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%I #14 Jan 04 2021 06:22:05
%S 1,3,3,6,9,18,9,36,36,27,108,108,27,162,324,216,81,486,972,648,81,648,
%T 1944,2592,1296,243,1944,5832,7776,3888,243,2430,9720,19440,19440,
%U 7776,729,7290,29160,58320,58320,23328,729,8748,43740,116640,174960,139968,46656
%N 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).
%C 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).
%C A sequence is palindromic if and only if its degree of asymmetry is 0.
%C Sum(kT(n,k),k>=0) = A274499(n).
%F T(n,k) = 2^k*3^ceiling(n/2)*binomial(floor(n/2),k).
%F G.f.: G(t,z) = (1 + 3z)/(1 - 3(1 + 2t)z^2).
%F 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.
%e From _Andrew Howroyd_, Jan 10 2018: (Start)
%e Triangle begins:
%e 1;
%e 3;
%e 3, 6;
%e 9, 18;
%e 9, 36, 36;
%e 27, 108, 108;
%e 27, 162, 324, 216;
%e 81, 486, 972, 648;
%e 81, 648, 1944, 2592, 1296;
%e ...
%e (End)
%e T(2,0) = 3 because we have 00, 11, and 22.
%e T(2,1) = 6 because we have 01, 02, 10, 12, 20, and 21.
%p 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
%t T[n_, k_] := 2^k 3^Ceiling[n/2] Binomial[Floor[n/2], k];
%t Table[T[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten (* _Jean-François Alcover_, Jan 04 2021 *)
%o (PARI)
%o T(n,k) = 2^k*3^ceil(n/2)*binomial(floor(n/2),k);
%o for(n=0, 10, for(k=0, n\2, print1(T(n, k), ", ")); print); \\ _Andrew Howroyd_, Jan 10 2018
%Y Cf. A274496, A274497, A274499.
%K nonn,tabf
%O 0,2
%A _Emeric Deutsch_, Jul 27 2016
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