A103293 Number of ways to color n regions arranged in a line such that consecutive regions do not have the same color.

1, 1, 1, 2, 4, 11, 32, 117, 468, 2152, 10743, 58487, 340390, 2110219, 13830235, 95475556, 691543094, 5240285139, 41432986588, 341040317063, 2916376237350, 25862097486758, 237434959191057, 2253358057283035, 22076003468637450, 222979436690612445

Offset: 0

Author

Hugo van der Sanden, Mar 10 2005

Keywords

• nonn

Comments

From David W. Wilson, Mar 10 2005: (Start)

Let M(n) be a map of n regions in a row. The number of ways to color M(n) if same-color regions are allowed to touch is given by A000110(n).

For example, M(4) has A000110(4) = 15 such colorings: aaaa aaab aaba aabb aabc abaa abab abac abba abbb abbc abca abcb abcc abcd.

The number of colorings of M(n) that are equivalent to their reverse is given by A080107(n). For example, M(4) has A080107(4) = 7 colorings that are equivalent to their reversal: aaaa aabb abab abba abbc abca abcd.

The number of distinct colorings when reversals are counted as equivalent is given by (A000110(n) + A080107(n))/2, which is essentially the present sequence. M(4) has 11 colorings that are distinct up to reversal: aaaa aaab aaba aabb aabc abab abac abba abbc abca abcd.

We can redo the whole analysis, this time forbidding same-color regions to touch. When we do, we get the same sequences, each with an extra 1 at the beginning. (End)

Note that A056325 gives the number of reversible string structures with n beads using a maximum of six different colors ... and, of course, any limit on the number of colors will be the same as this sequence above up to that number.

If the two ends of the line are distinguishable, so that 'abcb' and 'abac' are distinct, we get the Bell numbers, A000110(n - 1).

With a different offset, number of set partitions of [n] up to reflection (i<->n+1-i). E.g., there are 4 partitions of [3]: 123, 1-23, 13-2, 1-2-3 but not 12-3 because it is the reflection of 1-23. - David Callan, Oct 10 2005

Examples

			For n=4, possible arrangements are 'abab', 'abac', 'abca', 'abcd'; we do not include 'abcb' since it is equivalent to 'abac' (if you reverse and renormalize).
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..400
• Allan Bickle, How to Count k-Paths, J. Integer Sequences, 25 (2022) Article 22.5.6.
• Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
• Matthew Bolan, Jose Brox, Mario Carneiro, Martin Dvořák, Andrés Goens, Harald Husum, Zoltan Kocsis, Alex Meiburg, Pietro Monticone, David Renshaw, Jérémy Scanvic, Shreyas Srinivas, Terence Tao, Anand Rao Tadipatri, Vlad Tsyrklevich, Daniel Weber, and Fan Zheng, The equational theories project: using Lean and Github to complete an implication graph in universal algebra, Equational Theories Project 2024. See p. 41.

Crossrefs

The numbers of unlabeled k-paths for k = 2..7 are given in A005418, A001998, A056323, A056324, A056325, and A345207, respectively (these are also columns of the array in A320750). The sequences counting the unlabeled k-paths converge to this sequence when k goes to infinity.

Row sums of A284949.

Cf. A000110, A056325.

Programs

Maple

with(combinat): b:= n-> coeff(series(exp((exp(2*x)-3)/2+exp(x)), x, n+1), x,n)*n!: a:= n-> `if`(n=0, 1, (bell(n-1) +`if`(modp(n,2)=1, b((n-1)/2), add(binomial(n/2-1,k) *b(k), k=0..n/2-1)))/2): seq(a(n), n=0..30); # Alois P. Heinz, Sep 05 2008

Mathematica

b[n_] := SeriesCoefficient[Exp[(Exp[2*x] - 3)/2 + Exp[x]], {x, 0, n}]*n!; a[n_] := If[n == 0, 1, (BellB[n - 1] + If[Mod[n, 2] == 1, b[(n - 1)/2], Sum[Binomial[n/2 - 1, k] *b[k], {k, 0, n/2 - 1}]])/2]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jan 17 2016, after Alois P. Heinz *)
Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0],
   k Ach[n-2, k] + Ach[n-2, k-1] + Ach[n-2, k-2]] (* achiral *)
Table[Sum[(StirlingS2[n-1, k] + Ach[n-1, k])/2, {k, 0, n-1}], {n, 1, 30}]
(* with a(0) omitted - Robert A. Russell, May 19 2018 *)

Python

from functools import lru_cache
from sympy.functions.combinatorial.numbers import stirling
def A103293(n):
    if n == 0: return 1
    @lru_cache(maxsize=None)
    def ach(n,k): return (n==k) if n<2 else k*ach(n-2,k)+ach(n-2,k-1)+ach(n-2,k-2)
    return sum(stirling(n-1,k,kind=2)+ach(n-1,k)>>1 for k in range(n)) # Chai Wah Wu, Oct 15 2024

Formula

a(n) = Sum_{k=0..n-1} (Stirling2(n-1,k) + Ach(n-1,k))/2 for n>0, where Ach(n,k) = [n>1] * (k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)) + [n<2 & n>=0 & n==k]. - Robert A. Russell, May 19 2018

Extensions

More terms from David W. Wilson, Mar 10 2005