%I
%S 2,3,2,3,4,2,3,2,3,4,2,3,4,5,2,3,2,3,4,2,3,2,3,4,2,3,4,5,2,3,2,3,4,2,
%T 3,4,5,2,3,4,5,6,2,3,2,3,4,2,3,2,3,4,2,3,4,5,2,3,2,3,4,2,3,2,3,4,2,3,
%U 4,5,2,3,2,3,4,2,3,4,5,2,3,4,5,6,2,3,2,3,4,2,3,2,3,4,2,3,4,5,2,3,2,3,4,2,3
%N Limiting sequence if we start with 2 and successively replace n with 2,3,4,...,n,n+1.
%C We get 2, 23, 23234, 23234232342345 and so on. The lengths are 1,2,5,14,42,... which are the Catalan numbers: A000108. The sums of the numbers in these strings are also the Catalan numbers.
%C In A071159 the ndigit terms follow the 2, 3, 2, 3, 4, ... rule: the number of terms in which the first n1 digits are the same is 2, 3, 2, 3, 4, ... and the last digits of the terms are 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 4, ..., A007001. For example, 1111, 1112, 1121, 1122, 1123, 1211, 1212, 1221, 1222, 1223, 1231, 1232, 1233, 1234.
%C a(A000108(n)) = n+1 and a(m) < n+1 for m < A000108(n).  _Reinhard Zumkeller_, Feb 17 2012
%C Let (T(1) < T(2) < ... < T(A000108(m))) denote the sequence of Young tableaux of shape (2^m) ordered lexicographically with respect to their columns, and let f(T(i), T(j)) denote the first label of disagreement among T(i) and T(j). Then, empirically, the reverse of the list (f(T(1), T(2)), f(T(1), T(3)), ..., f(T(1), T(A000108(m)))) agrees with the first A000108(m)  1 terms in this sequence, for all m > 1, as illustrated in the below example.  _John M. Campbell_, Sep 07 2018
%H Reinhard Zumkeller, <a href="/A076050/b076050.txt">Table of n, a(n) for n = 1..10000</a>
%H Italo J. Dejter, <a href="https://www.researchgate.net/publication/245576352_On_a_lexical_tree_for_the_middlelevels_graph_problem">The role of restricted growth strings in the two middle levels of the Boolean lattice B_(2k+1)</a>, University of Puerto Rico, 2018.
%e From _John M. Campbell_, Sep 07 2018: (Start)
%e There are A000108(3) = 5 Young tableaux of shape (2^3) = (2, 2, 2), which are listed below lexicographically.
%e [3 6] [4 6] [4 6] [5 6] [5 6]
%e [2 5] < [2 5] < [3 5] < [2 4] < [3 4]
%e [1 4] [1 3] [1 2] [1 3] [1 2]
%e As above, let (T(1), T(2), ..., T(5)) denote this list. The first label of disagreement between T(1) and T(5) is 2; that between T(1) and T(4) is 3; that between T(1) and T(3) is 2; that between T(1) and T(2) is 3. The sequence (2, 3, 2, 3) agrees with the first 4 terms in this sequence. If we repeat this process using Young tableaux of shape (2^4), we obtain the sequence (2, 3, 2, 3, 4, 2, 3, 2, 3, 4, 2, 3, 4). (End)
%o (PARI) a(n)=local(v,w); if(n<1,0,v=[1]; while(#v<n,w=[]; for(i=1,#v,w=concat(w,vector(v[i]+1,j,j))); v=w); 1+v[n])
%o (Haskell)
%o a076050 n = a076050_list !! (n1)
%o a076050_list = 2 : f [2] where
%o f xs = (drop (length xs) xs') ++ (f xs') where
%o xs' = concatMap ((enumFromTo 2) . (+ 1)) xs
%o  _Reinhard Zumkeller_, Feb 17 2012
%Y Cf. A000108, A071159. a(n)=A007001(n)+1.
%K easy,nonn,nice,changed
%O 1,1
%A _Miklos Kristof_, Oct 30 2002
