A347824 Number of ways to write n as x^4 + y^4 + (z^2 + 23*w^2)/16, where x,y,z,w are nonnegative integers with x <= y.

1, 2, 3, 3, 3, 2, 2, 1, 2, 3, 3, 2, 2, 3, 3, 1, 3, 4, 6, 4, 4, 1, 1, 2, 4, 7, 6, 4, 5, 6, 2, 2, 5, 5, 4, 3, 4, 3, 4, 3, 6, 8, 3, 4, 4, 2, 2, 3, 8, 5, 6, 2, 6, 5, 5, 6, 7, 2, 3, 4, 2, 2, 2, 4, 7, 5, 4, 1, 5, 3, 4, 7, 4, 6, 5, 4, 2, 1, 5, 5, 7, 7, 7, 6, 5, 3, 5, 4, 7, 7, 5, 4, 2, 5, 11, 7, 6, 9, 11, 5, 5

Offset: 0

Author

Zhi-Wei Sun, Jan 23 2022

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,....

This has been verified for n up to 2*10^6. See also A347827 for a further refinement.

It seems that a(n) = 1 only for n = 0, 7, 15, 21, 22, 67, 77, 137, 252, 291, 437, 471, 477, 597, 1161, 4692, 7107.

For m = 32, 48, we also conjecture that every n = 0,1,2,... can be written as x^4 + y^4 + (z^2 + 23*w^2)/m, where x,y,z,w are nonnegative integers.

Examples

			a(7) = 1 with 7 = 0^4 + 1^4 + (2^2 + 23*2^2)/16.
a(15) = 1 with 15 = 1^4 + 1^4 + (1^2 + 23*3^2)/16.
a(67) = 1 with 67 = 1^4 + 2^4 + (15^2 + 23*5^2)/16.
a(477) = 1 with 477 = 0^4 + 2^4 + (27^2 + 23*17^2)/16.
a(597) = 1 with 597 = 2^4 + 4^4 + (5^2 + 23*15^2)/16.
a(1161) = 1 with 1161 = 2^4 + 2^4 + (89^2 + 23*21^2)/16.
a(4692) = 1 with 4692 = 2^4 + 5^4 + (248^2 + 23*12^2)/16.
a(7107) = 1 with 7107 = 1^4 + 5^4 + (239^2 + 23*45^2)/16.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, Sums of four rational squares with certain restrictions, arXiv:2010.05775 [math.NT], 2020-2022.

Crossrefs

Cf. A000290, A000583, A004831, A347562, A347827, A349942, A349956.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[16(n-x^4-y^4)-23z^2],r=r+1],{x,0,(n/2)^(1/4)},{y,x,(n-x^4)^(1/4)},{z,0,Sqrt[16(n-x^4-y^4)/23]}];tab=Append[tab,r],{n,0,100}];Print[tab]