A117625 Maximum number of regions defined by n zigzag-lines in the plane when a zigzag-line is defined as consisting of two parallel infinite half-lines joined by a straight line segment.

1, 2, 12, 31, 59, 96, 142, 197, 261, 334, 416, 507, 607, 716, 834, 961, 1097, 1242, 1396, 1559, 1731, 1912, 2102, 2301, 2509, 2726, 2952, 3187, 3431, 3684, 3946, 4217, 4497, 4786, 5084, 5391, 5707, 6032, 6366, 6709, 7061, 7422, 7792

Offset: 0

Author

Peter C. Heinig (algorithms(AT)gmx.de), Apr 08 2006

Keywords

• easy
• nonn

Comments

Here is the sketched solution from Concrete Mathematics, second edition, p. 499. Consider n straight lines in general position in the plane. As shown in Section 1.2 of the book, this divides the plane into r(n) = n*(n+1)/2 + 1 regions, the maximum possible (cf. A000124). There are n*(n-1)/2 intersection points. Replace these n lines by extremely narrow zig-zags with segments sufficiently long that there are nine intersections between each pair of zigzags. Each of the n*(n-1)/2 intersection points now gives eight new regions. So a(n) = n*(n+1)/2 + 1 + 8*n*(n-1)/2 = 9*n^2/2 - 7*n/2 + 1. - N. J. A. Sloane, May 19 2025

Note that the requirements imposed on the zigzag-line are neither the weakest nor the strongest imaginable. To relax the conditions, one might allow non-parallel half-lines. To strengthen them, one might demand the connecting line segment to be perpendicular to both half lines but still allow an arbitrary length of it, or go even further and additionally demand that all line segments be of equal length. The two latter cases would lend the problem a metrical nature.

Examples

			a(0)= 1 because the plane is one region.
		

References

• R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics, 2nd Edition, Chapter 1, Problem 13, pages 19 and 499, Addison-Wesley Publishing

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000124.

Programs

Magma

[(9*n^2-7*n+2)/2: n in [0..50]]; // Vincenzo Librandi, Jul 08 2012

Maple

seq((9*k^2-7*k+2)/2, k=0..42);

Mathematica

CoefficientList[Series[(1-x+9*x^2)/(1-x)^3,{x,0,50}],x] (* Vincenzo Librandi, Jul 08 2012 *)

PARI

a(n)=n*(9*n-7)/2+1 \\ Charles R Greathouse IV, Jun 17 2017

Formula

Recurrence: a(n) = a(n-1) + 9*n - 8 for n > 0.

Closed Form: a(n) = 9*n^2/2 - 7*n/2 + 1.

O.g.f: -(1-x+9*x^2)/(-1+x)^3 = -17/(-1+x)^2-9/(-1+x)^3-9/(-1+x) . - R. J. Mathar, Dec 05 2007

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 08 2012

E.g.f.: exp(x)*(2 + 2*x + 9*x^2)/2. - Stefano Spezia, May 20 2025