A056330 Number of reversible string structures with n beads using exactly six different colors.

0, 0, 0, 0, 0, 1, 12, 142, 1346, 11511, 89974, 662674, 4662574, 31724735, 210361046, 1367510326, 8752976610, 55343947975, 346541488998, 2153041587538, 13292844257198, 81652683550119, 499484958151630

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 six different elements. An unoriented row is equivalent to its reverse. - Robert A. Russell, Oct 14 2018

Examples

			For a(7)=12, the color patterns are ABCDEFA, ABCDEBF, ABCDCEF, AABCDEF, ABACDEF, ABCADEF, ABCDAEF, ABBCDEF, ABCBDEF, ABCDBEF, and ABCCDEF. The first three 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

• Index entries for linear recurrences with constant coefficients, signature (21, -159, 399, 1085, -8085, 9555, 34125, -98644, 5544, 253764, -248724, -136800, 317520, -129600).

Crossrefs

Column 6 of A284949.

Cf. A056313.

Cf. A000770 (oriented), A320529 (chiral), A304976 (achiral).

Programs

Mathematica

k=6; Table[(StirlingS2[n,k] + If[EvenQ[n], StirlingS2[n/2+3,6] - 3StirlingS2[n/2+2,6] - 8StirlingS2[n/2+1,6] + 16StirlingS2[n/2,6], 3StirlingS2[(n+5)/2,6] - 17StirlingS2[(n+3)/2,6] + 20StirlingS2[(n+1)/2,6]])/2, {n,30}] (* Robert A. Russell, Oct 14 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 = 6; Table[(StirlingS2[n, k] + Ach[n, k])/2, {n, 1, 30}] (* Robert A. Russell, Oct 14 2018 *)
LinearRecurrence[{21, -159, 399, 1085, -8085, 9555, 34125, -98644, 5544, 253764, -248724, -136800, 317520, -129600}, {0, 0, 0, 0, 0, 1, 12, 142, 1346, 11511, 89974, 662674, 4662574, 31724735}, 40] (* Robert A. Russell, Oct 14 2018 *)

Formula

a(n) = A056325(n) - A056324(n).

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

a(n) = (S2(n,k) + A(n,k))/2, where k=6 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].

G.f.: (x^6 / Product_{k=1..6} (1 - k*x) + x^6 (1+x) (1-4x^2) (1+2x-x^2-4x^3) / Product_{k=1..6} (1 - k*x^2)) / 2.

a(n) = (A000770(n) + A304976(n)) / 2 = A000770(n) - A320529(n) = A320529(n) + A304976(n). (End)