A305751 Number of achiral color patterns (set partitions) in a row or cycle of length n with 5 or fewer colors (subsets).

1, 1, 2, 3, 7, 12, 30, 55, 141, 266, 688, 1313, 3407, 6532, 16970, 32595, 84721, 162846, 423348, 813973, 2116227, 4069352, 10580110, 20345735, 52898501, 101726626, 264488408, 508629033, 1322433847, 2543136972, 6612152850, 12715668475

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 ABCDE are equivalent, as are AABCDE and BBCDEA. A color pattern is achiral if it is equivalent to its reversal. Rotations of the colors of a cycle are equivalent, so for cycles AABBCDE = BBCDEAA = CDEAABB.

Examples

			For a(5) = 12, the achiral patterns for both rows and cycles are AAAAA, AABAA, ABABA, ABBBA, AABCC, ABACA, ABBBC, ABCAB, ABCBA, ABCBD, ABCDA, and ABCDE.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1,7,-7,-10,10).

Crossrefs

Fifth column of A305749.

Cf. A056272 (oriented), A056324 (unoriented), A320935 (chiral), for rows.

Cf. A056293 (oriented), A056355 (unoriented), A320745 (chiral), for cycles.

Programs

Maple

seq(coeff(series((1-2*x)*(1+2*x-2*x^2-3*x^3+x^4)/((1-x)*(1-2*x^2)*(1-5*x^2)),x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Oct 30 2018

Mathematica

Table[If[EvenQ[n], StirlingS2[(n+10)/2, 5] - 13 StirlingS2[(n+8)/2, 5] + 62 StirlingS2[(n+6)/2, 5] - 130 StirlingS2[(n+4)/2, 5] + 110 StirlingS2[(n+2)/2, 5] - 24 StirlingS2[n/2, 5], StirlingS2[(n+9)/2, 5] - 12 StirlingS2[(n+7)/2, 5] + 52 StirlingS2[(n+5)/2, 5] - 95 StirlingS2[(n+3)/2, 5] + 60 StirlingS2[(n+1)/2, 5]], {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=5; Table[Sum[Ach[n, j], {j, 0, k}], {n, 0, 40}]
CoefficientList[Series[(1-2x)(1+2x-2x^2-3x^3+x^4) / ((1- x)(1-2x^2)(1-5x^2)), {x,0,40}], x]
Join[{1},LinearRecurrence[{1,7,-7,-10,10},{1,2,3,7,12},40]]
Join[{1}, Table[If[EvenQ[n], (15 + 20 2^(n/2) + 13 5^(n/2)) / 60, (3 + 2 2^((n+1)/2) + 5^((n+1)/2)) / 12], {n,40}]]

Formula

a(n) = Sum_{j=0..5} 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 - 2x)*(1+2x-2x^2-3x^3+x^4) / ((1-x)*(1-2x^2)*(1-5x^2)).

a(2m) = S2(m+5,5) - 13*S2(m+4,5) + 62*S2(m+3,5) - 130*S2(m+2,5) + 110*S2(m+1,5) - 24*S2(m,5);

a(2m-1) = S2(m+4,5) - 12*S2(m+3,5) + 52*S2(m+2,5) - 95*S2(m+1,5) + 60*S2(m,5), where S2(n,k) is the Stirling subset number A008277.

For n>0, a(2m) = (15 + 20*2^m + 13*5^m) / 60.

a(2m-1) = (3 + 2*2^m + 5^m) / 12.

a(n) = 2*A056324(n) - A056272(n) = A056272(n) - 2*A320935(n) = A056324(n) - A320935(n).

a(n) = 2*A056355(n) - A056293(n) = A056293(n) - 2*A320745(n) = A056355(n) - A320745(n).

a(n) = A057427(n) + A052551(n-2) + A304973(n) + A304974(n) + A304975(n).

a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - 10*a(n-4) + 10*a(n-5). - Muniru A Asiru, Oct 30 2018