2).
a(56) = 1 since 56 = 0^3 + 0*1 + 4*(3*4+2).
a(101) = 1 since 101 = 2^3 + 8*9 + (-3)*(3*(-3)+2).
MATHEMATICA
pQ[x_]:=pQ[x]=x>0&&IntegerQ[Sqrt[3x+1]]
Do[r=0; Do[If[pQ[n-x^3-y(y+1)], r=r+1], {x, 0, n^(1/3)}, {y, 0, (Sqrt[4(n-x^3)+1]-1)/2}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 17 2016
STATUS
approved