OFFSET
0,2
COMMENTS
Conjecture: We have {a*w^3+b*x^3+c*y^2+d*z^2: w,x,y,z = 0,1,2,...} = {0,1,2,...} if (a,b,c,d) is among the following 63 quadruples:
(1,1,1,2),(1,1,2,4),(1,2,1,1),(1,2,1,2),(1,2,1,3),(1,2,1,4),(1,2,1,6),(1,2,1,13),(1,2,2,3),(1,2,2,4),(1,2,2,5),(1,3,1,1),(1,3,1,2),(1,3,1,3),(1,3,1,5),(1,3,1,6),(1,3,2,3),(1,3,2,4),(1,3,2,5),(1,4,1,1),(1,4,1,2),(1,4,1,3),(1,4,2,2),(1,4,2,3),(1,4,2,5),(1,5,1,1),(1,5,1,2),(1,6,1,1),(1,6,1,3),(1,7,1,2),(1,8,1,2),(1,9,1,2),(1,9,2,4),(1,10,1,2),(1,11,1,2),(1,11,2,4),(1,12,1,2),(1,14,1,2),(1,15,1,2),(2,3,1,1),(2,3,1,2),(2,3,1,3),(2,3,1,4),(2,4,1,1),(2,4,1,2),(2,4,1,6),(2,4,1,8),(2,4,1,10),(2,5,1,3),(2,6,1,1),(2,7,1,3),(2,8,1,1),(2,8,1,4),(2,10,1,1),(2,13,1,1),(3,4,1,2),(3,5,1,2),(3,7,1,2),(3,9,1,2),(4,5,1,2),(4,6,1,2),(4,8,1,2),(4,11,1,2).
Conjecture verified up to 10^11 for all quadruples. - Mauro Fiorentini, Jul 18 2023
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
EXAMPLE
a(7) = 2 since 7 = 1^3 + 2*0^3 + 2^2 + 2*1^2 = 1^3 + 2*1^3 + 2^2 + 2*0^2.
a(15) = 1 since 15 = 1^3 + 2*1^3 + 2^2 + 2*2^2.
MATHEMATICA
SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^3-2y^3-2z^2], r=r+1], {x, 0, n^(1/3)}, {y, 0, ((n-x^3)/2)^(1/3)}, {z, 0, Sqrt[(n-x^3-2y^3)/2]}]; Print[n, " ", r]; Continue, {n, 0, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 03 2015
STATUS
approved