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A101452
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Triangle read by rows: T(n,k) is number of noncrossing trees with n edges and having k branches.
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0
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1, 2, 1, 4, 4, 4, 8, 12, 24, 11, 16, 32, 96, 88, 41, 32, 80, 320, 440, 410, 146, 64, 192, 960, 1760, 2460, 1752, 564, 128, 448, 2688, 6160, 11480, 12264, 7896, 2199, 256, 1024, 7168, 19712, 45920, 65408, 63168, 35184, 8835, 512, 2304, 18432, 59136, 165312, 294336, 379008, 316656, 159030, 35989
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OFFSET
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1,2
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COMMENTS
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T(n,k) = 2^(n-k)*binomial(n-1,k-1)*A030981(k).
Row sums yield the ternary numbers (A001764).
The average number of branchnodes over all noncrossing trees with n edges is n(n-1)(19n^2-23n+10)/(3(3n-1)(3n-2)) ~ 19n/27 (see A045738).
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LINKS
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FORMULA
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T(n, k) = [2^(n-k)/k]binomial(n-1, k-1)*Sum_{i=1..k} (-2)^(k-i)*binomial(k, i)*binomial(3i, i-1).
G.f.: G(t, z) = 1/(1-F), where F satisfies F = z(t + 2tF^2/(1-F) + tF^2/(1-F)^2 + 2F).
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EXAMPLE
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T(2,1)=2 because we have /_ and _\; T(2,2)=1 because we have /\
Triangle begins:
1;
2, 1;
4, 4, 4;
8, 12, 24, 11;
16, 32, 96, 88, 41;
...
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MAPLE
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T:=(n, k)->(2^(n-k)/k)*binomial(n-1, k-1)*sum((-2)^(k-i)*binomial(k, i)*binomial(3*i, i-1), i=1..k):for n from 1 to 10 do seq(T(n, k), k=1..n) od; # yields sequence in triangular form
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MATHEMATICA
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T[n_, k_] := 2^(n-k)/k Binomial[n-1, k-1] Sum[(-2)^(k-i) Binomial[k, i] Binomial[3i, i-1], {i, 1, k}];
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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