|
%I #63 Apr 20 2019 08:08:13
%S 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,
%T 0,0,245,245,1176,1764,1603,8,0,0,448,560,3584,8064,14656,13000,9,0,0,
%U 756,1134,9072,27216,74196,131625,118872,10,0,0,1200,2100,20160,75600,274800,731250,1320800,1202880
%N Triangle read by rows: T(n, k) = n*binomial(n, k)*A000757(k), 0 <= k <= n.
%C 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).
%C Row sums give A000142.
%H Luis Manuel Rivera Martínez, <a href="/A233440/b233440.txt">Rows n = 0..30 of triangle, flattened</a>
%H R. Moreno and L. M. Rivera, <a href="http://arxiv.org/abs/1306.5708">Blocks in cycles and k-commuting permutations</a>, arXiv:1306.5708 [math.CO], 2013-2014.
%H Luis Manuel Rivera, <a href="http://arxiv.org/abs/1406.3081">Integer sequences and k-commuting permutations</a>, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
%F T(n,k) = n*C(n,k)*A000757(k), 0 <= k <= n.
%F Bivariate e.g.f.: G(z, u) = z*exp(z*(1-u))*(u/(1-z*u)+(1-log(1-z*u))*(1-u)).
%F T(n, 0) = A001477(n), n>=0;
%F T(n, 1) = A000004(n), n>=1;
%F T(n, 2) = A000004(n), n>=2;
%F T(n, 3) = A004320(n-2), n>=3;
%F T(n, 4) = A027764(n-1), n>=4;
%F T(n, 5) = A027765(n-1)*A000757(5), n>=5;
%F T(n, 6) = A027766(n-1)*A000757(6), n>=6;
%F T(n, 7) = A027767(n-1)*A000757(7), n>=7;
%F T(n, 8) = A027768(n-1)*A000757(8), n>=8;
%F T(n, 9) = A027769(n-1)*A000757(9), n>=9;
%F T(n, 10) = A027770(n-1)*A000757(10), n>=10;
%F T(n, 11) = A027771(n-1)*A000757(11), n>=11;
%F T(n, 12) = A027772(n-1)*A000757(12), n>=12;
%F T(n, 13) = A027773(n-1)*A000757(13), n>=13;
%F T(n, 14) = A027774(n-1)*A000757(14), n>=14;
%F T(n, 15) = A027775(n-1)*A000757(15), n>=15;
%F T(n, 16) = A027776(n-1)*A000757(16), n>=16. - _Luis Manuel Rivera Martínez_, Feb 08 2014
%F T(n, 0)+T(n, 3) = n*A050407(n+1), for n>=0. - _Luis Manuel Rivera Martínez_, Mar 06 2014
%e 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).
%t T[n_, k_] := n Binomial[n, k] ((-1)^k+Sum[(-1)^j k!/(k-j)/j!, {j, 0, k-1}]);
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Nov 03 2018 *)
%Y Cf. A007318, A000757.
%K nonn,tabl
%O 0,4
%A _Luis Manuel Rivera Martínez_, Dec 09 2013
|
