A055503 Take n points in general position in the plane; draw all the (infinite) straight lines joining them; sequence gives number of connected regions formed.

1, 1, 2, 7, 18, 41, 85, 162, 287, 478, 756, 1145, 1672, 2367, 3263, 4396, 5805, 7532, 9622, 12123, 15086, 18565, 22617, 27302, 32683, 38826, 45800, 53677, 62532, 72443, 83491, 95760, 109337, 124312, 140778, 158831, 178570, 200097, 223517, 248938, 276471

Offset: 0

Author

N. J. A. Sloane, Jul 10 2000; Jul 03 2012

Keywords

• nonn
• nice

Comments

Jul 02 2012: Duane DeTemple points out that one could argue that a(1) should be 0, not 1, since if the single point is removed from the plane, the result is not simply connected (and then the formula given below applies for all n). However, the sequence as described by Comtet only specifies "connected", not "simply connected", so I prefer to have a(1)=1. - N. J. A. Sloane, Jul 03 2012

n points in general position determine "n choose 2" lines, so a(n) <= A000124(n(n-1)/2). If n > 3, the lines are not in general position and so a(n) < A000124(n(n-1)/2). - Jonathan Sondow, Dec 01 2015

Examples

			For n=2: draw three vertices forming a triangle and the three infinite straight lines joining them. There are a(3) = 7 connected regions.
		

References

• L. Comtet, Advanced Combinatorics, Reidel, 1974, Problem 1, p. 72; and Problem 8, p. 74.

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000
• Olympiad Portal 74, Face-to-face rounds of the Chelyabinsk Region Open Olympiad 2008 (in Russian). See Mathematics 11 grade, problem 4.
• Michal Opler, Pavel Valtr, and Tung Anh Vu, On the Arrangement of Hyperplanes Determined by n Points, EuroCG (39th European Workshop on Computational Geometry, Barcelona, Spain 2023) Session 7B, Talk 1, Vol. 54, No. 6.
• Index entries for linear recurrences with constant coefficients, signature (5, -10, 10, -5, 1).

Crossrefs

Cf. A000124, A263883. Subsequence of A177862.

3-dimensional equivalent: A153445.

Programs

Maple

A055503 := n->(1/8)*(n^4-6*n^3+23*n^2-26*n+8); [for n >1]

Mathematica

Join[{1,1},Table[(1/8)(n-1)(n^3-5n^2+18n-8),{n,2,80}]] (* Harvey P. Dale, May 06 2011 *)

Formula

a(n) = (1/8)*(n-1)*(n^3-5*n^2+18*n-8) for n>1.

For n>1: a(0)=2, a(1)=7, a(2)=18, a(3)=41, a(4)=85, a(n)=5a(n-1)- 10a(n-2)+ 10a(n-3)-5a(n-4)+a(n-5). [Harvey P. Dale, May 06 2011]

For n>1, G.f.: (-2+3x-3x^2-x^3)/(-1+x)^5. [Harvey P. Dale, May 06 2011]

Extensions

a(1) changed from 0 to 1 by N. J. A. Sloane, Dec 07 2008