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A270469
Number of ordered ways to write n = x^3 + y*(y+1) + z*(3*z+2), where x and y are nonnegative integers and z is a nonzero integer.
19
1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 3, 1, 2, 2, 3, 2, 2, 3, 2, 1, 4, 4, 2, 2, 2, 2, 1, 5, 4, 2, 2, 2, 3, 4, 4, 5, 2, 2, 3, 3, 5, 2, 5, 3, 2, 4, 5, 4, 2, 3, 3, 3, 3, 4, 3, 1, 2, 5, 3, 4, 3, 4, 4, 4, 5, 4, 3, 4, 4, 3, 6, 5, 5, 3, 3, 3, 6, 6, 2, 4
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.
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}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 17 2016
STATUS
approved