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A119851 Triangle read by rows: T(n,k) is the number of ternary words of length n containing k 012's (n >= 0, 0 <= k <= floor(n/3)). 1
1, 3, 9, 26, 1, 75, 6, 216, 27, 622, 106, 1, 1791, 387, 9, 5157, 1350, 54, 14849, 4566, 267, 1, 42756, 15102, 1179, 12, 123111, 49113, 4833, 90, 354484, 157622, 18798, 536, 1, 1020696, 500520, 70317, 2775, 15, 2938977, 1575558, 255231, 13068, 135 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

Row n has 1+floor(n/3) terms.

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

Sum_{k>=0} k*T(n,k) = (n-2)*3^(n-3) = A027741(n-1).

LINKS

Table of n, a(n) for n=0..44.

FORMULA

T(n,k) = Sum_{j=0..n-3k} binomial(-(k-1),j) * binomial(j,(n-3k-j)/2) * (-3)^((3j+3k-n)/2). - Max Alekseyev

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

Recurrence relation: T(n,k) = 3*T(n-1,k) - T(n-3,k) + T(n-3,k-1) for n >= 3.

EXAMPLE

T(4,1)=6 because we have 0012, 0120, 0121, 0122, 1012 and 2012.

Triangle starts:

    1;

    3;

    9;

   26,   1;

   75,   6;

  216,  27;

  622, 106, 1;

MAPLE

G:=1/(1-3*z+z^3-t*z^3): Gser:=simplify(series(G, z=0, 20)): P[0]:=1: for n from 1 to 15 do P[n]:=sort(coeff(Gser, z^n)) od: for n from 0 to 15 do seq(coeff(P[n], t, j), j=0..floor(n/3)) od; # yields sequence in triangular form

PROG

(PARI) { T(n, k) = sum(j=0, n-3*k, if((n-3*k-j)%2, 0, binomial(-(k-1), j) *

binomial(j, (n-3*k-j)/2) * (-3)^((3*j+3*k-n)/2) )) } \\ Max Alekseyev

CROSSREFS

Cf. A000244, A076264 (=T(n,0)), A119852 (=T(n,1)), A027741.

Sequence in context: A074440 A006204 A013572 * A119825 A235538 A218916

Adjacent sequences:  A119848 A119849 A119850 * A119852 A119853 A119854

KEYWORD

nonn,tabf

AUTHOR

Emeric Deutsch, May 26 2006

STATUS

approved

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Last modified November 17 14:03 EST 2018. Contains 317276 sequences. (Running on oeis4.)