A087645 -id:A087645 - cp's OEIS frontend

A071223 Triangle T(n,k) (n >= 2, 1 <= k <= n) read by rows: number of linearly inducible orderings of n points in k-dimensional Euclidean space.

2, 2, 6, 2, 12, 24, 2, 20, 72, 120, 2, 30, 172, 480, 720, 2, 42, 352, 1512, 3600, 5040, 2, 56, 646, 3976, 14184, 30240, 40320, 2, 72, 1094, 9144, 45992, 143712, 282240, 362880, 2, 90, 1742, 18990, 128288, 557640, 1575648, 2903040, 3628800, 2, 110, 2642, 36410, 318188, 1840520, 7152048, 18659520, 32659200, 39916800

Offset: 2

N. J. A. Sloane, Oct 26 2003

This can also be regarded as the lower triangular part of an infinite square array - see Example section and A198889.

Second and third columns are A002378 and A087645.

			Triangle begins:
  2
  2  6
  2 12  24
  2 20  72 120
  2 30 172 480 720
  ...
This triangle is the lower triangular part of a square array which begins
  2   2   2   2   2 ...
  2   6   6   6   6 ...
  2  12  24  24  24 ...
  2  20  72 120 120 ...
  2  30 172 480 720 ...
  ...

T. M. Cover, The number of linearly inducible orderings of points in d-space, SIAM J. Applied Math., 15 (1967), 434-439.

Cf. A087644, A002378, A087645.

T:=proc(n,k) if k>=n then 0 elif k=1 and n>=2 then 2 elif n=2 and k>=1 then 2 elif k=n-1 then n! else T(n-1,k)+(n-1)*T(n-1,k-1) fi end:seq(seq(T(n,k),k=1..n-1),n=2..12);

T[n_ /; n >= 2, 1] = 2; T[2, k_ /; k >= 1] = 2;
T[n_, k_] := T[n, k] = T[n-1, k] + (n-1)*T[n-1, k-1];
T[n_, k_] /; k >= n-1 = n!;
Flatten[Table[T[n, k], {n, 2, 11}, {k, 1, n-1}]]  (* Jean-François Alcover, Apr 27 2011 *)

T(n, 1) = 2 for n >= 2, T(2, k) = 2 for k >= 1, T(n+1, k) = T(n, k) + n*T(n, k-1). Also T(n, k) = n! for k >= n-1.

More terms from Emeric Deutsch, May 24 2004

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).

Original entry on oeis.org

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

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

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.

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

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).

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.

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

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

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

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)

cp's OEIS Frontend

A071223 Triangle T(n,k) (n >= 2, 1 <= k <= n) read by rows: number of linearly inducible orderings of n points in k-dimensional Euclidean space.

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