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.

Showing 1-3 of 3 results.

A056323 Number of reversible string structures with n beads using a maximum of four different colors.

Original entry on oeis.org

1, 1, 2, 4, 11, 31, 107, 379, 1451, 5611, 22187, 87979, 350891, 1400491, 5597867, 22379179, 89500331, 357952171, 1431743147, 5726775979, 22906841771, 91626580651, 366505274027, 1466017950379, 5864067607211
Offset: 0

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Author

Keywords

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 four or fewer nonempty subsets. - Robert A. Russell, Oct 28 2018
There are nonrecursive formulas, generating functions, and computer programs for A124303 and A305750, which can be used in conjunction with the formula. - Robert A. Russell, Oct 28 2018
From Allan Bickle, Jun 02 2022: (Start)
a(n) is the number of (unlabeled) 4-paths with n+6 vertices. (A 4-path with order n at least 6 can be constructed from a 5-clique by iteratively adding a new 4-leaf (vertex of degree 4) adjacent to an existing 4-clique containing an existing 4-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. A032121.
Column 4 of A320750.
Cf. A124303 (oriented), A320934 (chiral), A305750 (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=4; Table[Sum[StirlingS2[n,j]+Ach[n,j],{j,0,k}]/2,{n,0,40}] (* Robert A. Russell, Oct 28 2018 *)
    LinearRecurrence[{5, 0, -20, 16}, {1, 1, 2, 4, 11}, 40] (* Robert A. Russell, Oct 28 2018 *)

Formula

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.
For n > 0, a(n) = (16 + (-2)^n + 15*2^n + 4^n)/48. - Colin Barker, Nov 24 2012
G.f.: (1 - 4x - 3x^2 + 14x^3 - 5x^4) / ((1-x)*(1-4x)*(1-4x^2)). - Colin Barker, Nov 24 2012 [Adapted to offset 0 by Robert A. Russell, Nov 09 2018]
From Robert A. Russell, Oct 28 2018: (Start)
a(n) = (A124303(n) + A305750(n)) / 2.
a(n) = A124303(n) - A320934(n) = A320934(n) + A305750(n).
a(n) = Sum_{j=0..k} (S2(n,j) + Ach(n,j)) / 2, where k=4 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). (End)

Extensions

a(0)=1 prepended by Robert A. Russell, Nov 09 2018

A320751 Array read by antidiagonals: T(n,k) is the number of chiral pairs of color patterns (set partitions) in a row of length n using k or fewer colors (subsets).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 1, 4, 6, 0, 0, 0, 1, 4, 16, 12, 0, 0, 0, 1, 4, 20, 52, 28, 0, 0, 0, 1, 4, 20, 80, 169, 56, 0, 0, 0, 1, 4, 20, 86, 336, 520, 120, 0, 0, 0, 1, 4, 20, 86, 400, 1344, 1600, 240, 0, 0, 0, 1, 4, 20, 86, 409, 1852, 5440, 4840, 496, 0
Offset: 1

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Author

Robert A. Russell, Oct 27 2018

Keywords

Comments

Two color patterns are equivalent if the colors are permuted.
A chiral row is not equivalent to its reverse.
T(n,k)=Xi_k(P_n) which is the number of non-equivalent distinguishing partitions of the path on n vertices, with at most k parts. Two partitions P1 and P2 of a graph G are said to be equivalent if there is a nontrivial automorphism of G which maps P1 onto P2. A distinguishing partition is a partition of the vertex set of G such that no nontrivial automorphism of G can preserve it. - Bahman Ahmadi, Sep 02 2019

Examples

			Array begins with T(1,1):
0   0     0      0       0       0       0       0       0       0 ...
0   0     0      0       0       0       0       0       0       0 ...
0   1     1      1       1       1       1       1       1       1 ...
0   2     4      4       4       4       4       4       4       4 ...
0   6    16     20      20      20      20      20      20      20 ...
0  12    52     80      86      86      86      86      86      86 ...
0  28   169    336     400     409     409     409     409     409 ...
0  56   520   1344    1852    1976    1988    1988    1988    1988 ...
0 120  1600   5440    8868   10168   10388   10404   10404   10404 ...
0 240  4840  21760   42892   54208   57108   57468   57488   57488 ...
0 496 14641  87296  210346  299859  331705  337595  338155  338180 ...
0 992 44044 349184 1038034 1699012 2012202 2091458 2102518 2103348 ...
For T(4,2)=2, the chiral pairs are AAAB-ABBB and AABA-ABAA.
For T(4,3)=4, the above, AABC-ABCC, and ABAC-ABCB.
		

Crossrefs

Columns 1-6 are A000004, A122746(n-3), A107767(n-1), A320934, A320935, A320936.
As k increases, columns converge to A320937.
Cf. transpose of A278984 (oriented), A320750 (unoriented), A305749 (achiral).
Partial column sums of A320525.

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 *)
    Table[Sum[StirlingS2[n,j] - Ach[n,j], {j,k-n+1}]/2, {k,15}, {n,k}] // Flatten

Formula

T(n,k) = Sum_{j=1..k} (S2(n,j) - Ach(n,j)) / 2, where 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)).
T(n,k) = (A278984(k,n) - A305749(n,k)) / 2 = A278984(k,n) - A320750(n,k) = A320750(n,k) - A305749(n,k).
T(n,k) = Sum_{j=1..k} A320525(n,j).

A305750 Number of achiral color patterns (set partitions) in a row or cycle of length n with 4 or fewer colors (subsets).

Original entry on oeis.org

1, 1, 2, 3, 7, 11, 27, 43, 107, 171, 427, 683, 1707, 2731, 6827, 10923, 27307, 43691, 109227, 174763, 436907, 699051, 1747627, 2796203, 6990507, 11184811, 27962027, 44739243, 111848107, 178956971, 447392427, 715827883, 1789569707, 2863311531, 7158278827
Offset: 0

Views

Author

Robert A. Russell, Jun 09 2018

Keywords

Comments

An equivalent color pattern is obtained when we permute the colors. Thus all permutations of ABCD are equivalent, as are AABCD and BBCDA. A color pattern is achiral if it is equivalent to its reversal. Rotations of the colors of a cycle are equivalent, so for cycles AABBCD = BBCDAA = CDAABB.

Examples

			For a(4) = 7, the achiral row patterns are AAAA, AABB, ABAB, ABBA, ABBC, ABCA, and ABCD. The cycle patterns are AAAA, AAAB, AABB, ABAB, AABC, ABAC, and ABCD.
		

Crossrefs

Fourth column of A305749.
Cf. A124303 (oriented), A056323 (unoriented), A320934 (chiral), for rows.
Cf. A056292 (oriented), A056354 (unoriented), A320744 (chiral), for cycles.

Programs

  • GAP
    a:=[1,2,3];; for n in [4..40] do a[n]:=a[n-1]+4*a[n-2]-4*a[n-3]; od; Concatenation([1],a); # Muniru A Asiru, Oct 28 2018
  • Maple
    seq(coeff(series((1-3*x^2+x^3)/((1-x)*(1-4*x^2)),x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Oct 28 2018
  • Mathematica
    Table[If[EvenQ[n], StirlingS2[(n+8)/2, 4] - 8 StirlingS2[(n+6)/2, 4] + 22 StirlingS2[(n+4)/2, 4] - 23 StirlingS2[(n+2)/2, 4] + 6 StirlingS2[n/2, 4], StirlingS2[(n+7)/2, 4] - 7 StirlingS2[(n+5)/2, 4] + 16 StirlingS2[(n+3)/2, 4] - 12 StirlingS2[(n+1)/2, 4]], {n, 0, 40}]
    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=4; Table[Sum[Ach[n, j], {j, 0, k}], {n, 0, 40}]
    (* or *)
    CoefficientList[Series[(1-3x^2+x^3)/((1-x)(1-4x^2)), {x,0,40}], x]
    (* or *)
    Join[{1},LinearRecurrence[{1,4,-4},{1,2,3},40]]
    (* or *)
    Join[{1},Table[If[EvenQ[n], (4+5 4^(n/2))/12, (2+4^((n+1)/2))/6], {n,40}]]

Formula

a(n) = Sum_{j=0..4} Ach(n,j), where Ach(n,k) = [n>1] * (k*T(n-2,k) + T(n-2,k-1) + T(n-2,k-2)) + [0<=n<2 & n==k].
G.f.: (1 - 3x^2 + x^3) / ((1-x) * (1-4x^2)).
a(2m) = S2(m+4,4) - 8*S2(m+3,4) + 22*S2(m+2,4) - 23*S2(m+1,4) + 6*S2(m,4);
a(2m-1) = S2(m+3,4) - 7*S2(m+2,4) + 16*S2(m+1,4) - 12*S2(m,4), where S2(n,k) is the Stirling subset number A008277.
For m>0, a(2m) = (4 + 5*4^m) / 12.
a(2m-1) = (2 + 4^m) / 6.
a(n) = 2*A056323(n) - A124303(n) = A124303(n) - 2*A320934(n) = A056323(n) - A320934(n).
a(n) = 2*A056354(n) - A056292(n) = A056292(n) - 2*A320744(n) = A056354(n) - A320744(n).
a(n) = A057427(n) + A052551(n-2) + A304973(n) + A304974(n).
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3). - Muniru A Asiru, Oct 28 2018
Showing 1-3 of 3 results.