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A216649 Triangle T(n,k) in which n-th row lists in increasing order all positive integers with a representation as totally balanced 2n digit binary string such that all consecutive totally balanced substrings are in nondecreasing order; n>=1, 1<=k<=A000081(n+1).

%I A216649 #13 Jan 10 2015 11:10:41
%S A216649 2,10,12,42,44,52,56,170,172,180,184,204,212,216,232,240,682,684,692,
%T A216649 696,716,724,728,744,752,820,824,852,856,872,880,920,936,944,976,992,
%U A216649 2730,2732,2740,2744,2764,2772,2776,2792,2800,2868,2872,2900,2904,2920,2928
%N A216649 Triangle T(n,k) in which n-th row lists in increasing order all positive integers with a representation as totally balanced 2n digit binary string such that all consecutive totally balanced substrings are in nondecreasing order; n>=1, 1<=k<=A000081(n+1).
%C A216649 There is a simple bijection between the elements of row n and the rooted trees with n+1 nodes.  The tree has a root node.  Each matching pair (1,0) in the binary string representation encodes an additional node, the totally balanced substrings encode lists of subtrees.
%H A216649 Alois P. Heinz, <a href="/A216649/b216649.txt">Rows n = 1..11, flattened</a>
%F A216649 T(n,k) = A216648(n+1,k)/2 - 2^(2*n).
%e A216649 172 is element of row 4, the binary string representation (with totally balanced substrings enclosed in parentheses) is (10)(10)(1(10)0).  The encoded rooted tree is:
%e A216649 .    o
%e A216649 .   /|\
%e A216649 .  o o o
%e A216649 .      |
%e A216649 .      o
%e A216649 Triangle T(n,k) begins:
%e A216649 2;
%e A216649 10,     12;
%e A216649 42,     44,   52,   56;
%e A216649 170,   172,  180,  184,  204,  212,  216,  232,  240;
%e A216649 682,   684,  692,  696,  716,  724,  728,  744,  752,  820,  824, ...
%e A216649 2730, 2732, 2740, 2744, 2764, 2772, 2776, 2792, 2800, 2868, 2872, ...
%e A216649 Triangle T(n,k) in binary:
%e A216649 10;
%e A216649 1010,       1100;
%e A216649 101010,     101100,     110100,     111000;
%e A216649 10101010,   10101100,   10110100,   10111000,   11001100,   11010100, ...
%e A216649 1010101010, 1010101100, 1010110100, 1010111000, 1011001100, 1011010100, ...
%p A216649 F:= proc(n) option remember; `if`(n=1, [10], sort(map(h->
%p A216649       parse(cat(1, sort(h)[], 0)), g(n-1, n-1)))) end:
%p A216649 g:= proc(n, i) option remember; `if`(i=1, [[10$n]], [seq(seq(seq(
%p A216649       [seq (F(i)[w[t]-t+1], t=1..j),v[]], w=combinat[choose](
%p A216649       [$1..nops(F(i))+j-1], j)), v=g(n-i*j, i-1)), j=0..n/i)])
%p A216649     end:
%p A216649 b:= proc(n) local h, i, r; h, r:= n/10, 0; for i from 0
%p A216649       while h>1 do r:= r+2^i*irem(h, 10, 'h') od; r
%p A216649     end:
%p A216649 T:= proc(n) option remember; map(b, F(n+1))[] end:
%p A216649 seq(T(n), n=1..6);
%Y A216649 First column gives: A020988.
%Y A216649 Last elements of rows give: A020522.
%Y A216649 Row lengths are: A000081(n+1).
%Y A216649 Subsequence of A014486, A031443.
%Y A216649 Cf. A061773, A216349, A216350, A216648.
%K A216649 nonn,tabf
%O A216649 1,1
%A A216649 _Alois P. Heinz_, Sep 12 2012