A233440 Triangle read by rows: T(n, k) = n*binomial(n, k)*A000757(k), 0 <= k <= n.

0, 1, 0, 2, 0, 0, 3, 0, 0, 3, 4, 0, 0, 16, 4, 5, 0, 0, 50, 25, 40, 6, 0, 0, 120, 90, 288, 216, 7, 0, 0, 245, 245, 1176, 1764, 1603, 8, 0, 0, 448, 560, 3584, 8064, 14656, 13000, 9, 0, 0, 756, 1134, 9072, 27216, 74196, 131625, 118872, 10, 0, 0, 1200, 2100, 20160, 75600, 274800, 731250, 1320800, 1202880

Offset

0,4

Comments

For n >= 0, 0 <= k <= n, T(n, k) is the number of permutations of n symbols that k-commute with an n-cycle (we say that two permutations f and g k-commute if H(fg, gf) = k, where H(, ) denotes the Hamming distance between permutations).

Row sums give A000142.

Links

Luis Manuel Rivera Martínez, Rows n = 0..30 of triangle, flattened

R. Moreno and L. M. Rivera, Blocks in cycles and k-commuting permutations, arXiv:1306.5708 [math.CO], 2013-2014.

Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.

Formula

T(n,k) = n*C(n,k)*A000757(k), 0 <= k <= n.

Bivariate e.g.f.: G(z, u) = z*exp(z*(1-u))*(u/(1-z*u)+(1-log(1-z*u))*(1-u)).

T(n, 0) = A001477(n), n>=0;

T(n, 1) = A000004(n), n>=1;

T(n, 2) = A000004(n), n>=2;

T(n, 3) = A004320(n-2), n>=3;

T(n, 4) = A027764(n-1), n>=4;

T(n, 5) = A027765(n-1)*A000757(5), n>=5;

T(n, 6) = A027766(n-1)*A000757(6), n>=6;

T(n, 7) = A027767(n-1)*A000757(7), n>=7;

T(n, 8) = A027768(n-1)*A000757(8), n>=8;

T(n, 9) = A027769(n-1)*A000757(9), n>=9;

T(n, 10) = A027770(n-1)*A000757(10), n>=10;

T(n, 11) = A027771(n-1)*A000757(11), n>=11;

T(n, 12) = A027772(n-1)*A000757(12), n>=12;

T(n, 13) = A027773(n-1)*A000757(13), n>=13;

T(n, 14) = A027774(n-1)*A000757(14), n>=14;

T(n, 15) = A027775(n-1)*A000757(15), n>=15;

T(n, 16) = A027776(n-1)*A000757(16), n>=16. - Luis Manuel Rivera Martínez, Feb 08 2014

T(n, 0)+T(n, 3) = n*A050407(n+1), for n>=0. - Luis Manuel Rivera Martínez, Mar 06 2014

Example

For n = 4 and k = 4, T(4, 4) = 4 because all the permutations of 4 symbols that 4-commute with permutation (1, 2, 3, 4) are (1, 3), (2, 4), (1, 2)(3, 4) and (1, 4)(2, 3).

Mathematica

T[n_, k_] := n Binomial[n, k] ((-1)^k+Sum[(-1)^j k!/(k-j)/j!, {j, 0, k-1}]);
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 03 2018 *)

Crossrefs

Cf. A007318, A000757.

Sequence in context: A049597 A210951 A348127 * A280728 A175676 A035377

Adjacent sequences: A233437 A233438 A233439 * A233441 A233442 A233443

Keyword

nonn,tabl

Author

Luis Manuel Rivera Martínez, Dec 09 2013

Status

approved