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A217585
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Number of triangles with endpoints of the form (x,x^2), x in {-n,...,n}, having at least one angle of 45 degrees.
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1
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0, 1, 5, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 111, 115, 119, 123, 127, 131, 135, 139, 143, 147, 151, 155, 159, 163, 167, 171, 175, 179, 183, 187, 191, 195, 199, 203, 207, 211, 215, 219, 223, 227, 231, 235, 239, 243, 247
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OFFSET
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0,3
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LINKS
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FORMULA
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a(n) = 4n-1 for n > 2.
a(n) = 2*a(n-1) - a(n-2) for n > 4.
G.f.: -x*(2*x^3 - 2*x^2 - 3*x - 1) / (x-1)^2. (End)
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EXAMPLE
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For n=1, there is one such triangle, having its points at abscissas [a,b,c] = [-1,0,1].
For n=2, there are 4 more such triangles, with [a,b,c] = [-2, 1, 2], [-2, -1, 2], [-1, 1, 2] and [-2, -1, 1].
For n=3, there are 6 more such triangles, namely [a,b,c] = [-3, 0, 2]*, [-3, 2, 3], [-2, 2, 3] and the symmetric of these. The first one (marked *) is the first instance where the 45-degree angle is at the bottom and not on the left or right as for all others so far.
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PROG
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(PARI) A217585(n, dump=0)={my(c=0, C=180/45/Pi, S=100+n^2); forvec(v=vector(3, i, [if(i<2, -n), if(i>1, n)]), bestappr(abs(atan(v[2]+v[1])-atan(v[3]+v[1]))*C, S)==1 &c++ &!(dump&print1(v", ")) &(v[2]||c-=1/2) &next; bestappr(abs(atan(1/(v[2]-v[1]))+atan(1/(v[2]+v[3])))*C, S)==1 &c++ &!(dump&print1(v"*, ")) &(v[2]||c-=1/2) &next, 2); c*2}
(PARI) concat(0, Vec(-x*(2*x^3-2*x^2-3*x-1)/(x-1)^2 + O(x^100))) \\ Colin Barker, Sep 20 2014
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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