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

Author

Zhi-Wei Sun, Dec 05 2021

Keywords

• nonn

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.

Examples

			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.
		

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.

Crossrefs

Cf. A000290, A000583, A349943.

Programs

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]