A348890 Number of ways to write n as 5*w^4 + x^4 + y^2 + z^2, where w,x,y,z are nonnegative integers with y <= z.

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

Offset: 0

Author

Zhi-Wei Sun, Jan 28 2022

Keywords

• nonn

Comments

Conjecture: a(n) = 0 only for n == 12 (mod 16).

This has been verified for n up to 10^8.

Now we show that a(n) = 0 whenever n == 12 (mod 16). If 16*q + 12 = 5*w^4 + x^4 + y^2 + z^2 with q,w,x,y,z integers, then the equality modulo 8 yields that w,x,y,z are all even, hence 4*q + 3 == 20*(w/2)^4 + 4*(x/2)^4 + (y/2)^2 + (z/2)^2 and thus (y/2)^2 + (z/2)^2 == 3 (mod 4) which is impossible.

It seems that a(n) = 1 only for n = 0, 3, 4, 27, 43, 48, 49, 63, 67, 72, 75, 192, 215, 303, 1092.

Examples

			a(192) = 1 with 192 = 5*1^4 + 3^4 + 5^2 + 9^2.
a(215) = 1 with 215 = 5*1^4 + 2^4 + 5^2 + 13^2.
a(303) = 1 with 303 = 5*1^4 + 0^4 + 3^2 + 17^2.
a(1092) = 1 with 1092 = 5*0^4 + 2^4 + 20^2 + 26^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34 (2017), no.2, 97-120.
• Zhi-Wei Sun, Sums of four rational squares with certain restrictions, arXiv:2010.05775 [math.NT], 2020-2022.

Crossrefs

Cf. A000290, A000583, A214891, A346643, A347865, A350857.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[n-5x^4-y^4-z^2],r=r+1],{x,0,(n/5)^(1/4)},{y,0,(n-5x^4)^(1/4)},
{z,0,Sqrt[(n-5x^4-y^4)/2]}];tab=Append[tab,r],{n,0,100}];Print[tab]