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%I #26 May 14 2021 08:16:09
%S 0,0,0,1,2,1,1,1,3,2,2,5,3,2,5,1,5,5,2,8,5,3,9,5,3,8,4,7,7,6,11,1,8,5,
%T 4,14,6,2,5,8,9,6,8,11,8,10,5,5,13,5,7,18,17,6,9,7,5,7,6,14,11,12,7,1,
%U 12,10,14,9,13,6,10,14,14,11,10,9,7,6,10,8,8,12,7,12,12,10,11,11,8,10,10,25,15,7,18,5,11,13,13,12
%N Number of ways to write n as x + y + z with x*y + 2*y*z + 3*z*x a square, where x,y,z are positive integers with x or y a power of two (including 2^0 = 1).
%C Conjecture: a(n) > 0 for all n > 3.
%C We have verified a(n) > 0 for all n = 4..10^5. Clearly, a(2*n) > 0 if a(n) > 0.
%H Zhi-Wei Sun, <a href="/A344058/b344058.txt">Table of n, a(n) for n = 1..2000</a>
%H Chao Huang and Zhi-Wei Sun, <a href="http://arxiv.org/abs/2105.03416">On partitions of integers with restrictions involving squares</a>, arXiv:2105.03416 [math.NT], 2021.
%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. See also <a href="http://arxiv.org/abs/1604.06723">arXiv:1604.06723 [math.NT]</a>.
%e a(6) = 1 with 6 = 3 + 2^0 + 2 and 3*2^0 + 2*2^0*2 + 3*2*3 = 5^2.
%e a(7) = 1 with 7 = 3 + 2^0 + 3 and 3*2^0 + 2*2^0*3 + 3*3*3 = 6^2.
%e For each k > 1, we have a(2^k) = 1 with 2^k = 2^(k-2) + 2^(k-1) + 2^(k-2) and 2^(k-2)*2^(k-1) + 2*2^(k-1)*2^(k-2) + 3*2^(k-2)*2^(k-2) = (3*2^(k-2))^2.
%t SQ[n_]:=IntegerQ[Sqrt[n]];
%t Pow[x_]:=x>0&&IntegerQ[Log[2,x]];
%t tab={};Do[r=0;Do[If[(Pow[x]||Pow[y])&&SQ[x*y+(2y+3x)*(n-x-y)],r=r+1],{x,1,n-2},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y Cf. A000290, A271644, A340274, A343862, A343897, A343950.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, May 08 2021
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