A271719 - OEIS

%I #14 May 01 2016 13:25:06

%S 0,0,0,0,0,2,0,0,0,0,1,1,3,1,2,4,3,1,3,4,3,2,2,1,4,2,5,4,5,2,6,4,7,4,

%T 6,4,8,5,7,7,10,3,9,7,10,9,10,4,8,5,6,4,9,1,8,5,7,6,12,4,17,11,15,10,

%U 15,8,21,12,15,9

%N Number of ordered ways to write n as x + y + z with x >= y > 0, z > 0 and gcd(x,y,z) = 1 such that x^2 + (2*y+z)^2 is a square.

%C Conjecture: a(n) > 0 for all n > 10, and a(n) = 1 only for n = 11, 12, 14, 18, 24, 54.

%C See also A271714 for a similar conjecture refining Lagrange's four-square theorem.

%H Zhi-Wei Sun, <a href="/A271719/b271719.txt">Table of n, a(n) for n = 1..2000</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1604.06723">Refining Lagrange's four-square theorem</a>, arXiv:1604.06723, 2016.

%e a(6) = 2 since 6 = 3 + 1 + 2 with 3 > 1, gcd(3,1,2) = 1 and 3^2 + (2*1+2)^2 = 5^2, and also 6 = 4 + 1 + 1 with 4 > 1, gcd(4,1,1) = 1 and 4^2 + (2*1+1)^2 = 5^2.

%e a(11) = 1 since 11 = 6 + 3 + 2 with 6 > 3, gcd(6,3,2) = 1 and 6^2 + (2*3+2)^2 = 10^2.

%e a(12) = 1 since 12 = 5 + 5 + 2 with 5 = 5, gcd(5,5,2) = 1 and 5^2 + (2*5+2)^2 = 13^2.

%e a(14) = 1 since 14 = 5 + 3 + 6 with 5 > 3, gcd(5,3,6) = 1 and 5^2 + (2*3+6)^2 = 13^2.

%e a(18) = 1 since 18 = 8 + 5 + 5 with 8 > 5, gcd(8,5,5) = 1 and 8^2 + (2*5+5)^2 = 17^2.

%e a(24) = 1 since 24 = 7 + 7 + 10 with 7 = 7, gcd(7,7,10) = 1 and 7^2 + (2*7+10)^2 = 25^2.

%e a(54) = 1 since 54 = 28 + 19 + 7 with 28 > 19, gcd(28,19,7) = 1 and 28^2 + (2*19+7)^2 = 53^2.

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

%t Do[r=0;Do[If[GCD[x,y,n-x-y]==1&&SQ[x^2+(2y+(n-x-y))^2],r=r+1],{x,1,n-2},{y,1,Min[x,n-1-x]}];Print[n," ",r];Label[aa];Continue,{n,1,70}]

%Y Cf. A000290, A271714.

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Apr 12 2016