A293181 Irregular triangle read by rows: T(n,k) is the number of k-partitions of {1..2n} that are invariant under a permutation consisting of n 2-cycles (1 <= k <= 2n).

1, 1, 1, 3, 2, 1, 1, 7, 10, 9, 3, 1, 1, 15, 38, 53, 34, 18, 4, 1, 1, 31, 130, 265, 261, 195, 80, 30, 5, 1, 1, 63, 422, 1221, 1700, 1696, 1016, 515, 155, 45, 6, 1, 1, 127, 1330, 5369, 10143, 13097, 10508, 6832, 2926, 1120, 266, 63, 7, 1

Offset: 1

Author

Andrew Howroyd, Oct 01 2017

Keywords

• nonn
• tabf

Comments

See A002872 for detailed description.

T(m,k) is the number of achiral color patterns in a row or loop of length 2m using exactly k different colors. Two color patterns are equivalent if we permute the colors. - Robert A. Russell, Apr 24 2018

T(n,k) = coefficient of x^k for A(2,n)(x) in Gilbert and Riordan's article. - Robert A. Russell, Jun 14 2018

Examples

			Triangle begins:
  1,   1;
  1,   3,    2,    1;
  1,   7,   10,    9,     3,     1;
  1,  15,   38,   53,    34,    18,     4,    1;
  1,  31,  130,  265,   261,   195,    80,   30,    5,    1;
  1,  63,  422, 1221,  1700,  1696,  1016,  515,  155,   45,   6,  1;
  1, 127, 1330, 5369, 10143, 13097, 10508, 6832, 2926, 1120, 266, 63, 7, 1;
  ...
For T(2,2)=3, the row patterns are AABB, ABAB, and ABBA.  The loop patterns are AAAB, AABB, and ABAB. - _Robert A. Russell_, Apr 24 2018
		

Links

• Alois P. Heinz, Rows n = 1..100, flattened (first 30 rows from Andrew Howroyd)
• Ira Gessel, What is the number of achiral color patterns for a row of n colors containing k different colors?, mathoverflow, Jan 30 2018.
• E. N. Gilbert and J. Riordan, Symmetry types of periodic sequences, Illinois J. Math., 5 (1961), 657-665.

Crossrefs

Row sums are A002872.

Maximum row values are A002873.

Number of achiral color patterns of length odd n in A140735.

Column k=3 gives A056182.

Programs

Mathematica

(* Ach[n, k] is the number of achiral color patterns for a row or loop of n
  colors containing k different colors *)
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, 2, 14, 2}, {k, 1, n}] // Flatten
(* Robert A. Russell, Feb 06 2018 *)
Table[Drop[MatrixPower[Table[Switch[j-i, 0, i-1, 1, 1, 2, 1, _, 0],
  {i, 1, 2n+1}, {j, 1, 2n+1}], n][[1]], 1], {n, 1, 10}] // Flatten
(* Robert A. Russell, Apr 14 2018 *)
Aeven[m_, k_] := Aeven[m, k] = If[m>0, k Aeven[m-1, k] + Aeven[m-1, k-1]
  + Aeven[m-1, k-2], Boole[m == 0 && k == 0]]
Table[Aeven[m, k], {m, 1, 10}, {k, 1, 2m}] // Flatten (* Robert A. Russell, Apr 24 2018 *)

PARI

\\ see A056391 for Polya enumeration functions
T(n,k) = 2*NonequivalentStructsExactly(CylinderPerms(2,n),k) - stirling(2*n,k,2);

PARI

seq(n)={Vec(serlaplace(exp(y*(exp(x + O(x*x^n))-1)+(1/2)*y^2*(exp(2*x + O(x*x^n))-1))) - 1)}
{my(T=seq(10)); for(n=1, #T, for(k=1, 2*n, print1(polcoeff(T[n], k), ", ")); print)} \\ Andrew Howroyd, Jan 31 2018

Formula

T(n,k) = coefficient of t^k x^n/n! in exp(t*(exp(x)-1)+(1/2)*t^2*(exp(2*x)-1)). - Ira M. Gessel, Jan 30 2018

T(m,k) = [m>0]*(k*T(m-1,k)+T(m-1,k-1)+T(m-1,k-2)) + [m==0]*[k==0]. - Robert A. Russell, Apr 24 2018

Conjecture: T(n,k) = R(n,k)-R(n,k-1), with R(n,k) = Sum_{m=0..k} m^n*A000085(m)*A038205(k-m)/(m!*(k-m)!). - Mikhail Kurkov, Jun 26 2018