%I #22 Jan 26 2022 13:09:49

%S 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,

%T 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,

%U 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

%N 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.

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

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

%C 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.

%C 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.

%H Zhi-Wei Sun, <a href="/A347824/b347824.txt">Table of n, a(n) for n = 0..10000</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/2010.05775">Sums of four rational squares with certain restrictions</a>, arXiv:2010.05775 [math.NT], 2020-2022.

%e a(7) = 1 with 7 = 0^4 + 1^4 + (2^2 + 23*2^2)/16.

%e a(15) = 1 with 15 = 1^4 + 1^4 + (1^2 + 23*3^2)/16.

%e a(67) = 1 with 67 = 1^4 + 2^4 + (15^2 + 23*5^2)/16.

%e a(477) = 1 with 477 = 0^4 + 2^4 + (27^2 + 23*17^2)/16.

%e a(597) = 1 with 597 = 2^4 + 4^4 + (5^2 + 23*15^2)/16.

%e a(1161) = 1 with 1161 = 2^4 + 2^4 + (89^2 + 23*21^2)/16.

%e a(4692) = 1 with 4692 = 2^4 + 5^4 + (248^2 + 23*12^2)/16.

%e a(7107) = 1 with 7107 = 1^4 + 5^4 + (239^2 + 23*45^2)/16.

%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];

%t 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]

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

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, Jan 23 2022