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A007001
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Trajectory of 1 under the morphism 1 -> 12, 2 -> 123, 3 -> 1234, etc.
(Formerly M0108)
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16
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1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2
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OFFSET
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1,2
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COMMENTS
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Records in this sequence occur at positions: 1, 2, 5, 14, 42, 132, 429, 1430, ... (which appear to be the Catalan numbers A000108). - Robert G. Wilson v, May 07 2005
The records do occur at Catalan numbers. Of the first C(n) numbers, the number that are equal to k is A033184(n,k), with the one n last. - Franklin T. Adams-Watters, Mar 29 2009
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, if we take away the zeros from (f(T(1), T(A000108(m) - i + 1)) - f(T(A000108(m) - i), T(A000108(m) - i + 1)), i=1..A000108(m)-1), we obtain the first A000108(m - 1) - 1 terms in this sequence. This is illustrated in the below example. - John M. Campbell, Sep 07 2018
The average of the first k terms tends to 3 as k tends to infinity. - Andrew Slattery, Jan 19 2021
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
J. West, Generating trees and forbidden subsequences, Proc. 6th FPSAC [ Conference on Formal Power Series and Algebraic Combinatorics ] (1994), pp. 441-450 (see p. 443).
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LINKS
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FORMULA
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EXAMPLE
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Letting m = 5, as above let (T(1) < T(2) < ... < T(42)) denote the lexicographic sequence of Young tableaux of shape (2, 2, 2, 2, 2). In this case, the sequence (f(T(1), T(43 - i)) - f(T(42 - i), T(43 - i)), i=1..41) is equal to (0, 1, 0, 0, 2, 0, 1, 0, 0, 2, 0, 0, 0, 3, 0, 1, 0, 0, 2, 0, 1, 0, 0, 2, 0, 0, 0, 3, 0, 1, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 0). Removing the zeroes from this tuple, we obtain (1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3), which gives us the first 13 = A000108(m - 1) - 1 terms in this sequence. For example, the first term in the preceding tuple is 0 since T(1) and T(42) are respectively
[ 5 10] [ 9 10]
[ 4 9 ] [ 7 8 ]
[ 3 8 ] [ 5 6 ]
[ 2 7 ] [ 3 4 ]
[ 1 6 ] [ 1 2 ]
and T(41) is equal to
[ 9 10]
[ 7 8 ]
[ 5 6 ]
[ 2 4 ]
[ 1 3 ]
so that the first letter of disagreement between T(1) and T(42) is 2, and that between T(41) and T(42) is also 2. (End)
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MATHEMATICA
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Nest[ Flatten[ # /. a_Integer -> Range[a + 1]] &, {1}, 6] (* Robert G. Wilson v, Jan 24 2006 *)
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PROG
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(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); v[n])
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CROSSREFS
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KEYWORD
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easy,nonn,nice
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AUTHOR
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EXTENSIONS
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More terms from Larry Reeves (larryr(AT)acm.org), Sep 22 2000
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STATUS
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approved
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