cp's OEIS Frontend

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).

%I A274498 #18 Mar 17 2025 11:25:57
%S A274498 1,3,3,6,9,18,9,36,36,27,108,108,27,162,324,216,81,486,972,648,81,648,
%T A274498 1944,2592,1296,243,1944,5832,7776,3888,243,2430,9720,19440,19440,
%U A274498 7776,729,7290,29160,58320,58320,23328,729,8748,43740,116640,174960,139968,46656
%N 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).
%C A274498 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 A274498 A sequence is palindromic if and only if its degree of asymmetry is 0.
%C A274498 Sum(kT(n,k),k>=0) = A274499(n).
%H A274498 S. Elizalde, E. Deutsch, <a href="https://doi.org/10.54550/ECA2022V2S1R7">The degree of asymmetry of a sequence</a>, Enum. Combinat. Applic. 2 (2022) no 1 #S2R7 corollary 2.2 at m=3 and dropping each 2nd row.
%F A274498 T(n,k) = 2^k*3^ceiling(n/2)*binomial(floor(n/2),k).
%F A274498 G.f.: G(t,z) = (1 + 3z)/(1 - 3(1 + 2t)z^2).
%F A274498 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 A274498 From _Andrew Howroyd_, Jan 10 2018: (Start)
%e A274498 Triangle begins:
%e A274498    1;
%e A274498    3;
%e A274498    3,   6;
%e A274498    9,  18;
%e A274498    9,  36,   36;
%e A274498   27, 108,  108;
%e A274498   27, 162,  324,  216;
%e A274498   81, 486,  972,  648;
%e A274498   81, 648, 1944, 2592, 1296;
%e A274498   ...
%e A274498 (End)
%e A274498 T(2,0) = 3 because we have 00, 11, and 22.
%e A274498 T(2,1) = 6 because we have 01, 02, 10, 12, 20, and 21.
%p A274498 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 A274498 T[n_, k_] := 2^k 3^Ceiling[n/2] Binomial[Floor[n/2], k];
%t A274498 Table[T[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten (* _Jean-François Alcover_, Jan 04 2021 *)
%o A274498 (PARI)
%o A274498 T(n,k) = 2^k*3^ceil(n/2)*binomial(floor(n/2),k);
%o A274498 for(n=0, 10, for(k=0, n\2, print1(T(n, k), ", ")); print); \\ _Andrew Howroyd_, Jan 10 2018
%Y A274498 Cf. A274496, A274497, A274499.
%K A274498 nonn,tabf
%O A274498 0,2
%A A274498 _Emeric Deutsch_, Jul 27 2016