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For a row of n adjacent rectangles the only polygons formed when dividing all possible rectangles along their diagonals are 3-gons (triangles) and 4-gons (quadrilaterals). Hence the only possible edge-sharing contacts are 3-gons with 3-gons, 3-gons with 4-gons, and 4-gons with 4-gons. This sequence lists the number of these three possible combinations for a row of n adjacent rectangles. Note that the edges along the outside of the n adjacent rectangles are not counted as they are only in one n-gon.

These are graphs T(1,n) described in A331452. - N. J. A. Sloane, Aug 03 2020

This sequence and A006533 are two equivalent ways of presenting the same sequence.

A quasipolynomial of order 2520. - Charles R Greathouse IV, Jan 15 2013

Also the circuit rank of the n-polygon diagonal intersection graph. - Eric W. Weisstein, Mar 08 2018

This sequence only counts polygons, in contrast to A006533 which also counts the n segments of the circumscribed circle delimited by the edges of the regular n-gon. Therefore a(n) = A006533(n) - n. See also A006561 which counts the intersection points, and A350000 which considers iterated "cutting along diagonals". - M. F. Hasler, Dec 13 2021

The Petrie polygon orthographic projection of a regular n-simplex is a regular (n+1)-gon with all diagonals drawn. Hence a(n+1) is the number of regions in the Petrie polygon of a regular n-simplex. - Mohammed Yaseen, Nov 05 2022

Also (1/4) * number of ways to select 3 distinct points forming a triangle of unsigned area = 1/2 from a square of grid points with side length n. Diagonal of triangle A320541. - Hugo Pfoertner, Oct 22 2018

From Chai Wah Wu, Aug 18 2021: (Start)

Theorem: a(n) = n^2 + Sum_{i=2..n} (n+1-i)*(2*n+2-i)*phi(i).

Proof: Since gcd(n,n) = 1 if and only if n = 1, Sum_{i=1..n, j=1..n, gcd(i,j)=1} (n+1-i)*(n+1-j) = n^2 + Sum_{i=1..n, j=1..n, gcd(i,j)=1, (i,j) <> (1,1)} (n+1-i)*(n+1-j)

= n^2 + Sum_{i=2..n, j=1..i, gcd(i,j)=1} (n+1-i)*(n+1-j) + Sum_{j=2..n, i=1..j, gcd(i,j)=1} (n+1-i)*(n+1-j) = n^2 + 2*Sum_{i=2..n, j=1..i, gcd(i,j)=1} (n+1-i)*(n+1-j), i.e., the diagonal is not double-counted.

This is equal to n^2 + 2*Sum_{i=2..n, j is a totative of i} (n+1-i)*(n+1-j). Since Sum_{j is a totative of i} 1 = phi(i) and for i > 1, Sum_{j is a totative of i} j = i*phi(i)/2, the conclusion follows.

Similar argument holds for corresponding formulas for A088658, A114043, A114146, A115005, etc.

(End)

I.e., vertex count of the n-polygon diagonal intersection graph. - Eric W. Weisstein, Mar 08 2018

Also the circumference of the n-polygon diagonal intersection graph (since these graphs are Hamiltonian). - Eric W. Weisstein, Mar 08 2018

a(n) = n + sum of row n of triangle A292105. - N. J. A. Sloane, Jun 01 2025

Assuming that the rectangles have vertices at (k,0) and (k,1), k=0..n, the projective map (x,y) -> ((1-y)/(x+1),y/(x+1)) maps their partition to the partition of the right isosceles triangle described by Alekseyev et al. (2015), for which Theorem 13 gives the number of regions, line segments, and intersection points. - Max Alekseyev, Apr 10 2019

The figure is made up of A324042 triangles and A324043 quadrilaterals. - N. J. A. Sloane, Mar 03 2020

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

A line segment (or edge) is considered to end at any vertex where two or more chords meet.

I.e., edge count of the n-polygon diagonal intersection graph. - Eric W. Weisstein, Mar 08 2018

There are n+1 points on each side of the square, but that counts the four corners twice, so there are a total of 4n points on the perimeter. - N. J. A. Sloane, Jan 23 2020

a(n) is always divisible by 4, by symmetry. If n is odd, a(n) is divisible by 8.

From Michael De Vlieger, Feb 19-20 2015: (Start)

For n > 0, the vertices of the bounding square generate diametrical bisectors that cross at the center. Thus each diagram has fourfold symmetry.

For n > 0, an orthogonal n X n grid is produced by corresponding horizontal and vertical points on opposite sides.

Terms {1, 3, 9} are not congruent to 0 (mod 8).

Number of edges: {0, 8, 92, 596, 1936, 6020, 11088, 26260, 42144, 72296, 107832, ...}. See A331448. (End)

Triangles A331452, A331453, A331454 do not have formulas, except for column 1. The column 2 sequences, A331763, A331765, A331766, are therefore the next ones to attack.

See A331452 for other illustrations.