A056327 Number of reversible string structures with n beads using exactly three different colors.

0, 0, 1, 4, 15, 50, 160, 502, 1545, 4730, 14356, 43474, 131145, 395150, 1188580, 3572902, 10732065, 32225810, 96733636, 290322394, 871200825, 2614097750, 7843255300, 23531775502, 70599259185, 211805902490

Offset: 1

Author

Marks R. Nester

Keywords

• nonn
• easy

Comments

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

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

Examples

			For a(4)=4, the color patterns are ABCA, ABBC, AABC, and ABAC. The first two are achiral. - _Robert A. Russell_, Oct 14 2018
		

References

• 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]

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (6,-6,-24,49,6,-66,36).

Crossrefs

Column 3 of A284949.

Cf. A056310.

Cf. A000392 (oriented), A320526 (chiral), A304973 (achiral).

Programs

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

Mathematica

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 *)
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]]
k=3; Table[(StirlingS2[n, k] + Ach[n, k])/2, {n,30}] (* Robert A. Russell, Oct 15 2018 *)
LinearRecurrence[{6, -6, -24, 49, 6, -66, 36}, {0, 0, 1, 4, 15, 50, 160}, 30] (* Robert A. Russell, Oct 15 2018 *)

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

Formula

a(n) = A001998(n-1) - A005418(n).

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

From Robert A. Russell, Oct 14 2018: (Start)

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].

a(n) = (A000392(n) + A304973(n)) / 2 = A000392(n) - A320526(n) = A320526(n) + A304973(n). (End)