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A076050 Limiting sequence if we start with 2 and successively replace n with 2,3,4,...,n,n+1.

%I A076050 #41 Mar 05 2024 07:18:41
%S A076050 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 A076050 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 A076050 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 A076050 Limiting sequence if we start with 2 and successively replace n with 2,3,4,...,n,n+1.
%C A076050 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 A076050 In A071159 the n-digit terms follow the 2, 3, 2, 3, 4, ... rule: the number of terms in which the first n-1 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 A076050 a(A000108(n)) = n+1 and a(m) < n+1 for m < A000108(n). - _Reinhard Zumkeller_, Feb 17 2012
%C A076050 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 A076050 Reinhard Zumkeller, <a href="/A076050/b076050.txt">Table of n, a(n) for n = 1..10000</a>
%H A076050 Italo J. Dejter, <a href="https://www.researchgate.net/publication/245576352_On_a_lexical_tree_for_the_middle-levels_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.
%H A076050 Italo J. Dejter, <a href="https://arxiv.org/abs/1911.02100">Reinterpreting Mütze's Theorem via Natural Enumeration of Ordered Rooted Trees</a>, arXiv:1911.02100 [math.CO], 2019.
%F A076050 a(n) = A007001(n) + 1.
%e A076050 From _John M. Campbell_, Sep 07 2018: (Start)
%e A076050 There are A000108(3) = 5 Young tableaux of shape (2^3) = (2, 2, 2), which are listed below lexicographically.
%e A076050    [3 6]   [4 6]   [4 6]   [5 6]   [5 6]
%e A076050    [2 5] < [2 5] < [3 5] < [2 4] < [3 4]
%e A076050    [1 4]   [1 3]   [1 2]   [1 3]   [1 2]
%e A076050 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)
%t A076050 Nest[Flatten[Map[Range[2, #+1] &, #]] &, {2}, 5] (* _Paolo Xausa_, Mar 04 2024 *)
%o A076050 (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 A076050 (Haskell)
%o A076050 a076050 n = a076050_list !! (n-1)
%o A076050 a076050_list = 2 : f [2] where
%o A076050    f xs = (drop (length xs) xs') ++ (f xs') where
%o A076050      xs' = concatMap ((enumFromTo 2) . (+ 1)) xs
%o A076050 -- _Reinhard Zumkeller_, Feb 17 2012
%Y A076050 Cf. A000108, A071159, A007001.
%K A076050 easy,nonn
%O A076050 1,1
%A A076050 _Miklos Kristof_, Oct 30 2002