%I A073346
%S A073346 1,1,0,0,0,0,1,2,0,0,0,0,0,0,0,0,2,4,0,0,0,0,2,4,0,0,0,0,1,0,8,8,0,0,0,
%T A073346 0,0,0,12,16,0,0,0,0,0,0,2,12,40,16,0,0,0,0,0,0,2,12,80,48,0,0,0,0,0,0,
%U A073346 0,0,12,136,144,32,0,0,0,0,0,0,0,2,20,224,384,128,0,0,0,0,0,0,0,0,0,16
%N A073346 Table T(n,k) (listed antidiagonalwise in order T(0,0), T(1,0), T(0,1), 
               T(2,0), T(1,1), ...) giving the number of rooted plane binary trees 
               of size n and "contracted height" k.
%C A073346 The height of binary trees is computed here otherwise as with A073245, 
               but whenever a complete binary tree of (2^k)-1 nodes with all its 
               leaves at the same level, i.e. one of the following trees:
%C A073346 ____________________\/\/\/\/_
%C A073346 _____________\/__\/__\/__\/__
%C A073346 ______________\__/____\_ /___
%C A073346 ____.____\/____\/______\/____ etc.
%C A073346 is encountered as a terminating subtree, it is regarded just a variant 
               of . (an empty tree, a single leaf) and contributes nothing to the 
               height of the tree.
%H A073346 H. Bottomley and A. Karttunen, <a href="a073345.txt">Notes concerning 
               diagonals of the square arrays A073345 and A073346.</a>
%F A073346 (See the Maple code below. Note that here we use the same convolution 
               recurrence as with A073345, but only the initial conditions for the 
               first two rows (k=0 and k=1) are different. Is there a nicer formula?)
%e A073346 The top-left corner of this square array:
%e A073346 1 1 0 1 0 0 0 1 ...
%e A073346 0 0 2 0 2 2 0 0 ...
%e A073346 0 0 0 4 4 8 12 12 ...
%e A073346 0 0 0 0 8 16 40 80 ...
%p A073346 A073346 := n -> A073346bi(A025581(n), A002262(n));
%p A073346 A073346bi := proc(n,k) option remember; local i,j; if(0 = k) then RETURN(A036987(n)); 
               fi; if(0 = n) then RETURN(0); fi; 2 * add(A073346bi(n-i-1,k-1) * 
               add(A073346bi(i,j),j=0..(k-1)),i=0..floor((n-1)/2)) + 2 * add(A073346bi(n-i-1,
               k-1) * add(A073346bi(i,j),j=0..(k-2)),i=(floor((n-1)/2)+1)..(n-1)) 
               - (`mod`(n,2))*(A073346bi(floor((n-1)/2),k-1)^2) - (`if`((1=k),1,
               0))*A036987(n); end;
%p A073346 A025581 := n -> binomial(1+floor((1/2)+sqrt(2*(1+n))),2) - (n+1);
%p A073346 A002262 := n -> n - binomial(floor((1/2)+sqrt(2*(1+n))),2);
%Y A073346 Variant: A073345. The first row: A036987. Column sums: A000108. Diagonals: 
               T(n, n) = A000007(n), T(n+1, n) = A000079(n), T(n+2, n) = A058922(n), 
               T(n+3, n) = A074092(n) - [see the attached notes.].
%Y A073346 A073430 gives the upper triangular region of this array. Used to compute 
               A073431. Entries on the row k are all divisible by 2^k, thus dividing 
               them out yields the array/triangle A074079/A074080.
%Y A073346 Sequence in context: A138045 A004530 A084863 this_sequence A114099 A028613 
               A024362
%Y A073346 Adjacent sequences: A073343 A073344 A073345 this_sequence A073347 A073348 
               A073349
%K A073346 nonn,tabl
%O A073346 0,8
%A A073346 Antti Karttunen Jul 31 2002