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Search: id:A120907
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| A120907 |
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Triangle read by rows: T(n,k) is the number of ternary words of length n on {0,1,2} having sum of the lengths of the drops equal to k (n>=0, k>=0). The drops of a ternary word on {0,1,2} are the subwords 10,20, and 21, their lengths being the differences 1, 2, and 1, respectively. |
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+0
2
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| 1, 3, 6, 2, 1, 10, 10, 7, 15, 30, 31, 4, 1, 21, 70, 105, 36, 11, 28, 140, 294, 184, 76, 6, 1, 36, 252, 714, 696, 396, 78, 15, 45, 420, 1554, 2160, 1666, 566, 141, 8, 1, 55, 660, 3102, 5808, 5918, 2990, 995, 136, 19, 66, 990, 5775, 13992, 18348, 12746, 5615, 1280, 226
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Row n has 2*floor(n/2)+1 terms (i.e. each of the rows 2n and 2n+1 has 2n+1 terms). Row sums are the powers of 3 (A000244). T(n,0)=A000217(n+1) (the triangular numbers). Sum(k*T(n,k),k>=0)=4(n-1)3^(n-2)=A120908(n)=4*A027471(n).
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FORMULA
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G.f.=G(t,z)=1/[(1-z+tz)(1-2z+z^2-tz-tz^2)].
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EXAMPLE
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T(4,3)=4 because we have 1020,2010,2021, and 2120.
Triangle starts:
1;
3;
6,2,1;
10,10,7;
15,30,31,4,1;
21,70,105,36,11;
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MAPLE
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G:=1/(1-z+t*z)/(1-2*z+z^2-t*z-t*z^2): Gser:=simplify(series(G, z=0, 15)): P[0]:=1: for n from 1 to 12 do P[n]:=sort(coeff(Gser, z^n)) od: for n from 0 to 12 do seq(coeff(P[n], t, j), j=0..2*floor(n/2)) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A000244, A000217, A120908, A027471, A120906.
Adjacent sequences: A120904 A120905 A120906 this_sequence A120908 A120909 A120910
Sequence in context: A058178 A058078 A016551 this_sequence A133358 A058099 A124085
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 15 2006
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