login
A262857
Number of ordered ways to write n as w^3 + 2*x^3 + y^2 + 2*z^2, where w, x, y and z are nonnegative integers.
11
1, 2, 3, 4, 4, 3, 3, 2, 3, 5, 5, 6, 6, 3, 4, 1, 4, 6, 7, 10, 7, 5, 4, 2, 5, 8, 8, 9, 9, 6, 6, 2, 6, 10, 8, 13, 9, 6, 7, 5, 5, 8, 6, 9, 10, 6, 9, 4, 5, 9, 6, 13, 10, 7, 11, 6, 8, 10, 8, 10, 12, 9, 9, 7, 8, 13, 10, 16, 12, 6, 12, 8, 10, 13, 12, 13, 12, 8, 11, 7, 10, 16, 15, 17, 16, 6, 11, 7, 12, 16, 11, 16, 9, 10, 5, 6, 10, 15, 17, 18, 16
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
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}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 03 2015
STATUS
approved