|
Search: id:A119851
|
|
|
| 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)). |
|
+0
2
|
|
| 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; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
Row n has 1+floor(n/3) terms. Sum of entries in row n is 3^n (A000244). T(n,0)=A076264(n). T(n,1)=A119852(n). Sum(k*T(n,k),k>=0)=(n-2)*3^(n-3)=A027741(n-1).
|
|
FORMULA
|
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
|
|
CROSSREFS
|
Cf. A000244, A076264, A119852, A027741.
Adjacent sequences: A119848 A119849 A119850 this_sequence A119852 A119853 A119854
Sequence in context: A074440 A006204 A013572 this_sequence A119825 A037260 A035313
|
|
KEYWORD
|
nonn,tabf
|
|
AUTHOR
|
Emeric Deutsch (deutsch(AT)duke.poly.edu), May 26 2006
|
|
|
Search completed in 0.002 seconds
|
