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Search: id:A119914
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| A119914 |
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Triangle read by rows: T(n,k) is number of ternary words of length n and having k runs of 0's of odd length (0<=k<=ceil(n/2); a run of 0's is a subsequence of consecutive 0's of maximal length). |
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+0
3
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| 1, 2, 1, 5, 4, 12, 13, 2, 29, 40, 12, 70, 117, 52, 4, 169, 332, 196, 32, 408, 921, 678, 172, 8, 985, 2512, 2216, 768, 80, 2378, 6761, 6952, 3064, 512, 16, 5741, 18004, 21144, 11328, 2640, 192, 13860, 47525, 62762, 39624, 11920, 1424, 32, 33461, 124536
(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 1+ceil(n/2) terms. Sum of entries in row n is 3^n (A000244). T(n,0)=A000129(n+1) (Pell numbers). T(n,1)=A119915(n). Sum(k*T(n,k), k>=0)=A119916(n).
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FORMULA
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G.f.=G(t,z)=(1+tz)/(1-2z-z^2-2tz^2). T(n,k)=2T(n-1,k)+T(n-2,k)+2T(n-2,k-1) (n>=2).
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EXAMPLE
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T(4,2)=12 because we have 0101, 0102, 0110, 0120, 0201, 0202, 0210, 0220, 1010, 1020, 2010, and 2020.
Triangle starts:
1;
2,1;
5,4;
12,13,2;
29,40,12;
70,117,52,4;
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MAPLE
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G:=(1+t*z)/(1-2*z-z^2-2*t*z^2): Gser:=simplify(series(G, z=0, 14)): 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..ceil(n/2)) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A000244, A000129, A119915, A119916.
Sequence in context: A019473 A056605 A091802 this_sequence A120924 A079285 A124660
Adjacent sequences: A119911 A119912 A119913 this_sequence A119915 A119916 A119917
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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), May 29 2006
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