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 1-8 of 8 results.

A304972 Triangle read by rows of achiral color patterns (set partitions) for a row or loop of length n. T(n,k) is the number using exactly k colors (sets).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 3, 5, 2, 1, 1, 7, 10, 9, 3, 1, 1, 7, 19, 16, 12, 3, 1, 1, 15, 38, 53, 34, 18, 4, 1, 1, 15, 65, 90, 95, 46, 22, 4, 1, 1, 31, 130, 265, 261, 195, 80, 30, 5, 1, 1, 31, 211, 440, 630, 461, 295, 100, 35, 5, 1, 1, 63, 422, 1221, 1700, 1696, 1016, 515, 155, 45, 6, 1, 1, 63, 665, 2002
Offset: 1

Views

Author

Robert A. Russell, May 22 2018

Keywords

Comments

Two color patterns are equivalent if we permute the colors. Achiral color patterns must be equivalent if we reverse the order of the pattern.

Examples

			Triangle begins:
1;
1,   1;
1,   1,    1;
1,   3,    2,    1;
1,   3,    5,    2,     1;
1,   7,   10,    9,     3,     1;
1,   7,   19,   16,    12,     3,     1;
1,  15,   38,   53,    34,    18,     4,    1;
1,  15,   65,   90,    95,    46,    22,    4,    1;
1,  31,  130,  265,   261,   195,    80,   30,    5,    1;
1,  31,  211,  440,   630,   461,   295,  100,   35,    5,   1;
1,  63,  422, 1221,  1700,  1696,  1016,  515,  155,   45,   6,  1
1,  63,  665, 2002,  3801,  3836,  3156, 1556,  710,  185,  51,  6, 1;
1, 127, 1330, 5369, 10143, 13097, 10508, 6832, 2926, 1120, 266, 63, 7, 1;
For T(4,2)=3, the row patterns are AABB, ABAB, and ABBA.  The loop patterns are AAAB, AABB, and ABAB.
For T(5,3)=5, the color patterns for both rows and loops are AABCC, ABACA, ABBBC, ABCAB, and ABCBA.
		

Crossrefs

Columns 1-6 are A057427, A052551(n-2), A304973, A304974, A304975, A304976.
A305008 has coefficients that determine the function and generating function for each column.
Row sums are A080107.

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]]
    Table[Ach[n, k], {n, 1, 15}, {k, 1, n}] // Flatten
    Ach[n_, k_] := Ach[n, k] = Which[0==k, Boole[0==n], 1==k, Boole[n>0],
      OddQ[n], Sum[Binomial[(n-1)/2, i] Ach[n-1-2i, k-1], {i, 0, (n-1)/2}],
      True, Sum[Binomial[n/2-1, i] (Ach[n-2-2i, k-1]
      + 2^i Ach[n-2-2i, k-2]), {i, 0, n/2-1}]]
    Table[Ach[n, k], {n, 1, 15}, {k, 1, n}] // Flatten
  • PARI
    Ach(n)={my(M=matrix(n,n,i,k,i>=k)); for(i=3, n, for(k=2, n, M[i,k]=k*M[i-2,k] + M[i-2,k-1] + if(k>2, M[i-2,k-2]))); M}
    { my(A=Ach(10)); for(n=1, #A, print(A[n,1..n])) } \\ Andrew Howroyd, Sep 18 2019

Formula

T(n,k) = [n>1] * (k*T(n-2,k) + T(n-2,k-1) + T(n-2,k-2)) + [n<2 & n==k & n>=0].
T(2m-1,k) = A140735(m,k).
T(2m,k) = A293181(m,k).
T(n,k) = [k==0 & n==0] + [k==1 & n>0]
+ [k>1 & n==1 mod 2] * Sum_{i=0..(n-1)/2} (C((n-1)/2, i) * T(n-1-2i, k-1))
+ [k>1 & n==0 mod 2] * Sum_{i=0..(n-2)/2} (C((n-2)/2, i) * (T(n-2-2i, k-1)
+ 2^i * T(n-2-2i, k-2))) where C(n,k) is a binomial coefficient.

A056328 Number of reversible string structures with n beads using exactly four different colors.

Original entry on oeis.org

0, 0, 0, 1, 6, 37, 183, 877, 3930, 17185, 73095, 306361, 1267266, 5198557, 21182343, 85910917, 347187210, 1399451545, 5629911015, 22616256721, 90754855026, 363890126677, 1458172596903, 5840531635357, 23385650196090
Offset: 1

Views

Author

Keywords

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

Examples

			For a(5)=6, the color patterns are ABCDA, ABCBD, AABCD, ABACD, ABCAD, and ABBCD. 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]

Crossrefs

Column 4 of A284949.
Cf. A056311.
Cf. A000453 (oriented), A320527 (chiral), A304974 (achiral).

Programs

  • Mathematica
    k=4; Table[(StirlingS2[n,k] + If[EvenQ[n], StirlingS2[n/2+2,4] - StirlingS2[n/2+1,4] - 2StirlingS2[n/2,4], 2StirlingS2[(n+3)/2,4] - 4StirlingS2[(n+1)/2,4]])/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 = 4; Table[(StirlingS2[n, k] + Ach[n, k])/2, {n, 1, 30}] (* Robert A. Russell, Oct 14 2018 *)
    LinearRecurrence[{8, -12, -44, 121, 12, -228, 144}, {0, 0, 0, 1, 6, 37, 183}, 30] (* Robert A. Russell, Oct 14 2018 *)

Formula

a(n) = A056323(n) - A001998(n-1).
Empirical g.f.: -x^4*(3*x^3 + x^2 - 2*x + 1) / ((x-1)*(2*x-1)*(2*x+1)*(3*x-1)*(4*x-1)*(3*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=4 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) = (A000453(n) + A304974(n)) / 2 = A000453(n) - A320527(n) = A320527(n) + A304974(n). (End)

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

A305008 Triangle read by rows of coefficients for functions and generating functions for the number of achiral color patterns (set partitions) for a row or loop of varying length using exactly n colors (sets).

Original entry on oeis.org

1, 1, 1, 1, 1, 0, 1, 2, -1, -2, 1, 2, -1, -4, -2, 1, 3, -3, -11, 0, 6, 1, 3, -3, -17, -8, 20, 16, 1, 4, -6, -32, 1, 64, 20, -20, 1, 4, -6, -44, -19, 140, 136, -120, -132, 1, 5, -10, -70, 5, 301, 152, -396, -280, 28, 1, 5, -10, -90, -35, 541, 608, -1228, -1752, 800, 1216, 1, 6, -15, -130, 15, 966, 643, -2798
Offset: 0

Views

Author

Robert A. Russell, May 23 2018

Keywords

Comments

Triangle begins with T(0,0).
Two color patterns are equivalent if we permute the colors. Achiral color patterns must be equivalent if we reverse the order of the pattern.
The generating function for exactly n colors (column n of A304972) is
x^n * Sum_{k=0..n} (T(n, k) * x^k) / Product_{k=1..n} (1 - k*x^2).
Both the numerator and denominator of this g.f. have factors of (1+x) and (1-(n-2)*x^2) when n > 2.
Letting S2(m,n) be the Stirling subset number A008277(m,n), the function for exactly n colors for a row or loop of length m, A304972(m,n), n even, is
[m==0 mod 2] * Sum_{k=0..n/2} T(n, 2k) * S2((m+n)/2-k, n) +
[m==1 mod 2] * Sum_{k=1..n/2} T(n, 2k-1) * S2((m+n+1)/2-k, n).
When n is odd, the function for A304972(m,n) is
[m==0 mod 2] * Sum_{k=0..(n-1)/2} T(n, 2k+1) * S2((m+n-1)-k, n) +
[m==1 mod 2] * Sum_{k=0..(n-1)/2} T(n, 2k) * S2((m+n)/2-k, n).

Examples

			Triangle begins:
1;
1, 1;
1, 1,   0;
1, 2,  -1,   -2;
1, 2,  -1,   -4,  -2;
1, 3,  -3,  -11,   0,   6;
1, 3,  -3,  -17,  -8,  20,  16;
1, 4,  -6,  -32,   1,  64,  20,   -20;
1, 4,  -6,  -44, -19, 140, 136,  -120,  -132;
1, 5, -10,  -70,   5, 301, 152,  -396,  -280,   28;
1, 5, -10,  -90, -35, 541, 608, -1228, -1752,  800, 1216;
1, 6, -15, -130,  15, 966, 643, -2798, -3028, 2236, 3600, 936;
		

Crossrefs

Coefficients for functions and generating functions of A304973, A304974, A304975, A304976, which are columns 3-6 of A304972.

Programs

  • Mathematica
    Coef[n_, -1] := Coef[n, -1] = 0; Coef[n_, 0] := Coef[n, 0] = Boole[n>=0];
    Coef[n_, k_] := Coef[n, k] = If[k > n, 0, Coef[n-1, k-1] + Coef[n-2, k] - (n-1) Coef[n-2, k-2]]
    Table[Coef[n, k], {n, 0, 30}, {k, 0, n}] // Flatten

Formula

T(n,k) = [1 <= k <= n] * (T(n-1, k-1) + T(n-2, k) - (n-1) * T(n-2, k-2)) + [k==0 & n>=0].

A305751 Number of achiral color patterns (set partitions) in a row or cycle of length n with 5 or fewer colors (subsets).

Original entry on oeis.org

1, 1, 2, 3, 7, 12, 30, 55, 141, 266, 688, 1313, 3407, 6532, 16970, 32595, 84721, 162846, 423348, 813973, 2116227, 4069352, 10580110, 20345735, 52898501, 101726626, 264488408, 508629033, 1322433847, 2543136972, 6612152850, 12715668475
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 ABCDE are equivalent, as are AABCDE and BBCDEA. A color pattern is achiral if it is equivalent to its reversal. Rotations of the colors of a cycle are equivalent, so for cycles AABBCDE = BBCDEAA = CDEAABB.

Examples

			For a(5) = 12, the achiral patterns for both rows and cycles are AAAAA, AABAA, ABABA, ABBBA, AABCC, ABACA, ABBBC, ABCAB, ABCBA, ABCBD, ABCDA, and ABCDE.
		

Crossrefs

Fifth column of A305749.
Cf. A056272 (oriented), A056324 (unoriented), A320935 (chiral), for rows.
Cf. A056293 (oriented), A056355 (unoriented), A320745 (chiral), for cycles.

Programs

  • Maple
    seq(coeff(series((1-2*x)*(1+2*x-2*x^2-3*x^3+x^4)/((1-x)*(1-2*x^2)*(1-5*x^2)),x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Oct 30 2018
  • Mathematica
    Table[If[EvenQ[n], StirlingS2[(n+10)/2, 5] - 13 StirlingS2[(n+8)/2, 5] + 62 StirlingS2[(n+6)/2, 5] - 130 StirlingS2[(n+4)/2, 5] + 110 StirlingS2[(n+2)/2, 5] - 24 StirlingS2[n/2, 5], StirlingS2[(n+9)/2, 5] - 12 StirlingS2[(n+7)/2, 5] + 52 StirlingS2[(n+5)/2, 5] - 95 StirlingS2[(n+3)/2, 5] + 60 StirlingS2[(n+1)/2, 5]], {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=5; Table[Sum[Ach[n, j], {j, 0, k}], {n, 0, 40}]
    CoefficientList[Series[(1-2x)(1+2x-2x^2-3x^3+x^4) / ((1- x)(1-2x^2)(1-5x^2)), {x,0,40}], x]
    Join[{1},LinearRecurrence[{1,7,-7,-10,10},{1,2,3,7,12},40]]
    Join[{1}, Table[If[EvenQ[n], (15 + 20 2^(n/2) + 13 5^(n/2)) / 60, (3 + 2 2^((n+1)/2) + 5^((n+1)/2)) / 12], {n,40}]]

Formula

a(n) = Sum_{j=0..5} 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 - 2x)*(1+2x-2x^2-3x^3+x^4) / ((1-x)*(1-2x^2)*(1-5x^2)).
a(2m) = S2(m+5,5) - 13*S2(m+4,5) + 62*S2(m+3,5) - 130*S2(m+2,5) + 110*S2(m+1,5) - 24*S2(m,5);
a(2m-1) = S2(m+4,5) - 12*S2(m+3,5) + 52*S2(m+2,5) - 95*S2(m+1,5) + 60*S2(m,5), where S2(n,k) is the Stirling subset number A008277.
For n>0, a(2m) = (15 + 20*2^m + 13*5^m) / 60.
a(2m-1) = (3 + 2*2^m + 5^m) / 12.
a(n) = 2*A056324(n) - A056272(n) = A056272(n) - 2*A320935(n) = A056324(n) - A320935(n).
a(n) = 2*A056355(n) - A056293(n) = A056293(n) - 2*A320745(n) = A056355(n) - A320745(n).
a(n) = A057427(n) + A052551(n-2) + A304973(n) + A304974(n) + A304975(n).
a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - 10*a(n-4) + 10*a(n-5). - Muniru A Asiru, Oct 30 2018

A305752 Number of achiral color patterns (set partitions) in a row or cycle of length n with 6 or fewer colors (subsets).

Original entry on oeis.org

1, 1, 2, 3, 7, 12, 31, 58, 159, 312, 883, 1774, 5103, 10368, 30067, 61414, 178815, 366168, 1068259, 2190190, 6395919, 13120944, 38335123, 78665590, 229890591, 471814344, 1378985155, 2830350526, 8272839855, 16980500640, 49633834099, 101878204486
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 ABCDE are equivalent, as are AABCDEF and BBCDEFA. A color pattern is achiral if it is equivalent to its reversal. Rotations of the colors of a cycle are equivalent, so for cycles AABCCDEF = BCCDEFAA = CCDEFAAB.

Examples

			For a(5) = 12, the achiral patterns for both rows and cycles are AAAAA, AABAA, ABABA, ABBBA, AABCC, ABACA, ABBBC, ABCAB, ABCBA, ABCBD, ABCDA, and ABCDE.
		

Crossrefs

Sixth column of A305749.
Cf. A056273 (oriented), A056325 (unoriented), A320936 (chiral), for rows.
Cf. A056294 (oriented), A056356 (unoriented), A320746 (chiral), for cycles.

Programs

  • Maple
    seq(coeff(series((1-10*x^2+x^3+29*x^4-6*x^5-25*x^6+8*x^7)/((1-x)*(1-2*x^2)*(1-3*x^2)*(1-6*x^2)),x,n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Oct 30 2018
  • Mathematica
    Table[If[EvenQ[n], StirlingS2[(n+12)/2, 6] - 19 StirlingS2[(n+10)/2, 6] + 140 StirlingS2[(n+8)/2, 6] - 501 StirlingS2[(n+6)/2, 6] + 887 StirlingS2[(n+4)/2, 6] - 692 StirlingS2[(n+2)/2, 6] + 160 StirlingS2[n/2, 6], StirlingS2[(n+11)/2, 6] - 18 StirlingS2[(n+9)/2, 6] + 124 StirlingS2[(n+7)/2, 6] - 404 StirlingS2[(n+5)/2, 6] + 613 StirlingS2[(n+3)/2, 6] - 340 StirlingS2[(n+1)/2, 6]], {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=6; Table[Sum[Ach[n, j], {j, 0, k}], {n, 0, 40}]
    CoefficientList[Series[(1-10x^2+x^3+29x^4-6x^5-25x^6+8x^7) / ((1-x)(1-2x^2)(1-3x^2)(1-6 x^2)), {x, 0, 40}], x]
    LinearRecurrence[{1,11,-11,-36,36,36,-36},{1,1,2,3,7,12,31,58},40]
    Join[{1}, Table[If[EvenQ[n], (36 + 45 2^(n/2) + 40 3^(n/2) + 19 6^(n/2)) / 180, (72 + 45 2^((n+1)/2) + 40 3^((n+1)/2) + 13 6^((n+1)/2)) / 360], {n,40}]]

Formula

a(n) = Sum_{j=0..6} 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.: (1-10x^2+x^3+29x^4-6x^5-25x^6+8x^7) / ((1-x)*(1-2x^2)*(1-3x^2)*(1-6x^2)).
a(2m) = S2(m+6,6) - 19*S2(m+5,6) + 140*S2(m+4,6) - 501*S2(m+3,6) + 887*S2(m+2,6) - 692*S2(m+1,6) + 160*S2(m,6);
a(2m-1) = S2(m+5,6) - 18*S2(m+4,6) + 124*S2(m+3,6) - 404*S2(m+2,6) + 613*S2(m+1,6) - 340*S2(m,6), where S2(n,k) is the Stirling subset number A008277.
For n>0, a(2m) = (36 + 45*2^m + 40*3^m + 19*6^m) / 180.
a(2m-1) = (72 + 45*2^m + 40*3^m + 13*6^m) / 360.
a(n) = 2*A056325(n) - A056273(n) = A056273(n) - 2*A320936(n) = A056325(n) - A320936(n).
a(n) = 2*A056356(n) - A056294(n) = A056294(n) - 2*A320746(n) = A056356(n) - A320936(n).
a(n) = A057427(n) + A052551(n-2) + A304973(n) + A304974(n) + A304975(n) + A304976(n).
a(n) = a(n-1) + 11*a(n-2) - 11*a(n-3) - 36*a(n-4) + 36*a(n-5) + 36*a(n-6) - 36*a(n-7). - Muniru A Asiru, Oct 30 2018

A320527 Number of chiral pairs of color patterns (set partitions) in a row of length n using exactly 4 colors (subsets).

Original entry on oeis.org

0, 0, 0, 0, 4, 28, 167, 824, 3840, 16920, 72655, 305140, 1265264, 5193188, 21173607, 85887984, 347150080, 1399355440, 5629755935, 22615859180, 90754215024, 363888497148, 1458169977847, 5840524999144, 23385639542720, 93613165023560, 374664497695215, 1499293455643620, 5999080285068784, 24002040333605908
Offset: 1

Views

Author

Robert A. Russell, Oct 14 2018

Keywords

Comments

Two color patterns are equivalent if we permute the colors. Chiral color patterns must not be equivalent if we reverse the order of the pattern.

Examples

			For a(5)=4, the chiral pairs are AABCD-ABCDD, ABACD-ABCDC, ABBCD-ABCCD and ABCAD-ABCDB.
		

Crossrefs

Col. 4 of A320525.
Cf. A000453 (oriented), A056328 (unoriented), A304974 (achiral).

Programs

  • Mathematica
    k=4; Table[(StirlingS2[n,k] - If[EvenQ[n], StirlingS2[n/2+2,4] - StirlingS2[n/2+1,4] - 2StirlingS2[n/2,4], 2StirlingS2[(n+3)/2,4] - 4StirlingS2[(n+1)/2,4]])/2, {n,30}]
    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[(StirlingS2[n, k] - Ach[n, k])/2, {n, 1, 30}]
    LinearRecurrence[{8, -12, -44, 121, 12, -228, 144}, {0, 0, 0, 0, 4, 28, 167}, 30]

Formula

a(n) = (S2(n,k) - A(n,k))/2, where k=4 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^4 / Product_{k=1..4} (1 - k*x) - x^4*(1 + x)^2*(1 - 2 x^2) / Product_{k=1..4} (1 - k*x^2)) / 2.
a(n) = (A000453(n) - A304974(n)) / 2 = A000453(n) - A056328(n) = A056328(n) - A304974(n).

A320644 Number of chiral pairs of color patterns (set partitions) in a cycle of length n using exactly 4 colors (subsets).

Original entry on oeis.org

0, 0, 0, 0, 0, 2, 17, 84, 388, 1586, 6405, 24927, 96404, 368641, 1407515, 5357974, 20403120, 77699323, 296229485, 1130614092, 4321324766, 16539645539, 63397442097, 243352167691, 935420468092, 3600493932070, 13876442107403, 53546144395718, 206864753332164, 800067806813323, 3097590602034137, 12004772596768984, 46568647645538594, 180809553280920680
Offset: 1

Views

Author

Robert A. Russell, Oct 18 2018

Keywords

Comments

Two color patterns are the same if the colors are permuted. A chiral cycle is different from its reverse.
Adnk[d,n,k] in Mathematica program is coefficient of x^k in A(d,n)(x) in Gilbert and Riordan reference.
There are nonrecursive formulas, generating functions, and computer programs for A056297 and A304974, which can be used in conjunction with the first formula.

Examples

			For a(6)=2, the chiral pairs are AABACD-AABCAD and AABCBD-AABCDC.
		

Crossrefs

Column 4 of A320647.
Cf. A056297 (oriented), A056359 (unoriented), A304974 (achiral).

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 *)
    Adnk[d_,n_,k_] := Adnk[d,n,k] = If[n>0 && k>0, Adnk[d,n-1,k]k + DivisorSum[d,Adnk[d,n-1,k-#] &], Boole[n==0 && k==0]]
    k=4; Table[DivisorSum[n,EulerPhi[#]Adnk[#,n/#,k]&]/(2n) - Ach[n,k]/2,{n,40}]

Formula

a(n) = (A056297(n) - A304974(n)) / 2 = A056297(n) - A056359(n) = A056359(n) - A304974(n).
a(n) = -Ach(n,k)/2 + (1/2n)*Sum_{d|n} phi(d)*A(d,n/d,k), where k=4 is number of colors or sets, 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)), and A(d,n,k) = [n==0 & k==0] + [n>0 & k>0]*(k*A(d,n-1,k) + Sum_{j|d} A(d,n-1,k-j)).
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