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%I #31 Feb 17 2017 02:35:56
%S 1,2,3,2,2,3,3,2,1,3,4,2,1,2,2,2,2,3,5,2,3,3,2,1,1,4,5,4,2,2,4,3,3,3,
%T 6,2,6,5,3,3,3,7,6,2,2,5,4,1,2,3,7,6,8,4,5,5,2,4,5,2,3,5,3,4,2,5,9,4,
%U 5,4,5,1,3,5,4,5,5,4,2,3,3
%N Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and x + 24*y are squares.
%C Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 8, 12, 23, 24, 47, 71, 168, 344, 632, 1724).
%C By the linked JNT paper, any nonnegative integer can be written as the sum of a fourth power and three squares.
%C We have verified a(n) > 0 for all n = 0..10^7.
%C See also A281977, A282013 and A282014 for similar conjectures.
%C a(n) <= A273404(n). Starts to differ from A273404 at n=145. - _R. J. Mathar_, Feb 12 2017
%C Qing-Hu Hou at Tianjin Univ. has verified a(n) > 0 for all n = 0..10^10.
%C I would like to offer 2400 US dollars for the first proof of my conjecture that a(n) > 0 for any nonnegative integer n. - _Zhi-Wei Sun_, Feb 14 2017
%H Zhi-Wei Sun, <a href="/A281976/b281976.txt">Table of n, a(n) for n = 0..10000</a>
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.11.008">Refining Lagrange's four-square theorem</a>, J. Number Theory 175(2017), 167-190.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1701.05868">Restricted sums of four squares</a>, arXiv:1701.05868 [math.NT], 2017.
%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;d465bba.1702">The 24-conjecture with $2400 prize</a>, a message to Number Theory List, Feb. 14, 2017.
%e a(8) = 1 since 8 = 0^2 + 0^2 + 2^2 + 2^2 with 0 = 0^2 and 0 + 24*0 = 0^2.
%e a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1 = 1^2 and 1 + 24*1 = 5^2.
%e a(23) = 1 since 23 = 1^2 + 2^2 + 3^2 + 3^2 with 1 = 1^2 and 1 + 24*2 = 7^2.
%e a(24) = 1 since 24 = 4^2 + 0^2 + 2^2 + 2^2 with 4 = 2^2 and 4 + 24*0 = 2^2.
%e a(47) = 1 since 47 = 1^2 + 1^2 + 3^2 + 6^2 with 1 = 1^2 and 1 + 24*1 = 5^2.
%e a(71) = 1 since 71 = 1^2 + 5^2 + 3^2 + 6^2 with 1 = 1^2 and 1 + 24*5 = 11^2.
%e a(168) = 1 since 168 = 4^2 + 4^2 + 6^2 + 10^2 with 4 = 2^2 and 4 + 24*4 = 10^2.
%e a(344) = 1 since 344 = 4^2 + 0^2 + 2^2 + 18^2 with 4 = 2^2 and 4 + 24*0 = 2^2.
%e a(632) = 1 since 632 = 0^2 + 6^2 + 14^2 + 20^2 with 0 = 0^2 and 0 + 24*6 = 12^2.
%e a(1724) = 1 since 1724 = 25^2 + 1^2 + 3^2 + 33^2 with 25 = 5^2 and 25 + 24*1 = 7^2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[x^2+24y],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,Sqrt[(n-x^4-y^2)/2]}];Print[n," ",r];Continue,{n,0,80}]
%Y Cf. A000118, A000290, A000583, A270969, A273404, A281939, A281941, A281975, A281977, A281980, A282013, A282014.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Feb 04 2017
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