A120910 Triangle read by rows: T(n,k) is the number of ternary words of length n having k levels (n >= 1, 0 <= k <= n-1). A level is a pair of identical consecutive letters.

3, 6, 3, 12, 12, 3, 24, 36, 18, 3, 48, 96, 72, 24, 3, 96, 240, 240, 120, 30, 3, 192, 576, 720, 480, 180, 36, 3, 384, 1344, 2016, 1680, 840, 252, 42, 3, 768, 3072, 5376, 5376, 3360, 1344, 336, 48, 3, 1536, 6912, 13824, 16128, 12096, 6048, 2016, 432, 54, 3, 3072

Offset

1,1

Comments

Row sums are the powers of 3 (A000244).

T(n,k) = 3*A038207(n-1,k).

T(n,k) = A120909(n,n-k).

Sum_{k>=0} k*T(n,k) = (n-1)*3^(n-1) = A036290(n-1).

Links

Table of n, a(n) for n=1..56.

Formula

T(n,k) = 3*2^(n-k-1)*binomial(n-1,k).

G.f.: (1 - (y - 1)*x)/(1 - (y + 2)*x). Generally for the number of length n words with k levels on an m-ary alphabet (m>1): (1 - (y - 1)*x)/(1 - (y + m - 1)*x). - Geoffrey Critzer, May 19 2014

Example

T(3,1)=12 because we have 001,002,011,022,100,110,112,122,200,211,220 and 221.
Triangle starts:
   3;
   6,  3
  12, 12,  3;
  24, 36, 18,  3;
  48, 96, 72, 24,  3;

Maple

T:=(n, k)->3*2^(n-k-1)*binomial(n-1, k): for n from 1 to 11 do seq(T(n, k), k=0..n-1) od; # yields sequence in triangular form

Mathematica

sol=Solve[{a==v(z^2+a z), b==v(z^2+b z), c==v(z^2+c z)}, {a, b, c}]; f[z_, u_]:=1/(1-3z-a-b-c)/.sol/.v->u-1; nn=10; Drop[Map[Select[#, #>0&]&, Level[CoefficientList[Series[f[z, u], {z, 0, nn}], {z, u}], {2}]], 1]//Grid (* Geoffrey Critzer, May 19 2014 *)

Crossrefs

Cf. A000244, A038207, A120909, A036290.

Sequence in context: A337035 A203491 A085709 * A109044 A205844 A205865

Adjacent sequences: A120907 A120908 A120909 * A120911 A120912 A120913

Keyword

nonn,tabl

Author

Emeric Deutsch, Jul 15 2006

Status

approved