A217585 Number of triangles with endpoints of the form (x,x^2), x in {-n,...,n}, having at least one angle of 45 degrees.

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

Offset

0,3

Links

Table of n, a(n) for n=0..62.

Project Euler, Problem 397: Triangle on parabola, Oct 07 2012.

Index entries for linear recurrences with constant coefficients, signature (2,-1).

Formula

a(n) = 4n-1 for n > 2.

From Colin Barker, Sep 20 2014: (Start)

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)

Example

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.

Prog

(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

Crossrefs

Sequence in context: A314005 A164355 A287322 * A314006 A314007 A183227

Adjacent sequences: A217582 A217583 A217584 * A217586 A217587 A217588

Keyword

nonn,easy

Author

M. F. Hasler, Oct 07 2012

Status

approved