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%I #10 Jul 27 2016 03:13:16
%S 1,2,1,0,2,2,2,1,1,1,0,3,1,2,1,1,3,2,5,3,4,3,1,1,1,1,2,2,2,4,2,2,4,2,
%T 7,3,1,6,2,1,2,3,4,5,1,1,3,5,3,3,4,3,7,3,2,4,3,4,4,3,1,4,5,3,6,4,4,4,
%U 5,7,7,3,6,5,5,4,3,11,2,2,4
%N Number of ordered ways to write n as 2*w^2 + x^2 + y^2 + z^2 with w + x + 2*y + 4*z a square, where w,x,y,z are nonnegative integers.
%C Conjecture: (i) a(n) > 0 except for n = 3, 10, and a(n) = 1 only for n = 0, 2, 7, 8, 9, 12, 14, 15, 22, 23, 24, 25, 36, 39, 44, 45, 60, 87, 98, 106, 110, 111, 183.
%C (ii) Any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that x + 2*y + 3*z - 3*w is a square.
%C (iii) For each triple (a,b,c) = (1,2,1), (1,2,3), (1,3,1), (2,4,1), (2,4,2), (2,4,3), (2,4,4), (2,4,8), (8,9,5), any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that a*x + b*y - c*z is a square.
%C (iv) Any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that x + 2*y - 2*z is twice a nonnegative cube. Also, each natural number can be written as x^2 + y^2 + z^2 + 2*w^3 with x,y,z,w nonnegative integers such that x + 3*y - z is a square.
%C See also A275344 and A275301 for related conjectures. We are able to show that each natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w integers such that x + y + z = t^2 for some t = 0, 1, 2.
%H Zhi-Wei Sun, <a href="/A275409/b275409.txt">Table of n, a(n) for n = 0..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1604.06723">Refining Lagrange's four-square theorem</a>, arXiv:1604.06723 [math.GM], 2016.
%e a(2) = 1 since 2 = 2*1^2 + 0^2 + 0^2 + 0^2 with 1 + 0 + 2*0 + 4*0 = 1^2.
%e a(7) = 1 since 7 = 2*1^2 + 0^2 + 2^2 + 1^2 with 1 + 0 + 2*2 + 4*1 = 3^2.
%e a(8) = 1 since 8 = 2*1^2 + 2^2 + 1^2 + 1^2 with 1 + 2 + 2*1 + 4*1 = 3^2.
%e a(9) = 1 since 9 = 2*2^2 + 0^2 + 1^2 + 0^2 with 2 + 0 + 2*1 + 4*0 = 2^2.
%e a(12) = 1 since 12 = 2*2^2 + 2^2 + 0^2 + 0^2 with 2 + 2 + 2*0 + 4*0 = 2^2.
%e a(14) = 1 since 14 = 2*0^2 + 2^2 + 1^2 + 3^2 with 0 + 2 + 2*1 + 4*3 = 4^2.
%e a(15) = 1 since 15 = 2*1^2 + 2^2 + 3^2 + 0^2 with 1 + 2 + 2*3 + 4*0 = 3^2.
%e a(22) = 1 since 22 = 2*1^2 + 4^2 + 2^2 + 0^2 with 1 + 4 + 2*2 + 4*0 = 3^2.
%e a(23) = 1 since 23 = 2*3^2 + 2^2 + 0^2 + 1^2 with 3 + 2 + 2*0 + 4*1 = 3^2.
%e a(24) = 1 since 24 = 2*0^2 + 4^2 + 2^2 + 2^2 with 0 + 4 + 2*2 + 4*2 = 4^2.
%e a(25) = 1 since 25 = 2*0^2 + 4^2 + 0^2 + 3^2 with 0 + 4 + 2*0 + 4*3 = 4^2.
%e a(36) = 1 since 36 = 2*3^2 + 1^2 + 4^2 + 1^2 with 3 + 1 + 2*4 + 4*1 = 4^2.
%e a(39) = 1 since 39 = 2*1^2 + 6^2 + 1^2 + 0^2 with 1 + 6 + 2*1 + 4*0 = 3^2.
%e a(44) = 1 since 44 = 2*3^2 + 0^2 + 1^2 + 5^2 with 3 + 0 + 2*1 + 4*5 = 5^2.
%e a(45) = 1 since 45 = 2*0^2 + 5^2 + 2^2 + 4^2 with 0 + 5 + 2*2 + 4*4 = 5^2.
%e a(60) = 1 since 60 = 2*2^2 + 6^2 + 4^2 + 0^2 with 2 + 6 + 2*4 + 4*0 = 4^2.
%e a(87) = 1 since 87 = 2*3^2 + 2^2 + 8^2 + 1^2 with 3 + 2 + 2*8 + 4*1 = 5^2.
%e a(98) = 1 since 98 = 2*4^2 + 1^2 + 8^2 + 1^2 with 4 + 1 + 2*8 + 4*1 = 5^2.
%e a(106) = 1 since 106 = 2*2^2 + 8^2 + 3^2 + 5^2 with 2 + 8 + 2*3 + 4*5 = 6^2.
%e a(110) = 1 since 110 = 2*6^2 + 5^2 + 3^2 + 2^2 with 6 + 5 + 2*3 + 4*2 = 5^2.
%e a(111) = 1 since 111 = 2*5^2 + 3^2 + 6^2 + 4^2 with 5 + 3 + 2*6 + 4*4 = 6^2.
%e a(183) = 1 since 183 = 2*3^2 + 10^2 + 4^2 + 7^2 with 3 + 10 + 2*4 + 4*7 = 7^2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
%t Do[r=0;Do[If[SQ[n-2*w^2-x^2-y^2]&&SQ[w+x+2y+4*Sqrt[n-2*w^2-x^2-y^2]],r=r+1],{w,0,Sqrt[n/2]},{x,0,Sqrt[n-2*w^2]},{y,0,Sqrt[n-2*w^2-x^2]}];Print[n," ",r];Continue,{n,0,80}]
%Y Cf. A000290, A271518, A275297, A275301, A275344.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Jul 26 2016
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