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 A007001 Trajectory of 1 under the morphism 1 -> 12, 2 -> 123, 3 -> 1234, etc. (Formerly M0108) 15
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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 REFERENCES 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). LINKS T. D. Noe, Table of n, a(n) for n = 1..4862 (8 iterations) C. Banderier, A. Denise, P. Flajolet, M. Bousquet-Mélou et al., Generating Functions for Generating Trees, Discrete Mathematics 246(1-3), March 2002, pp. 29-55. A. Karttunen, Notes concerning A080237-tree and related sequences. S. Lehr, J. Shallit and J. Tromp, On the vector space of the automatic reals, Theoret. Comput. Sci. 163 (1996), no. 1-2, 193-210. James Propp and N. J. A. Sloane, Email, March 1994 FORMULA From n > 1 onward a(n) = A080237(A081291(n-1)). - Antti Karttunen, Jul 31 2003 EXAMPLE From John M. Campbell, Sep 07 2018: (Start) 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) MATHEMATICA Nest[ Flatten[ # /. a_Integer -> Range[a + 1]] &, {1}, 6] (* Robert G. Wilson v, Jan 24 2006 *) PROG (PARI) a(n)=local(v, w); if(n<1, 0, v=; while(#v

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Last modified November 13 17:34 EST 2019. Contains 329106 sequences. (Running on oeis4.)