A274498 Triangle read by rows: T(n,k) is the number of ternary words of length n having degree of asymmetry equal to k (n>=0; 0<=k<=n/2).

1, 3, 3, 6, 9, 18, 9, 36, 36, 27, 108, 108, 27, 162, 324, 216, 81, 486, 972, 648, 81, 648, 1944, 2592, 1296, 243, 1944, 5832, 7776, 3888, 243, 2430, 9720, 19440, 19440, 7776, 729, 7290, 29160, 58320, 58320, 23328, 729, 8748, 43740, 116640, 174960, 139968, 46656

Offset: 0

Author

Emeric Deutsch, Jul 27 2016

Keywords

• nonn
• tabf

Comments

The degree of asymmetry of a finite sequence of numbers is defined to be the number of pairs of symmetrically positioned distinct entries. Example: the degree of asymmetry of (2,7,6,4,5,7,3) is 2, counting the pairs (2,3) and (6,5).

A sequence is palindromic if and only if its degree of asymmetry is 0.

Sum(kT(n,k),k>=0) = A274499(n).

Examples

			From _Andrew Howroyd_, Jan 10 2018: (Start)
Triangle begins:
   1;
   3;
   3,   6;
   9,  18;
   9,  36,   36;
  27, 108,  108;
  27, 162,  324,  216;
  81, 486,  972,  648;
  81, 648, 1944, 2592, 1296;
  ...
(End)
T(2,0) = 3 because we have 00, 11, and 22.
T(2,1) = 6 because we have 01, 02, 10, 12, 20, and 21.
		

Links

• S. Elizalde, E. Deutsch, The degree of asymmetry of a sequence, Enum. Combinat. Applic. 2 (2022) no 1 #S2R7 corollary 2.2 at m=3 and dropping each 2nd row.

Crossrefs

Cf. A274496, A274497, A274499.

Programs

Maple

T := proc(n,k) options operator, arrow: 2^k*3^ceil((1/2)*n)*binomial(floor((1/2)*n), k) end proc: for n from 0 to 15 do seq(T(n, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form

Mathematica

T[n_, k_] := 2^k 3^Ceiling[n/2] Binomial[Floor[n/2], k];
Table[T[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten (* Jean-François Alcover, Jan 04 2021 *)

PARI

T(n,k) = 2^k*3^ceil(n/2)*binomial(floor(n/2),k);
for(n=0, 10, for(k=0, n\2, print1(T(n, k), ", ")); print); \\ Andrew Howroyd, Jan 10 2018

Formula

T(n,k) = 2^k*3^ceiling(n/2)*binomial(floor(n/2),k).

G.f.: G(t,z) = (1 + 3z)/(1 - 3(1 + 2t)z^2).

The row generating polynomials P[n] satisfy P[n] = 3(1 + 2t)P[n-2] (n>=2). Easy to see if we note that the ternary words of length n (n>=2) are 0w0, 0w1, 0w2, 1w0, 1w1, 1w2, 2w0, 2w1, 2w2, where w is a ternary word of length n - 2.