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A271714
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Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 such that (10*w+5*x)^2 + (12*y+36*z)^2 is a square, where w is a positive integer and x,y,z are nonnegative integers.
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39
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1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 2, 1, 3, 1, 2, 1, 2, 3, 1, 4, 4, 2, 2, 1, 3, 3, 5, 2, 2, 5, 2, 1, 2, 3, 3, 3, 2, 3, 2, 3, 4, 4, 2, 3, 9, 2, 3, 1, 1, 6, 2, 3, 4, 6, 4, 1, 2, 5, 3, 3, 4, 3, 5, 1, 4, 5, 1, 3, 6, 6, 1, 3, 4, 5, 12, 2, 4, 6, 2, 4
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OFFSET
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1,5
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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 = 7, 9, 19, 49, 133, 589, 2^k, 2^k*3, 4^k*q (k = 0,1,2,... and q = 14, 67, 71, 199).
(ii) If P(y,z) is one of 2y-3z, 2y-8z and 4y-6z, then any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w-x)^2 + P(y,z)^2 is a square.
(iii) For each triple (a,b,c) = (1,4,4), (1,12,12), (2,4,8), (2,6,6), (2,12,12), (3,4,4), (3,4,8), (3,8,8), (3,12,12), (3,12,36), (5,4,4), (5,4,8), (5,8,16), (5,36,36), (6,4,4), (7,12,12), (7,20,20), (7,24,24), (9,4,4), (9,12,12),(9,36,36), (11,12,12), (13,4,4), (15,12,12), (16,12,12), (21,20,20), (21,24,24), (23,12,12), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that (w+a*x)^2 + (b*y-c*z)^2 is a square.
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LINKS
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EXAMPLE
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a(2) = 1 since 2 = 1^2 + 1^2 + 0^2 + 0^2 with (10*1+5*1)^2 + (12*0+36*0)^2 = 15^2 + 0^2 = 15^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with (10*1+5*1)^2 + (12*0+36*1)^2 = 15^2 + 36^2 = 39^2.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with (10*2+5*0)^2 + (12*0+36*0)^2 = 20^2 + 0^2 = 20^2.
a(6) = 1 since 6 = 2^2 + 0^2 + 1^2 + 1^2 with (10*2+5*0)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with (10*1+5*2)^2 + (12*1+36*1)^2 = 20^2 + 48^2 = 52^2.
a(9) = 1 since 9 = 3^2 + 0^2 + 0^2 + 0^2 with (10*3+5*0)^2 + (12*0+36*0)^2 = 30^2 + 0^2 = 30^2.
a(19) = 1 since 19 = 3^2 + 0^2 + 3^2 + 1^2 with (10*3+5*0)^2 + (12*3+36*1)^2 = 30^2 + 72^2 = 78^2.
a(49) = 1 since 49 = 7^2 + 0^2 + 0^2 + 0^2 with (10*7+5*0)^2 + (12*0+36*0)^2 = 70^2 + 0^2 = 70^2.
a(133) = 1 since 133 = 9^2 + 0^2 + 6^2 + 4^2 with (10*9+5*0)^2 + (12*6+36*4)^2 = 90^2 + 216^2 = 234^2.
a(589) = 1 since 589 = 17^2 + 10^2 + 2^2 + 14^2 with (10*17+5*10)^2 + (12*2+36*14)^2 = 220^2 + 528^2 = 572^2.
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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[(10*Sqrt[n-x^2-y^2-z^2]+5x)^2+(12y+36z)^2], r=r+1], {x, 0, Sqrt[n-1]}, {y, 0, Sqrt[n-1-x^2]}, {z, 0, Sqrt[n-1-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
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CROSSREFS
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Cf. A000118, A000290, A271510, A271513, A271518, A271608, A271644, A271665, A271719, A271721, A271724.
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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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