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Search: id:A119900
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| A119900 |
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Triangle read by rows: T(n,k) is the number of binary words of length n with k strictly increasing runs (0<=k<=n; for example, the binary word 1/0/01/01/1/1/01 has 7 strictly increasing runs). |
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+0
3
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| 1, 0, 2, 0, 1, 3, 0, 0, 4, 4, 0, 0, 1, 10, 5, 0, 0, 0, 6, 20, 6, 0, 0, 0, 1, 21, 35, 7, 0, 0, 0, 0, 8, 56, 56, 8, 0, 0, 0, 0, 1, 36, 126, 84, 9, 0, 0, 0, 0, 0, 10, 120, 252, 120, 10, 0, 0, 0, 0, 0, 1, 55, 330, 462, 165, 11, 0, 0, 0, 0, 0, 0, 12, 220, 792, 792, 220, 12, 0, 0, 0, 0, 0, 0, 1, 78
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OFFSET
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0,3
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COMMENT
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Sum of entries in row n is 2^n (A000079). Sum of entries in column k is A001906(k+1) (the even indexed Fibonacci numbers). Row n contains 1+floor(n/2) nonzero terms. Sum(k*T(n,k),k=0..n)=(3n+1)*2^(n-2)=A066373(n+1) for n>=1. T(n,k)=A034867(n,n-k).
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FORMULA
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T(n,k)=binom(n+1,2k-n). G.f.=1/[1-2tz-t(1-t)z^2].
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EXAMPLE
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T(5,3)=6 because we have 0/01/01, 01/0/01, 01/01/0, 01/1/01, 01/01/1, and 1/01/01 (the runs are separated by /).
Triangle starts:
1;
0,2;
0,1,3;
0,0,4,4;
0,0,1,10,5;
0,0,0,6,20,6;
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MAPLE
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T:=(n, k)->binomial(n+1, 2*k-n): for n from 0 to 12 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A000079, A001906, A066373, A034867.
Sequence in context: A062283 A136493 A132213 this_sequence A141097 A096335 A129503
Adjacent sequences: A119897 A119898 A119899 this_sequence A119901 A119902 A119903
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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 27 2006
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