A271381 Number of ordered ways to write n as u^2 + v^2 + x^3 + y^3, where u, v, x, y are nonnegative integers with 2 | u*v, u <= v and x <= y.

1, 2, 2, 1, 1, 2, 2, 1, 2, 4, 3, 1, 1, 3, 2, 1, 3, 5, 3, 1, 2, 3, 2, 0, 2, 5, 3, 3, 3, 4, 1, 2, 4, 5, 3, 2, 5, 4, 3, 2, 4, 6, 2, 3, 4, 5, 2, 2, 4, 3, 2, 2, 5, 6, 4, 3, 2, 3, 2, 2, 4, 4, 3, 3, 5, 7, 4, 5, 5, 6, 4, 2, 6, 9, 6, 2, 4, 5, 1, 3, 8

Offset

0,2

Comments

Conjecture: (i) a(n) > 0 except for n = 23, and a(n) = 1 only for n = 0, 3, 4, 7, 11, 12, 15, 19, 23, 30, 78, 203, 219.

(ii) Any natural number can be written as the sum of two squares and three fourth powers.

Links

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

Example

a(3) = 1 since 3 = 0^2 + 1^2 + 1^3 + 1^3 with 0 even.
a(4) = 1 since 4 = 0^2 + 2^2 + 0^3 + 0^3 with 0 and 2 even.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^3 + 1^3 with 2 even.
a(11) = 1 since 11 = 0^2 + 3^2 + 1^3 + 1^3 with 0 even.
a(12) = 1 since 12 = 0^2 + 2^2 + 0^3 + 2^3 with 0 and 2 even.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^3 + 1^3 with 2 even.
a(19) = 1 since 19 = 1^2 + 4^2 + 1^3 + 1^3 with 4 even.
a(30) = 1 since 30 = 2^2 + 5^2 + 0^3 + 1^3 with 2 even.
a(78) = 1 since 78 = 2^2 + 3^2 + 1^3 + 4^3 with 2 even.
a(203) = 1 since 203 = 7^2 + 10^2 + 3^3 + 3^3 with 10 even.
a(219) = 1 since 219 = 8^2 + 8^2 + 3^3 + 4^3 with 8 even.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^3-y^3-u^2]&&(Mod[u*Sqrt[n-x^3-y^3-u^2], 2]==0), r=r+1], {x, 0, (n/2)^(1/3)}, {y, x, (n-x^3)^(1/3)}, {u, 0, ((n-x^3-y^3)/2)^(1/2)}]; Print[n, " ", r]; Continue, {n, 0, 80}]

Crossrefs

Cf. A000290, A000578, A000583, A262813, A262824, A262827, A262857, A262880, A270488, A270559, A270969, A271325.

Sequence in context: A320473 A194884 A078401 * A086247 A116452 A219966

Adjacent sequences: A271378 A271379 A271380 * A271382 A271383 A271384

Keyword

nonn

Author

Zhi-Wei Sun, Apr 06 2016

Status

approved