cp's OEIS Frontend

A273107 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (8*x+12*y)^2 + (15*z)^2 a square, where x,y,z,w are nonnegative integers with x+y > 0 and z > 0.

%I A273107 #18 May 18 2016 23:24:38
%S A273107 0,1,2,1,0,1,1,1,1,1,3,4,2,3,2,1,2,3,3,1,2,3,1,1,1,1,4,3,3,4,1,1,5,3,
%T A273107 2,3,3,5,2,1,2,1,3,3,3,3,2,4,5,5,2,4,5,6,1,3,7,3,5,4,2,6,4,1,5,4,5,4,
%U A273107 7,7,4,3,5,4,5,6,2,10,3,1
%N A273107 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (8*x+12*y)^2 + (15*z)^2 a square, where x,y,z,w are nonnegative integers with x+y > 0 and z > 0.
%C A273107 Conjecture: a(n) > 0 for all n > 5, and a(n) = 1 only for n = 7, 9, 23, 25, 31, 55, 2^k*m (k = 1,2,... and m = 1, 5), 2^(2k+1)*m (k = 0,1,2,... and m = 3, 13, 21).
%C A273107 This conjecture implies that any integer n > 5 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 8*x+12*y and 15*z are the two legs of a right triangle with positive integer sides.
%C A273107 See also A271714, A273108, A273110 and A273134 for similar conjectures related to Pythagorean triples. For more conjectural refinements of Lagrange's four-square theorem, one may consult arXiv:1604.06723.
%H A273107 Zhi-Wei Sun, <a href="/A273107/b273107.txt">Table of n, a(n) for n = 1..10000</a>
%H A273107 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 A273107 a(2) = 1 since 2 = 1^2 + 0^2 + 1^2 + 0^2 with 1 + 0 > 0 < 1 and (8*1+12*0)^2 + (15*1)^2 = 17^2.
%e A273107 a(4) = 1 since 4 = 1^2 + 1^2 + 1^2 + 1^2 with 1 + 1 > 0 < 1 and (8*1+12*1)^2 + (15*1)^2 = 25^2.
%e A273107 a(6) = 1 since 6 = 1^2 + 0^2 + 1^2 + 2^2 with 1 + 0 > 0 < 1 and (8*1+12*0)^2 + (15*1)^2 = 17^2.
%e A273107 a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 + 1 > 0 < 1 and (8*1+12*1)^2 + (15*1)^2 = 25^2.
%e A273107 a(9) = 1 since 9 = 2^2 + 0^2 + 2^2 + 1^2 with 2 + 0 > 0 < 2 and (8*2+12*0)^2 + (15*2)^2 = 34^2.
%e A273107 a(10) = 1 since 10 = 0^2 + 3^2 + 1^2 + 0^2 with 0 + 3 > 0 < 1 and (8*0+12*3)^2 + (15*1)^2 = 39^2.
%e A273107 a(20) = 1 since 20 = 3^2 + 1^2 + 1^2 + 3^2 with 3 + 1 > 0 < 1 and (8*3+12*1)^2 + (15*1)^2 = 39^2.
%e A273107 a(23) = 1 since 23 = 2^2 + 1^2 + 3^2 + 3^2 with 2 + 1 > 0 < 3 and (8*2+12*1)^2 + (15*3)^2 = 53^2.
%e A273107 a(25) = 1 since 25 = 1^2 + 2^2 + 4^2 + 2^2 with 1 + 2 > 0 < 4 and (8*1+12*2)^2 + (15*4)^2 = 68^2.
%e A273107 a(26) = 1 since 26 = 0^2 + 3^2 + 1^2 + 4^2 with 0 + 3 > 0 < 1 and (8*0+12*3)^2 + (15*1)^2 = 39^2.
%e A273107 a(31) = 1 since 31 = 3^2 + 3^2 + 3^2 + 2^2 with 3 + 3 > 0 < 3 and (8*3+12*3)^2 + (15*3)^2 = 75^2.
%e A273107 a(42) = 1 since 42 = 2^2 + 2^2 + 5^2 + 3^2 with 2 + 2 > 0 < 5 and (8*2+12*2)^2 + (15*5)^2 = 85^2.
%e A273107 a(55) = 1 since 55 = 6^2 + 1^2 + 3^2 + 3^2 with 6 + 1 > 0 < 3 and (8*6+12*1)^2 + (15*3)^2 = 75^2.
%t A273107 SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
%t A273107 Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(8x+12y)^2+(15z)^2],r=r+1],{x,0,Sqrt[n-1]},{y,Max[0,1-x],Sqrt[n-1-x^2]},{z,1,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]
%Y A273107 Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273108, A273110, A273134.
%K A273107 nonn
%O A273107 1,3
%A A273107 _Zhi-Wei Sun_, May 15 2016