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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.

%I A059515 #14 Dec 02 2024 16:29:18
%S A059515 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,
%T A059515 294,1056,241,1,0,0,0,0,270,5298,7050,727,1,0,0,0,0,90,12780,70350,
%U A059515 44472,2185,1,0,0,0,0,0,16020,334710,817746,273378,6559,1,0,0,0,0,0
%N 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.
%C A059515 See A300729 for a triangular version of this array. - _Peter Bala_, Jun 13 2019
%H A059515 IBM Ponder This, <a href="https://research.ibm.com/haifa/ponderthis/challenges/January2001.html">Jan 01 2001</a>
%F A059515 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).
%e A059515 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.
%Y A059515 Sum of rows gives A059516. Columns include A000007, A057427, A058481, A059117. Final positive number in each row is A000680.
%Y A059515 Cf. A300729.
%K A059515 nonn,tabl
%O A059515 0,13
%A A059515 _Henry Bottomley_, Jan 19 2001