

A059515


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.


3



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
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OFFSET

0,13


LINKS

Table of n, a(n) for n=0..69.
IBM Ponder This, Jan 01 2001


FORMULA

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).


EXAMPLE

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 aA (positive length). T(3,2)=12 since possibilities are: aAbB, baAB, bBaA, bBaA, abBA, aAbB, abAB, abBA, abAB, baAB, abAB, baAB.


CROSSREFS

Sum of rows gives A059516. Columns include A000007, A057427, A058481, A059117. Final positive number in each row is A000680.
Sequence in context: A024094 A157307 A036949 * A136428 A271697 A226371
Adjacent sequences: A059512 A059513 A059514 * A059516 A059517 A059518


KEYWORD

nonn,tabl


AUTHOR

Henry Bottomley, Jan 19 2001


STATUS

approved



