A181371 - OEIS

%I #11 Jan 31 2019 22:02:49

%S 1,3,8,1,21,6,55,25,1,144,90,9,377,300,51,1,987,954,234,12,2584,2939,

%T 951,86,1,6765,8850,3573,480,15,17711,26195,12707,2305,130,1,46368,

%U 76500,43398,10008,855,18,121393,221016,143682,40426,4740,183,1,317811

%N Triangle read by rows: T(n,k) is the number of ternary words (i.e., finite sequences of 0's, 1's and 2's) of length n having k occurrences of 01's (0 <= k <= floor(n/2)).

%C Row n contains 1 + floor(n/2) entries.

%C Sum of entries in row n is 3^n = A000244(n).

%C T(n,0) = F(2n+2) = A001906(n+1) (even-subscripted Fibonacci numbers).

%C T(n,1) = A001871(n-2).

%C Sum_{k>=0}k*T(n,k) = (n-1)*3^(n-2) = A027471(n) (n>=1).

%H Alois P. Heinz, <a href="/A181371/b181371.txt">Rows n = 0..200, flattened</a>

%H Marilena Barnabei, Flavio Bonetti, and Niccolò Castronuovo, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL21/Barnabei/barnabei5.html">Motzkin and Catalan Tunnel Polynomials</a>, J. Int. Seq., Vol. 21 (2018), Article 18.8.8.

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

%e T(3,1)=6 because we have 010, 011, 012, 001, 101 and 201.

%e T(4,2)=1 because we have 0101.

%e Triangle starts:

%e     1;

%e     3;

%e     8,  1;

%e    21,  6;

%e    55, 25,  1;

%e   144, 90,  9;

%p G := 1/(1-3*z+z^2-t*z^2): Gser := simplify(series(G, z = 0, 15)): for n from 0 to 13 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 13 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form

%Y Cf. A000244, A001871, A001906, A027471.

%K nonn,tabf

%O 0,2

%A _Emeric Deutsch_, Oct 31 2010