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A273134
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Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (x+8*y+8*z+15*w)^2+(6*(x+y+z+w))^2 a square, where x,y,z,w are nonnegative integers with y < z.
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15
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1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 4, 2, 1, 1, 2, 1, 1, 3, 2, 3, 3, 1, 1, 2, 1, 3, 3, 1, 3, 3, 3, 1, 1, 2, 5, 3, 2, 3, 1, 2, 2, 3, 2, 2, 4, 2, 4, 3, 1, 3, 4, 2, 4, 3, 1, 3, 1, 2, 5, 4, 3, 2, 3, 1, 4, 5, 2, 3, 5, 3, 2, 2, 1
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OFFSET
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1,9
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 7, 11, 15, 21, 23, 35, 39, 71, 95, 4^k*m (k = 0,1,2,... and m = 1, 2, 5, 6, 10, 14, 29, 30, 46, 62, 94, 110, 142, 190, 238, 334, 446).
(ii) For each polynomial P(x,y,z,w) = (x+3y+6z+17w)^2 + (20x+4y+8z+4w)^2, (x+3y+9z+17w)^2 + (20x+4y+12z+4w)^2, (x+3y+11z+15w)^2 + (12x+4y+4z+20w)^2, (3*(x+2y+3z+4w))^2 + (4*(x+4y+3z+2w))^2, (3*(x+2y+3z+4w))^2 + (4*(x+5y+3z+w))^2, any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that P(x,y,z,w) is a square.
Part (i) of this conjecture implies that any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x+8*y+8*z+15*w and 6*(x+y+z+w) are the two legs of a right triangle with positive integer sides. If a nonnegative integer n is not of the form 4^k*(16m+14) (k,m = 0,1,2,...), then n can be written as w^2+x^2+y^2+z^2 with w = x and hence (x+8y+8z+15w)^2 + (6(x+y+z+w))^2 = (8(2x+y+z))^2 + (6(2x+y+z))^2 = (10(2x+y+z))^2. Similar remarks apply to part (ii) of the conjecture.
See also A271714, A273107, A273108 and A273110 for similar conjectures related to Pythagorean triples. For more conjectural refinements of Lagrange's four-square theorem, one may consult arXiv:1604.06723.
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LINKS
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EXAMPLE
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a(1) = 1 since 1 = 0^2 + 0^2 + 1^2 + 0^2 with 0 < 1 and (0+8*0+8*1+15*0)^2 + (6*(0+0+1+0))^2 = 10^2.
a(2) = 1 since 2 = 1^2 + 0^2 + 1^2 + 0^2 with 0 < 1 and (1+8*0+8*1+15*0)^2 + (6*(1+0+1+0))^2 = 15^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 0 < 1 and (1+8*0+8*1+15*1)^2 + (6*(1+0+1+1))^2 = 30^2.
a(5) = 1 since 5 = 0^2 + 1^2 + 2^2 + 0^2 with 1 < 2 and (0+8*1+8*2+15*0)^2 + (6*(0+1+2+0))^2 = 30^2.
a(6) = 1 since 6 = 1^2 + 0^2 + 2^2 + 1^2 with 0 < 2 and (1+8*0+8*2+15*1)^2 + (6*(1+0+2+1))^2 = 40^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 2^2 + 1^2 with 1 < 2 and (1+8*1+8*2+15*1)^2 + (6*(1+1+2+1))^2 = 50^2.
a(10) = 1 since 10 = 0^2 + 1^2 + 3^2 + 0^2 with 1 < 3 and (0+8*1+8*3+15*0)^2 + (6*(0+1+3+0))^2 = 40^2.
a(11) = 1 since 11 = 1^2 + 0^2 + 3^2 + 1^2 with 0 < 3 and (1+8*0+8*3+15*1)^2 + (6*(1+0+3+1))^2 = 50^2.
a(14) = 1 since 14 = 3^2 + 1^2 + 2^2 + 0^2 with 1 < 2 and (3+8*1+8*2+15*0)^2 + (6*(3+1+2+0))^2 = 45^2.
a(15) = 1 since 15 = 1^2 + 2^2 + 3^2 + 1^2 with 2 < 3 and (1+8*2+8*3+15*1)^2 + (6*(1+2+3+1))^2 = 70^2.
a(21) = 1 since 21 = 2^2 + 2^2 + 3^2 + 2^2 with 2 < 3 and (2+8*2+8*3+15*2)^2 + (6*(2+2+3+2))^2 = 90^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 1 < 2 and (3+8*1+8*2+15*3)^2 + (6*(3+1+2+3))^2 = 90^2.
a(29) = 1 since 29 = 0^2 + 2^2 + 5^2 + 0^2 with 2 < 5 and (0+8*2+8*5+15*0)^2 + (6*(0+2+5+0))^2 = 70^2.
a(30) = 1 since 30 = 5^2 + 0^2 + 2^2 + 1^2 with 0 < 2 and (5+8*0+8*2+15*1)^2 + (6*(5+0+2+1))^2 = 60^2.
a(35) = 1 since 35 = 3^2 + 1^2 + 4^2 + 3^2 with 1 < 4 and (3+8*1+8*4+15*3)^2 + (6*(3+1+4+3))^2 = 110^2.
a(39) = 1 since 39 = 1^2 + 1^2 + 6^2 + 1^2 with 1 < 6 and (1+8*1+8*6+15*1)^2 + (6*(1+1+6+1))^2 = 90^2.
a(46) = 1 since 46 = 6^2 + 0^2 + 3^2 + 1^2 with 0 < 3 and (6+8*0+8*3+15*1)^2 + (6*(6+0+3+1))^2 = 75^2.
a(62) = 1 since 62 = 6^2 + 1^2 + 5^2 + 0^2 with 1 < 5 and (6+8*1+8*5+15*0)^2 + (6*(6+1+5+0))^2 = 90^2.
a(71) = 1 since 71 = 3^2 + 2^2 + 7^2 + 3^2 with 2 < 7 and (3+8*2+8*7+15*3)^2 + (6*(3+2+7+3))^2 = 150^2.
a(94) = 1 since 94 = 9^2 + 0^2 + 3^2 + 2^2 with 0 < 3 and (9+8*0+8*3+15*2)^2 + (6*(9+0+3+2))^2 = 105^2.
a(95) = 1 since 95 = 5^2 + 3^2 + 6^2 + 5^2 with 3 < 6 and (5+8*3+8*6+15*5)^2 + (6*(5+3+6+5))^2 = 190^2.
a(110) = 1 since 110 = 10^2 + 0^2 + 1^2 + 3^2 with 0 < 1 and (10+8*0+8*1+15*3)^2 + (6*(10+0+1+3))^2 = 105^2.
a(142) = 1 since 142 = 11^2 + 1^2 + 4^2 + 2^2 with 1 < 4 and (11+8*1+8*4+15*2)^2 + (6*(11+1+4+2))^2 = 135^2.
a(190) = 1 since 190 = 12^2 + 3^2 + 6^2 + 1^2 with 3 < 6 and (12+8*3+8*6+15*1)^2 + (6*(12+3+6+1))^2 = 165^2.
a(238) = 1 since 238 = 13^2 + 2^2 + 8^2 + 1^2 with 2 < 8 and (13+8*2+8*8+15*1)^2 + (6*(13+2+8+1))^2 = 180^2.
a(334) = 1 since 334 = 4^2 + 2^2 + 5^2 + 17^2 with 2 < 5 and
(4+8*2+8*5+15*17)^2 + (6*(4+2+5+17))^2 = 357^2.
a(446) = 1 since 446 = 17^2 + 6^2 + 11^2 + 0^2 with 6 < 11 and (17+8*6+8*11+15*0)^2 + (6*(17+6+11+0))^2 = 255^2.
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(x+8*y+8*z+15*Sqrt[n-x^2-y^2-z^2])^2+36(x+y+z+Sqrt[n-x^2-y^2-z^2])^2], r=r+1], {x, 0, Sqrt[n-1]}, {y, 0, (Sqrt[2(n-x^2)-1]-1)/2}, {z, y+1, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
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CROSSREFS
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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, A273107, A273108, A273110.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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