A349942 Number of ways to write n as a^4 + b^2 + (c^4 + d^2)/25 with a,b,c,d nonnegative integers.

1, 4, 6, 4, 3, 5, 4, 1, 1, 4, 8, 7, 2, 4, 6, 2, 4, 12, 13, 6, 7, 9, 4, 1, 2, 11, 19, 11, 2, 10, 10, 2, 6, 12, 12, 9, 11, 9, 8, 4, 3, 16, 18, 7, 1, 13, 10, 1, 4, 7, 17, 15, 11, 11, 10, 2, 4, 12, 11, 9, 4, 13, 12, 5, 3, 15, 25, 10, 10, 12, 8, 3, 4, 9, 17, 17, 4, 14, 16, 3, 5, 20, 20, 14, 13, 12, 14, 4, 3, 12, 30, 22, 3, 12, 13, 4, 4, 16, 24, 20, 11

Offset

0,2

Comments

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 7, 8, 23, 44, 47).

We have verified this for n up to 3*10^5.

As m/n = (m*n^3)/n^4 for any nonnegative integers m and n > 0, the conjecture implies that each nonnegative rational number can be written as x^4 + 25*y^4 + z^2 + w^2 with x,y,z,w rational numbers.

See also A349943 for similar conjectures.

Links

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

Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167--190.

Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.

Example

a(0) = 1 with 0 = 0^4 + 0^2 + (0^4 + 0^2)/25.
a(7) = 1 with 7 = 1^4 + 2^2 + (1^4 + 7^2)/25.
a(8) = 1 with 8 = 0^4 + 2^2 + (0^4 + 10^2)/25.
a(23) = 1 with 23 = 1^4 + 3^2 + (1^4 + 18^2)/25.
a(44) = 1 with 44 = 1^4 + 3^2 + (5^4 + 15^2)/25.
a(47) = 1 with 47 = 1^4 + 6^2 + (3^4 + 13^2)/25.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[25(n-x^4-y^2)-z^4], r=r+1], {x, 0, n^(1/4)}, {y, 0, Sqrt[n-x^4]}, {z, 0, (25(n-x^4-y^2))^(1/4)}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]

Crossrefs

Cf. A000290, A000583, A349943.

Sequence in context: A204693 A204817 A199721 * A187147 A128633 A001482

Adjacent sequences: A349939 A349940 A349941 * A349943 A349944 A349945

Keyword

nonn

Author

Zhi-Wei Sun, Dec 05 2021

Status

approved