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%I #29 Sep 08 2022 08:45:01
%S 0,0,1,4,15,50,160,502,1545,4730,14356,43474,131145,395150,1188580,
%T 3572902,10732065,32225810,96733636,290322394,871200825,2614097750,
%U 7843255300,23531775502,70599259185,211805902490
%N Number of reversible string structures with n beads using exactly three 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 three 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 G. C. Greubel, <a href="/A056327/b056327.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (6,-6,-24,49,6,-66,36).
%F a(n) = A001998(n-1) - A005418(n).
%F G.f.: x^3*(3*x^4 - 8*x^3 + 3*x^2 + 2*x - 1)/((x-1)*(2*x-1)*(3*x-1)*(2*x^2-1)*(3*x^2-1)). - _Colin Barker_, Sep 23 2012
%F From _Robert A. Russell_, Oct 14 2018: (Start)
%F a(n) = (S2(n,k) + A(n,k))/2, where k=3 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) = (A000392(n) + A304973(n)) / 2 = A000392(n) - A320526(n) = A320526(n) + A304973(n). (End)
%e For a(4)=4, the color patterns are ABCA, ABBC, AABC, and ABAC. The first two are achiral. - _Robert A. Russell_, Oct 14 2018
%t k=3; Table[(StirlingS2[n,k] + If[EvenQ[n], 2StirlingS2[n/2+1,3] - 2StirlingS2[n/2,3], StirlingS2[(n+3)/2,3] - StirlingS2[(n+1)/2,3]])/2, {n,30}] (* _Robert A. Russell_, Oct 15 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=3; Table[(StirlingS2[n, k] + Ach[n, k])/2, {n,30}] (* _Robert A. Russell_, Oct 15 2018 *)
%t LinearRecurrence[{6, -6, -24, 49, 6, -66, 36}, {0, 0, 1, 4, 15, 50, 160}, 30] (* _Robert A. Russell_, Oct 15 2018 *)
%o (PARI) m=40; v=concat([0,0,1,4,15,50,160], vector(m-7)); for(n=8, m, v[n] = 6*v[n-1] -6*v[n-2] -24*v[n-3] +49*v[n-4] +6*v[n-5] -66*v[n-6] +36*v[n-7] ); v \\ _G. C. Greubel_, Oct 16 2018
%o (Magma) I:=[0,0,1,4,15,50,160]; [n le 7 select I[n] else 6*Self(n-1) -6*Self(n-2) -24*Self(n-3) +49*Self(n-4) +6*Self(n-5) -66*Self(n-6) +36*Self(n-7): n in [1..40]]; // _G. C. Greubel_, Oct 16 2018
%Y Column 3 of A284949.
%Y Cf. A056310.
%Y Cf. A000392 (oriented), A320526 (chiral), A304973 (achiral).
%K nonn,easy
%O 1,4
%A _Marks R. Nester_
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