A275301 - OEIS

%I #14 Jul 24 2016 09:20:33

%S 1,2,2,2,2,1,1,1,1,3,2,1,2,2,1,1,3,4,4,5,3,2,5,2,4,5,2,3,4,3,1,3,4,4,

%T 5,3,3,5,6,3,4,3,2,4,3,4,4,3,3,7,3,4,5,3,6,4,4,4,3,3,2,3,2,2,8,3,4,8,

%U 4,3,8,3,4,9,3,4,3,4,1,3,4

%N Number of ordered ways to write n as x^2 + y^2 + z^2 + 2*w^2 with x + 2*y a cube, where x,y,z,w are nonnegative integers.

%C Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 5, 6, 7, 8, 11, 14, 15, 30, 78, 90, 93, 106, 111, 117, 125, 223, 335.

%C (ii) Any 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 + 2*y is a square.

%C See also A275344 for a similar conjecture.

%H Zhi-Wei Sun, <a href="/A275301/b275301.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(0) = 1 since 0 = 0^2 + 0^2 + 0^2 + 2*0^2 with 0 + 2*0 = 0^3.

%e a(2) = 2 since 2 = 1^2 + 0^2 + 1^2 + 2*0^2 with 1 + 2*0 =1^3, and 2 = 0^2 + 0^2 + 0^2 + 2*1^2 with 0 + 2*0 = 0^3.

%e a(5) = 1 since 5 = 1^2 + 0^2 + 2^2 + 2*0^2 with 1 + 2*0 = 1^3.

%e a(6) = 1 since 6 = 0^2 + 0^2 + 2^2 + 2*1^2 with 0 + 2*0 = 0^3.

%e a(7) = 1 since 7 = 1^2 + 0^2 + 2^2 + 2*1^2 with 1 + 2*0 = 1^3.

%e a(8) = 1 since 8 = 0^2 + 0^2 + 0^2 + 2*2^2 with 0 + 2*0 = 0^3.

%e a(11) = 1 since 11 = 0^2 + 0^2 + 3^2 + 2*1^2 with 0 + 2*0 = 0^3.

%e a(14) = 1 since 14 = 2^2 + 3^2 + 1^2 + 2*0^2 with 2 + 2*3 = 2^3.

%e a(15) = 1 since 15 = 2^2 + 3^2 + 0^2 + 2*1^2 with 2 + 2*3 = 2^3.

%e a(30) = 1 since 30 = 2^2 + 3^2 + 3^2 + 2*2^2 with 2 + 2*3 = 2^3.

%e a(78) = 1 since 78 = 6^2 + 1^2 + 3^2 + 2*4^2 with 6 + 2*1 = 2^3.

%e a(90) = 1 since 90 = 1^2 + 0^2 + 9^2 + 2*2^2 with 1 + 2*0 = 1^3.

%e a(93) = 1 since 93 = 4^2 + 2^2 + 1^2 + 2*6^2 with 4 + 2*2 = 2^3.

%e a(106) = 1 since 106 = 4^2 + 2^2 + 6^2 + 2*5^2 with 4 + 2*2 = 2^3.

%e a(111) = 1 since 111 = 2^2 + 3^2 + 0^2 + 2*7^2 with 2 + 2*3 = 2^3.

%e a(117) = 1 since 117 = 4^2 + 2^2 + 5^2 + 2*6^2 with 4 + 2*2 = 2^3.

%e a(125) = 1 since 125 = 6^2 + 1^2 + 4^2 + 2*6^2 with 6 + 2*1 = 2^3.

%e a(223) = 1 since 223 = 11^2 + 8^2 + 6^2 + 2*1^2 with 11 + 2*8 = 3^3.

%e a(335) = 1 since 335 = 11^2 + 8^2 + 10^2 + 2*5^2 with 11 + 2*8 = 3^3.

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

%t CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]

%t Do[r=0;Do[If[SQ[n-2w^2-x^2-y^2]&&CQ[x+2*y],r=r+1],{w,0,(n/2)^(1/2)},{x,0,Sqrt[n-2w^2]},{y,0,Sqrt[n-2w^2-x^2]}];Print[n," ",r];Continue,{n,0,80}]

%Y Cf. A000290, A000578, A271518, A275297, A275300, A275344.

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, Jul 22 2016