cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A056325 Number of reversible string structures with n beads using a maximum of six different colors.

Original entry on oeis.org

1, 1, 2, 4, 11, 32, 117, 467, 2135, 10480, 55091, 301633, 1704115, 9819216, 57365191, 338134521, 2005134639, 11937364184, 71254895955, 426063226937, 2550552314219, 15280103807200, 91588104196415, 549159428968825
Offset: 0

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Comments

A string and its reverse are considered to be equivalent. Permuting the colors will not change the structure. Thus aabc, cbaa and bbac are all considered to be identical.
Number of set partitions of an unoriented row of n elements with six or fewer nonempty subsets. - Robert A. Russell, Oct 28 2018
There are nonrecursive formulas, generating functions, and computer programs for A056273 and A305752, which can be used in conjunction with the first formula. - Robert A. Russell, Oct 28 2018
From Allan Bickle, Jun 23 2022: (Start)
a(n) is the number of (unlabeled) 6-paths with n+8 vertices. (A 6-path with order n at least 8 can be constructed from a 6-clique by iteratively adding a new 6-leaf (vertex of degree 6) adjacent to an existing 6-clique containing an existing 6-leaf.)
Recurrences appear in the papers by Bickle, Eckhoff, and Markenzon et al. (End)

Examples

			For a(4)=11, the 7 achiral patterns are AAAA, AABB, ABAB, ABBA, ABCA, ABBC, and ABCD.  The 4 chiral pairs are AAAB-ABBB, AABA-ABAA, AABC-ABCC, and ABAC-ABCB.
		

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]

Crossrefs

Cf. A056308.
Column 6 of A320750.
Cf. A056273 (oriented), A320936 (chiral), A305752 (achiral).
The numbers of unlabeled k-paths for k = 2..7 are given in A005418, A001998, A056323, A056324, A056325, and A345207, respectively.
The sequences above converge to A103293(n+1).

Programs

  • Mathematica
    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]] (* A304972 *)
    k=6; Table[Sum[StirlingS2[n,j]+Ach[n,j],{j,0,k}]/2,{n,0,40}] (* Robert A. Russell, Oct 28 2018 *)
    LinearRecurrence[{16, -84, 84, 685, -2140, 180, 7200, -8244, -4176, 11664, -5184}, {1, 1, 2, 4, 11, 32, 117, 467, 2135, 10480, 55091, 301633}, 40] (* Robert A. Russell, Oct 28 2018 *)
  • PARI
    Vec((1 - 15*x + 70*x^2 - 28*x^3 - 654*x^4 + 1479*x^5 + 783*x^6 - 5481*x^7 + 3512*x^8 + 4640*x^9 - 5922*x^10 + 1530*x^11) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 4*x)*(1 - 6*x)*(1 - 2*x^2)*(1 - 3*x^2)*(1 - 6*x^2)) + O(x^30)) \\ Colin Barker, Apr 15 2020

Formula

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.
From Robert A. Russell, Oct 28 2018: (Start)
a(n) = (A056273(n) + A305752(n)) / 2.
a(n) = A056273(n) - A320936(n) = A320936(n) + A305752(n).
a(n) = Sum_{j=0..k} (S2(n,j) + Ach(n,j)) / 2, where k=6 is the maximum number of colors, S2 is the Stirling subset number A008277, and Ach(n,k) = [n>=0 & n<2 & n==k] + [n>1]*(k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)).
a(n) = A000007(n) + A057427(n) + A056326(n) + A056327(n) + A056328(n) + A056329(n) + A056330(n).
(End)
From Colin Barker, Mar 24 2020: (Start)
G.f.: (1 - 15*x + 70*x^2 - 28*x^3 - 654*x^4 + 1479*x^5 + 783*x^6 - 5481*x^7 + 3512*x^8 + 4640*x^9 - 5922*x^10 + 1530*x^11) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 4*x)*(1 - 6*x)*(1 - 2*x^2)*(1 - 3*x^2)*(1 - 6*x^2)).
a(n) = 16*a(n-1) - 84*a(n-2) + 84*a(n-3) + 685*a(n-4) - 2140*a(n-5) + 180*a(n-6) + 7200*a(n-7) - 8244*a(n-8) - 4176*a(n-9) + 11664*a(n-10) - 5184*a(n-11) for n>11.
(End)
From Allan Bickle, Jun 23 2022: (Start)
a(n) = (1/1440)*6^n + (1/96)*4^n + (1/36)*3^n + (3/32)*2^n + (19/360)*6^(n/2) + (1/9)*3^(n/2) + (1/8)*2^(n/2) + 17/60 for n > 0 even;
a(n) = (1/1440)*6^n + (1/96)*4^n + (1/36)*3^n + (3/32)*2^n + (13/720)*6^((n+1)/2) + (1/18)*3^((n+1)/2) + (1/16)*2^((n+1)/2) + 17/60 for n > 0 odd. (End)

Extensions

Another term from Robert A. Russell, Oct 29 2018
a(0)=1 prepended by Robert A. Russell, Nov 09 2018