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%I #22 May 10 2020 08:30:16
%S 1,1,2,3,7,12,30,55,141,266,688,1313,3407,6532,16970,32595,84721,
%T 162846,423348,813973,2116227,4069352,10580110,20345735,52898501,
%U 101726626,264488408,508629033,1322433847,2543136972,6612152850,12715668475
%N Number of achiral color patterns (set partitions) in a row or cycle of length n with 5 or fewer colors (subsets).
%C 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.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,7,-7,-10,10).
%F 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].
%F G.f.: (1 - 2x)*(1+2x-2x^2-3x^3+x^4) / ((1-x)*(1-2x^2)*(1-5x^2)).
%F 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);
%F 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.
%F For n>0, a(2m) = (15 + 20*2^m + 13*5^m) / 60.
%F a(2m-1) = (3 + 2*2^m + 5^m) / 12.
%F a(n) = 2*A056324(n) - A056272(n) = A056272(n) - 2*A320935(n) = A056324(n) - A320935(n).
%F a(n) = 2*A056355(n) - A056293(n) = A056293(n) - 2*A320745(n) = A056355(n) - A320745(n).
%F a(n) = A057427(n) + A052551(n-2) + A304973(n) + A304974(n) + A304975(n).
%F 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
%e 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.
%p 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
%t 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}]
%t 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 *)
%t k=5; Table[Sum[Ach[n, j], {j, 0, k}], {n, 0, 40}]
%t CoefficientList[Series[(1-2x)(1+2x-2x^2-3x^3+x^4) / ((1- x)(1-2x^2)(1-5x^2)), {x,0,40}], x]
%t Join[{1},LinearRecurrence[{1,7,-7,-10,10},{1,2,3,7,12},40]]
%t 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}]]
%Y Fifth column of A305749.
%Y Cf. A056272 (oriented), A056324 (unoriented), A320935 (chiral), for rows.
%Y Cf. A056293 (oriented), A056355 (unoriented), A320745 (chiral), for cycles.
%K nonn,easy
%O 0,3
%A _Robert A. Russell_, Jun 09 2018
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