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A076050
Limiting sequence if we start with 2 and successively replace n with 2,3,4,...,n,n+1.
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, 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, 2, 3, 4, 2, 3, 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
OFFSET
1,1
COMMENTS
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.
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.
a(A000108(n)) = n+1 and a(m) < n+1 for m < A000108(n). - Reinhard Zumkeller, Feb 17 2012
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
FORMULA
a(n) = A007001(n) + 1.
EXAMPLE
From John M. Campbell, Sep 07 2018: (Start)
There are A000108(3) = 5 Young tableaux of shape (2^3) = (2, 2, 2), which are listed below lexicographically.
[3 6] [4 6] [4 6] [5 6] [5 6]
[2 5] < [2 5] < [3 5] < [2 4] < [3 4]
[1 4] [1 3] [1 2] [1 3] [1 2]
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)
MATHEMATICA
Nest[Flatten[Map[Range[2, #+1] &, #]] &, {2}, 5] (* Paolo Xausa, Mar 04 2024 *)
PROG
(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])
(Haskell)
a076050 n = a076050_list !! (n-1)
a076050_list = 2 : f [2] where
f xs = (drop (length xs) xs') ++ (f xs') where
xs' = concatMap ((enumFromTo 2) . (+ 1)) xs
-- Reinhard Zumkeller, Feb 17 2012
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Miklos Kristof, Oct 30 2002
STATUS
approved