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Search: id:A059515
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| A059515 |
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Square array T(k,n) by antidiagonals, where T(k,n) is number of ways of placing n identifiable nonnegative intervals with a total of exactly k starting and/or finishing points. |
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4
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| 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 7, 1, 0, 0, 0, 12, 25, 1, 0, 0, 0, 6, 138, 79, 1, 0, 0, 0, 0, 294, 1056, 241, 1, 0, 0, 0, 0, 270, 5298, 7050, 727, 1, 0, 0, 0, 0, 90, 12780, 70350, 44472, 2185, 1, 0, 0, 0, 0, 0, 16020, 334710, 817746, 273378, 6559, 1, 0, 0, 0, 0, 0
(list; table; graph; listen)
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OFFSET
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0,13
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LINKS
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IBM Ponder This, Jan. 2001
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FORMULA
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T(k, n) = T(k - 2, n - 1) * k * (k - 1)/2 + T(k - 1, n - 1) * k^2 + T(k, n - 1) * k * (k + 1)/2 with T(0, 0) = 1 = lambda(k, n) + lambda(k + 1, n) where lambda is A059117(k, n).
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EXAMPLE
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Rows are: 1,0,0,0,0,..., 0,1,1,0,0,..., 0,1,7,12,6,..., 0,1,25,138,294,..., etc. T(1,1)=1 since if a is starting point of interval and A is end point then only possibility is aA (zero length). T(2,1)=1 since possibility is a-A (positive length). T(3,2)=12 since possibilities are: aA-b-B, b-aA-B, b-B-aA, bB-a-A, a-bB-A, a-A-bB, ab-A-B, ab-B-A, a-b-AB, b-a-AB, a-bA-B, b-a-AB.
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CROSSREFS
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Sum of rows gives A059516. Columns include A000007, A057427, A058481, A059117. Final positive number in each row is A000680.
Sequence in context: A046273 A024094 A036949 this_sequence A136428 A053878 A070672
Adjacent sequences: A059512 A059513 A059514 this_sequence A059516 A059517 A059518
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KEYWORD
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nonn,tabl
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AUTHOR
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Henry Bottomley (se16(AT)btinternet.com), Jan 19 2001
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