OFFSET
1,6
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 five different elements. An unoriented row is equivalent to its reverse. - 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 (13, -48, -36, 551, -683, -1542, 3546, 80, -4280, 2400).
FORMULA
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
From Robert A. Russell, Oct 14 2018: (Start)
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].
EXAMPLE
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
MATHEMATICA
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 *)
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=5; Table[(StirlingS2[n, k] + Ach[n, k])/2, {n, 1, 30}] (* Robert A. Russell, Oct 14 2018 *)
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved