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A133399
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Triangle T(n,k)=number of forests of labeled rooted trees with n nodes, containing exactly k trees of height one, all others having height zero (n>=0, 0<=k<=floor(n/2)).
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3
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1, 1, 1, 2, 1, 9, 1, 28, 12, 1, 75, 120, 1, 186, 750, 120, 1, 441, 3780, 2100, 1, 1016, 16856, 21840, 1680, 1, 2295, 69552, 176400, 45360, 1, 5110, 272250, 1224720, 705600, 30240, 1, 11253, 1026300, 7692300, 8316000, 1164240, 1, 24564, 3762132, 45018600
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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REFERENCES
| A. P. Heinz, Finding Two-Tree-Factor Elements of Tableau-Defined Monoids in Time O(n^3), Ed. S. G. Akl, F. Fiala, W. W. Koczkodaj: Advances in Computing and Information, ICCI90 Niagara Falls, LNCS 468, Springer-Verlag (1990), pp. 120-128.
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LINKS
| Alois P. Heinz, Rows n = 0..200, flattened
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FORMULA
| T(n,k) = C(n,k) * k! * stirling2(n-k+1,k+1).
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EXAMPLE
| Triangle begins:
1;
1;
1, 2;
1, 9;
1, 28, 12;
1, 75, 120;
1, 186, 750, 120;
1, 441, 3780, 2100;
1, 1016, 16856, 21840, 1680;
1, 2295, 69552, 176400, 45360;
1, 5110, 272250, 1224720, 705600, 30240;
...
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MAPLE
| with(combinat): T:= (n, k)-> binomial (n, k)*k!*stirling2 (n-k+1, k+1): for n from 0 to 10 do lprint (seq (T(n, k), k=0..floor(n/2))) od;
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CROSSREFS
| Columns k=1,2 give: A058877, A133386. Row sums give: A000248. T(2*n,n) = A001813(n), T(2*n+1,n) = A002691(n).
Sequence in context: A206243 A187549 A100945 * A128751 A129168 A194555
Adjacent sequences: A133396 A133397 A133398 * A133400 A133401 A133402
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KEYWORD
| nonn,tabf
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AUTHOR
| Alois P. Heinz (heinz(AT)hs-heilbronn.de), Nov 24 2007
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