A305750 Number of achiral color patterns (set partitions) in a row or cycle of length n with 4 or fewer colors (subsets).

1, 1, 2, 3, 7, 11, 27, 43, 107, 171, 427, 683, 1707, 2731, 6827, 10923, 27307, 43691, 109227, 174763, 436907, 699051, 1747627, 2796203, 6990507, 11184811, 27962027, 44739243, 111848107, 178956971, 447392427, 715827883, 1789569707, 2863311531, 7158278827

Offset: 0

Author

Robert A. Russell, Jun 09 2018

Keywords

• nonn
• easy

Comments

An equivalent color pattern is obtained when we permute the colors. Thus all permutations of ABCD are equivalent, as are AABCD and BBCDA. A color pattern is achiral if it is equivalent to its reversal. Rotations of the colors of a cycle are equivalent, so for cycles AABBCD = BBCDAA = CDAABB.

Examples

			For a(4) = 7, the achiral row patterns are AAAA, AABB, ABAB, ABBA, ABBC, ABCA, and ABCD. The cycle patterns are AAAA, AAAB, AABB, ABAB, AABC, ABAC, and ABCD.
		

Links

• Muniru A Asiru, Table of n, a(n) for n = 0..2000
• Index entries for linear recurrences with constant coefficients, signature (1,4,-4).

Crossrefs

Fourth column of A305749.

Cf. A124303 (oriented), A056323 (unoriented), A320934 (chiral), for rows.

Cf. A056292 (oriented), A056354 (unoriented), A320744 (chiral), for cycles.

Programs

GAP

a:=[1,2,3];; for n in [4..40] do a[n]:=a[n-1]+4*a[n-2]-4*a[n-3]; od; Concatenation([1],a); # Muniru A Asiru, Oct 28 2018

Maple

seq(coeff(series((1-3*x^2+x^3)/((1-x)*(1-4*x^2)),x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Oct 28 2018

Mathematica

Table[If[EvenQ[n], StirlingS2[(n+8)/2, 4] - 8 StirlingS2[(n+6)/2, 4] + 22 StirlingS2[(n+4)/2, 4] - 23 StirlingS2[(n+2)/2, 4] + 6 StirlingS2[n/2, 4], StirlingS2[(n+7)/2, 4] - 7 StirlingS2[(n+5)/2, 4] + 16 StirlingS2[(n+3)/2, 4] - 12 StirlingS2[(n+1)/2, 4]], {n, 0, 40}]
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]]; (* A304972 *)
k=4; Table[Sum[Ach[n, j], {j, 0, k}], {n, 0, 40}]
(* or *)
CoefficientList[Series[(1-3x^2+x^3)/((1-x)(1-4x^2)), {x,0,40}], x]
(* or *)
Join[{1},LinearRecurrence[{1,4,-4},{1,2,3},40]]
(* or *)
Join[{1},Table[If[EvenQ[n], (4+5 4^(n/2))/12, (2+4^((n+1)/2))/6], {n,40}]]

Formula

a(n) = Sum_{j=0..4} Ach(n,j), where Ach(n,k) = [n>1] * (k*T(n-2,k) + T(n-2,k-1) + T(n-2,k-2)) + [0<=n<2 & n==k].

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

a(2m) = S2(m+4,4) - 8*S2(m+3,4) + 22*S2(m+2,4) - 23*S2(m+1,4) + 6*S2(m,4);

a(2m-1) = S2(m+3,4) - 7*S2(m+2,4) + 16*S2(m+1,4) - 12*S2(m,4), where S2(n,k) is the Stirling subset number A008277.

For m>0, a(2m) = (4 + 5*4^m) / 12.

a(2m-1) = (2 + 4^m) / 6.

a(n) = 2*A056323(n) - A124303(n) = A124303(n) - 2*A320934(n) = A056323(n) - A320934(n).

a(n) = 2*A056354(n) - A056292(n) = A056292(n) - 2*A320744(n) = A056354(n) - A320744(n).

a(n) = A057427(n) + A052551(n-2) + A304973(n) + A304974(n).

a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3). - Muniru A Asiru, Oct 28 2018