OFFSET
0,4
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
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.
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
EXAMPLE
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).
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Hugo van der Sanden, Mar 10 2005
EXTENSIONS
More terms from David W. Wilson, Mar 10 2005
STATUS
approved