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%I #21 Jan 09 2024 16:56:47
%S 1,0,1,2,1,1,0,0,1,1,2,1,2,3,2,2,1,1,1,2,1,1,0,0,1,0,0,0,1,1,1,2,1,1,
%T 1,2,1,2,3,2,2,1,1,2,2,3,2,3,4,3,3,2,2,2,3,2,2,1,1,2,1,1,1,1,1,2,1,2,
%U 3,2,2,1,1,1,2,1,1,0,0,1,0,0,0,1,2,1,1,0,0,1,0,0,0,1,0,0,0,0
%N Irregular triangle read by rows: T(n,k) is the number of co-atoms contained in the k-th balanced string of left/right parentheses of length 2*n, where strings within a row are in reverse lexicographical order.
%C See A368750 for the definition of balanced strings and atoms/co-atoms.
%D Donald E. Knuth, The Art of Computer Programming, Vol. 4A: Combinatorial Algorithms, Part 1, Addison-Wesley, 2011, Section 7.2.1.6, exercise 60, p. 478.
%H Paolo Xausa, <a href="/A368751/b368751.txt">Table of n, a(n) for n = 1..17576</a> (rows 1..8 of the triangle, flattened).
%F T(n,k) = A368752(n,k) - A368750(n,k).
%e Triangle begins:
%e [1] 1 0;
%e [2] 1 2 1 1 0 0;
%e [3] 1 1 2 1 2 3 2 2 1 1 1 2 1 1 0 0 1 0 0 0;
%e ...
%e The strings corresponding to row 2, in reverse lexicographical order, are:
%e "))((" (1 co-atom),
%e ")()(" (2 co-atoms),
%e ")(()" (1 co-atom),
%e "())(" (1 co-atom),
%e "()()" (0 co-atoms) and
%e "(())" (0 co-atoms).
%t strings[n_]:=Permutations[PadLeft[PadLeft[{},n,1],2n,-1]];
%t Array[Map[SequenceCount[Accumulate[#],{-1,0}]&,strings[#]]&,5]
%Y Cf. A000346 (row sums), A000984 (row lengths), A362030 and A368804 (binary words).
%Y Cf. A368750 (atoms), A368752 (all atoms), A368753 (defects).
%K nonn,tabf
%O 1,4
%A _Paolo Xausa_, Jan 05 2024
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