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A119851
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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)).
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1
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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; internal format)
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OFFSET
| 0,2
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COMMENTS
| Row n has 1+floor(n/3) terms.
Sum of entries in row n is 3^n (A000244).
Sum(k*T(n,k),k>=0) = (n-2)*3^(n-3) = A027741(n-1).
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FORMULA
| T(n,k) = \sum_{j=0}^{n-3k} binomial(-(k-1),j) * binomial(j,(n-3k-j)/2) *
(-3)^((3j+3k-n)/2) [From 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.
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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;
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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
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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) )) } \\ From Max Alekseyev
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CROSSREFS
| Cf. A000244, A076264 (=T(n,0)), A119852 (=T(n,1)), A027741.
Sequence in context: A074440 A006204 A013572 * A119825 A037260 A035313
Adjacent sequences: A119848 A119849 A119850 * A119852 A119853 A119854
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KEYWORD
| nonn,tabf
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), May 26 2006
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