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A056329 Number of reversible string structures with n beads using exactly five different colors. 4

%I #22 Mar 03 2024 11:24:32

%S 0,0,0,0,1,9,76,542,3523,21393,123680,690550,3756151,20042589,

%T 105394296,548123982,2826435403,14479204833,73794961960,374603884910,

%U 1895632969351,9568915372269,48208452866816

%N Number of reversible string structures with n beads using exactly five different colors.

%C A string and its reverse are considered to be equivalent. Permuting the colors will not change the structure.

%C Number of set partitions for an unoriented row of n elements using exactly five different elements. An unoriented row is equivalent to its reverse. - _Robert A. Russell_, Oct 14 2018

%D M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (13, -48, -36, 551, -683, -1542, 3546, 80, -4280, 2400).

%F a(n) = A056324(n) - A056323(n).

%F Empirical g.f.: -x^5*(70*x^5 - 102*x^4 + 22*x^3 + 7*x^2 - 4*x + 1) / ((x-1)*(2*x-1)*(2*x+1)*(3*x-1)*(4*x-1)*(5*x-1)*(2*x^2-1)*(5*x^2-1)). - _Colin Barker_, Nov 25 2012

%F From _Robert A. Russell_, Oct 14 2018: (Start)

%F a(n) = (S2(n,k) + A(n,k))/2, where k=5 is the number of colors (sets), S2 is the Stirling subset number A008277 and A(n,k) = [n>1] * (k*A(n-2,k) + A(n-2,k-1) + A(n-2,k-2)) + [n<2 & n==k & n>=0].

%F a(n) = (A000481(n) + A304975(n)) / 2 = A000481(n) - A320528(n) = A320528(n) + A304975(n). (End)

%e For a(6)=9, the color patterns are ABCDEA, ABCDBA, ABCCDE, AABCDE, ABACDE, ABCADE, ABCDAE, ABBCDE, and ABCBDE. The first three are achiral. - _Robert A. Russell_, Oct 14 2018

%t k=5; Table[(StirlingS2[n,k] + If[EvenQ[n], 3StirlingS2[n/2+2,5] - 11StirlingS2[n/2+1,5] + 6StirlingS2[n/2,5], StirlingS2[(n+5)/2,5] - 3StirlingS2[(n+3)/2,5]])/2, {n,30}] (* _Robert A. Russell_, Oct 14 2018 *)

%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]]

%t k=5; Table[(StirlingS2[n, k] + Ach[n, k])/2, {n, 1, 30}] (* _Robert A. Russell_, Oct 14 2018 *)

%t LinearRecurrence[{13, -48, -36, 551, -683, -1542, 3546, 80, -4280, 2400}, {0, 0, 0, 0, 1, 9, 76, 542, 3523, 21393}, 30] (* _Robert A. Russell_, Oct 14 2018 *)

%Y Column 5 of A284949.

%Y Cf. A056312.

%Y Cf. A000481 (oriented), A320528 (chiral), A304975 (achiral).

%K nonn

%O 1,6

%A _Marks R. Nester_

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