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Number of ways to write n as w^4 + x^4 + (y^2 + 23*z^2)/16, where w is zero or a power of two (including 2^0 = 1), and x,y,z are nonnegative integers.
4

%I #17 Jan 24 2022 08:23:46

%S 1,3,4,4,4,3,2,2,2,4,5,2,2,5,4,1,4,8,8,8,6,2,2,3,6,12,9,5,9,9,4,2,8,8,

%T 6,5,4,6,4,4,11,11,6,7,6,3,3,5,11,8,7,3,9,10,5,11,9,3,4,5,3,2,3,7,10,

%U 10,6,2,7,5,8,10,5,9,7,6,4,1,6,9,9,9,10,7,5,4,6,5,13,11,6,5,3,6,16,11,6,15,15,7,5

%N Number of ways to write n as w^4 + x^4 + (y^2 + 23*z^2)/16, where w is zero or a power of two (including 2^0 = 1), and x,y,z are nonnegative integers.

%C 23-Conjecture: a(n) > 0 for all n = 0,1,2,....

%C This is stronger than the conjecture in A347824, and it has been verified for n up to 3*10^6. See also a similar conjecture in A347562.

%C It seems that a(n) = 1 only for n = 0, 15, 77, 231, 291, 437, 471, 1161, 1402, 4692, 7107, 9727.

%H Zhi-Wei Sun, <a href="/A347827/b347827.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(231) = 1 with 231 = 0^4 + 3^4 + (10^2 + 23*10^2)/16.

%e a(437) = 1 with 437 = 3^4 + 4^4 + (40^2 + 23*0^2)/16.

%e a(1402) = 1 with 1402 = 2^4 + 5^4 + (3^2 + 23*23^2)/16.

%e a(9727) = 1 with 9727 = 0^4 + 6^4 + (367^2 + 23*3^2)/16.

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

%Y Cf. A000079, A000290, A000583, A347562, A347824.

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, Jan 23 2022