

A103293


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


5



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
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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 samecolor 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 samecolor 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+1i). E.g., there are 4 partitions of [3]: 123, 123, 132, 123 but not 123 because it is the reflection of 123.  David Callan, Oct 10 2005


LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..400


FORMULA

a(n) = Sum_{k=0..n1} (Stirling2(n1,k) + Ach(n1,k))/2 for n>0, where Ach(n,k) = [n>1] * (k*Ach(n2,k) + Ach(n2,k1) + Ach(n2,k2)) + [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(n1) +`if`(modp(n, 2)=1, b((n1)/2), add(binomial(n/21, k) *b(k), k=0..n/21)))/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}] (* JeanFrançois Alcover, Jan 17 2016, after Alois P. Heinz *)
Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0],
k Ach[n2, k] + Ach[n2, k1] + Ach[n2, k2]] (* achiral *)
Table[Sum[(StirlingS2[n1, k] + Ach[n1, k])/2, {k, 0, n1}], {n, 1, 30}]
(* with a(0) omitted  Robert A. Russell, May 19 2018 *)


CROSSREFS

Cf. A000110, A056325.
Sequence in context: A124504 A056324 A056325 * A123418 A123412 A074408
Adjacent sequences: A103290 A103291 A103292 * A103294 A103295 A103296


KEYWORD

nonn,changed


AUTHOR

Hugo van der Sanden, Mar 10 2005


EXTENSIONS

More terms from David W. Wilson, Mar 10 2005


STATUS

approved



