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A290447
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Consider n equally spaced points along a line and join every pair of points by a semicircle above the line; a(n) is the number of intersection points.
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35
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0, 0, 0, 1, 5, 15, 35, 70, 124, 200, 300, 445, 627, 875, 1189, 1564, 2006, 2568, 3225, 4035, 4972, 6030, 7250, 8701, 10323, 12156, 14235, 16554, 19124, 22072, 25250, 28863, 32827, 37166, 41949, 47142, 52653, 58794, 65503, 72741, 80437
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OFFSET
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1,5
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COMMENTS
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Only intersection points above the line are counted.
a(n) <= binomial(n,4) (A000332), since that is the number of pairs of intersecting semicircles. See A290461 for the differences.
The first time a triple intersection occurs is for n=9. Two fourfold intersections occur for n=13. - Torsten Sillke, Jul 27 2017
If the line is the x-axis and the two semicircles are for (x_1,0),(x_2,0) and (x_3,0),(x_4,0) (with x_1 < x_2, x_3 < x_4, and x_1 < x_3) then they intersect if and only if x_1 < x_3 < x_2 < x_4, and the intersection point has coordinates (x,y) with x=(x_3*x_4 - x_1*x_2) / (x_3 + x_4 - x_1 - x_2) and y^2 = (x_3-x_1)*(x_4-x_1)*(x_2-x_3)*(x_4-x_2) / (x_3 + x_4 - x_1 - x_2)^2. This allows identification of distinct (and duplicate) intersection points using only rational arithmetic. - David Applegate, Aug 07 2017
Suppose x_i are integers in the range 0 <= x_i < n. Then (x,y) is an intersection point if and only if (n-1-x,y) is an intersection point. Suppose x_4 < n-1. If (x,y) is an intersection point, then (i+x,y) is an intersection point for i = 1,..,n-1-x_4. - Chai Wah Wu, Aug 09 2017
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REFERENCES
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Torsten Sillke, email to N. J. A. Sloane, Jul 27 2017 (giving values for a(1)-a(13)).
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LINKS
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N. J. A. Sloane, Three (No, 8) Lovely Problems from the OEIS, Experimental Mathematics Seminar, Rutgers University, Oct 05 2017, Part I, Part 2, Slides. (Mentions this sequence)
N. J. A. Sloane (in collaboration with Scott R. Shannon), Art and Sequences, Slides of guest lecture in Math 640, Rutgers Univ., Feb 8, 2020. Mentions this sequence.
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PROG
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(PARI) A290447(n, U=[])={for(A=1, n-3, for(C=A+1, n-2, for(B=C+1, n-1, for(D=B+1, n, U=setunion(U, [[(C*D-A*B)/(C+D-A-B), (C-A)*(D-A)*(C-B)*(D-B)/(C+D-A-B)^2]]))))); #U} \\ M. F. Hasler, Aug 07 2017
(Python)
from itertools import combinations
from fractions import Fraction
p, p2 = set(), set()
for b, c, d in combinations(range(1, n), 3):
e = b + d - c
f1, f2, g = Fraction(b*d, e), Fraction(b*d*(c-b)*(d-c), e**2), (n-1)*e - 2*b*d
for i in range(n-d):
if 2*i*e < g:
p2.add((i+f1, f2))
elif 2*i*e == g:
p.add(f2)
else:
break
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CROSSREFS
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See A006561 for an analogous problem on a circle.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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