

A280153


Number of ways to write n as x^3 + 2*y^3 + z^2 + 4^k, where x is a positive integer and y,z,k are nonnegative integers.


2



0, 1, 1, 1, 2, 2, 1, 2, 2, 1, 3, 2, 3, 2, 2, 2, 1, 5, 2, 3, 4, 2, 3, 1, 4, 3, 4, 5, 4, 5, 2, 4, 4, 6, 3, 1, 6, 1, 2, 4, 3, 4, 3, 6, 3, 3, 4, 3, 5, 2, 3, 1, 5, 3, 2, 5, 2, 3, 3, 6, 3, 1, 5, 3, 4, 6, 6, 8, 7, 4, 5, 6, 3, 5, 7, 5, 3, 3, 5, 4
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OFFSET

1,5


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 7, 10, 17, 24, 36, 38, 52, 62, 115, 136, 185, 990.
(ii) Each positive integer n can be written as x^2 + 2*y^2 + z^3 + 8^k with x,y,z,k nonnegative integers.
(iii) Let a,b,c be positive integers with b <= c. Then any positive integer can be written as a*x^3 + b*y^2 + c*z^2 + 4^k with x,y,z,k nonnegative integers, if and only if (a,b,c) is among the following triples: (1,1,1), (1,1,2), (1,1,3), (1,1,5), (1,1,6), (1,2,3), (1,2,5), (2,1,1), (2,1,2), (2,1,3), (2,1,6), (4,1,2), (5,1,2), (8,1,2), (9,1,2).
We have verified that a(n) > 0 for all n = 2..10^6, and that part (ii) of the conjecture holds for all n = 1..10^6.
For any positive integer n, it is easy to see that at least one of n1, n8, n64 is not of the form 4^k*(8m+7) with k and m nonnegative integers, thus, by the GaussLegendre theorem on sums of three squares, n = x^2 + y^2 + z^2 + 8^k for some nonnegative integers x,y,z and k < 3.
See also A280356 for a similar conjecture involving powers of two.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.


EXAMPLE

a(7) = 1 since 7 = 1^3 + 2*1^3 + 0^2 + 4^1.
a(10) = 1 since 10 = 2^3 + 2*0^3 + 1^2 + 4^0.
a(17) = 1 since 17 = 1^3 + 2*0^3 + 0^2 + 4^2.
a(24) = 1 since 24 = 2^3 + 2*0^3 + 0^2 + 4^2.
a(36) = 1 since 36 = 2^3 + 2*1^3 + 5^2 + 4^0.
a(38) = 1 since 38 = 1^3 + 2*0^3 + 6^2 + 4^0.
a(52) = 1 since 52 = 3^3 + 2*0^3 + 3^2 + 4^2.
a(62) = 1 since 62 = 2^3 + 2*1^3 + 6^2 + 4^2.
a(115) = 1 since 115 = 2^3 + 2*3^3 + 7^2 + 4^1.
a(136) = 1 since 136 = 2^3 + 2*0^3 + 8^2 + 4^3.
a(185) = 1 since 185 = 3^3 + 2*3^3 + 10^2 + 4^1.
a(990) = 1 since 990 = 7^3 + 2*3^3 + 23^2 + 4^3.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[n4^kx^32y^3], r=r+1], {k, 0, Log[4, n]}, {x, 1, (n4^k)^(1/3)}, {y, 0, ((n4^kx^3)/2)^(1/3)}]; Print[n, " ", r]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000290, A000302, A000578, A262813, A262827, A262857, A270533, A270559, A280356.
Sequence in context: A227156 A123369 A178306 * A023671 A117535 A270747
Adjacent sequences: A280150 A280151 A280152 * A280154 A280155 A280156


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Dec 27 2016


STATUS

approved



