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A343411
Number of ways to write n as x^3 + [y^3/2] + [z^3/3] + 2^w, where [.] is the floor function, x,y,z are positive integers, and w is a nonnegative integer.
7
0, 1, 1, 1, 2, 1, 2, 1, 4, 1, 4, 3, 2, 3, 4, 5, 2, 7, 2, 2, 2, 5, 4, 5, 5, 3, 3, 3, 3, 7, 6, 3, 5, 5, 6, 2, 11, 3, 6, 2, 6, 6, 8, 10, 2, 9, 2, 5, 5, 10, 5, 2, 6, 4, 4, 7, 5, 7, 2, 2, 4, 6, 7, 3, 12, 3, 7, 4, 9, 6, 5, 10, 4, 15, 4, 8, 5, 11, 4, 8, 14, 6, 4, 6, 10, 7, 8, 9, 5, 6, 4, 4, 13, 5, 7, 3, 10, 2, 7, 11
OFFSET
1,5
COMMENTS
Conjecture: a(n) > 0 for all n > 1.
We have verified a(n) > 0 for all 1 < n <= 3*10^5.
Conjecture verified up to 10^10. - Giovanni Resta, Apr 14 2021
LINKS
Zhi-Wei Sun, Natural numbers represented by [x^2/a] + [y^2/b] + [z^2/c], arXiv:1504.01608 [math.NT], 2015.
EXAMPLE
a(2) = 1 with 2 = 1^3 + [1^3/2] + [1^3/3] + 2^0.
a(3) = 1 with 3 = 1^3 + [1^3/2] + [1^3/3] + 2^1.
a(4) = 1 with 4 = 1^3 + [1^3/2] + [2^3/3] + 2^0.
a(6) = 1 with 6 = 1^3 + [2^3/2] + [1^3/3] + 2^0.
a(8) = 1 with 8 = 1^3 + [2^3/2] + [2^3/3] + 2^0.
a(10) = 1 with 10 = 2^3 + [1^3/2] + [1^3/3] + 2^1.
a(103) = 1 with 103 = 3^3 + [1^3/2] + [6^3/3] + 2^2.
MATHEMATICA
CQ[n_]:=CQ[n]=n>0&&IntegerQ[n^(1/3)];
tab={}; Do[r=0; Do[If[CQ[n-Floor[x^3/2]-Floor[y^3/3]-2^z], r=r+1], {x, 1, (2n-1)^(1/3)}, {y, 1, (3(n-Floor[x^3/2])-1)^(1/3)}, {z, 0, Log[2, n-Floor[x^3/2]-Floor[y^3/3]]}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 14 2021
STATUS
approved