A262824 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.

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, 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, 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

Offset

0,2

Comments

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.

(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,...}.

Conjectures (i) and (ii) verified up to 10^11. - Mauro Fiorentini, Jul 22 2023

See also A262827 and A262857 for similar conjectures.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Example

a(0) = 1 since 0 = 0^2 + 0^3 + 2*0^3 + 3*0^3.
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.
a(23) = 1 since 23 = 2^2 + 0^3 + 2*2^3 + 3*1^3.
a(37) = 1 since 37 = 6^2 + 1^3 + 2*0^3 + 3*0^3.
a(72) = 1 since 72 = 8^2 + 2^3 + 2*0^3 + 3*0^3.

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
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}]

Crossrefs

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

Sequence in context: A352636 A117114 A307981 * A242899 A347865 A306606

Adjacent sequences: A262821 A262822 A262823 * A262825 A262826 A262827

Keyword

nonn

Author

Zhi-Wei Sun, Oct 03 2015

Status

approved