A308305 a(n) = s(n,n) + s(n,n-1) + s(n,n-2), where s(n,k) are the unsigned Stirling numbers of the first kind (see A132393).

1, 2, 6, 18, 46, 101, 197, 351, 583, 916, 1376, 1992, 2796, 3823, 5111, 6701, 8637, 10966, 13738, 17006, 20826, 25257, 30361, 36203, 42851, 50376, 58852, 68356, 78968, 90771, 103851, 118297, 134201, 151658, 170766, 191626, 214342, 239021, 265773, 294711

Offset: 1

Author

Alvaro Carbonero, Elizabeth Castellano, Charles Kulick, Karie Schmitz, Jun 05 2019

Keywords

• nonn
• easy

Comments

Pairwise perpendicular bisectors divide the Euclidean plane into a maximum of a(n) regions. This maximum value a(n) occurs when no three points are collinear and no four points are concyclic in the plane, and with no perpendicular bisectors parallel or coinciding [Zaslavsky, Eq. (1.1)]. This count of regions in the plane is relevant for social science applications to voting preferences based on proximity to candidates on issues.

References

• T. Zaslavsky, Perpendicular dissections of space. Discrete Comput. Geom. 27 (2002), no. 3, 303-351.

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Alvaro Carbonero, Beth Anne Castellano, Gary Gordon, Charles Kulick, Karie Schmitz, and Brittany Shelton, Permutations of point sets in R_d, arXiv:2106.14140 [math.CO], 2021.
• I. J. Good and T. N. Tideman, Stirling numbers and a geometric structure from voting theory, J. Combinatorial Theory Ser. A 23 (1977), 34-45.
• T. Zaslavsky, Perpendicular dissections of space, arXiv:1001.4435 [math.CO], 2010. See equation (1.1) with d=2.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

The unsigned Stirling numbers of the first kind s(n,k) are given in A132393.

The division of space formulation can be generalized to higher dimensions with use of A008275 by Good and Tideman's work.

The maximum number of regions generated by pairwise perpendicular bisectors on a sphere is given by A087645.

Programs

Magma

[(1/24)*(24 - 14*n + 21*n^2 - 10*n^3 + 3*n^4): n in [1..40]]; // Vincenzo Librandi, Jun 30 2019

Mathematica

Table[(1/24)(24 - 14 i + 21 i^2 - 10 i^3 + 3 i^4), {i, 40}]

PARI

Vec(x*(1 - 3*x + 6*x^2 - 2*x^3 + x^4) / (1 - x)^5 + O(x^40)) \\ Colin Barker, Jun 30 2019

Formula

a(n) = s(n,n) + s(n,n-1) + s(n,n-2), where s(n,k) are the unsigned Stirling numbers of the first kind.

a(n) = (1/24)*(24 - 14*n + 21*n^2 - 10*n^3 + 3*n^4).

From Colin Barker, Jun 30 2019: (Start)

G.f.: x*(1 - 3*x + 6*x^2 - 2*x^3 + x^4) / (1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.

(End)