%I
%S 0,1,23211,2432211,2354543212221,335465432122211
%N A071158codes for the fixed points of Vaillé's 1997 bijection on Dyck paths.
%C Vaille gives the terms a(2)a(4) on the last page of the 1997 paper. Note that this sequence migh be finite, for two reasons: (a) there are no more fixed points after some limit (the next one after a(5) must have at least 19 digits. All the terms must be of odd length). (b) some of the fixed points would require digits higher than "9", in which case the factorial expansion can nomore presented unambiguously in decimal. However, the sequence A126311 can accommodate also those cases.
%H J. Vaillé, <a href="http://dx.doi.org/10.1006/eujc.1996.0089">Une Bijection Explicative de Plusieurs Propriétés Remarquables des Ponts</a>, European J. Combin. 18 (1997), no. 1, 117124.
%e This sequence consists of those terms of A071158 for which the first factorial digit is equal to the number of 1's in the term and the following algorithm results the remaining factorial digits of the same term: First, extract all maximal subsequences from the term (for d ranging from 1 to the largest digit present) that consist of digits d and d+1 and place them next to each other, from left to right. E.g. for the term 2354543212221 this yields the sequence: 2212221,2332222,3443,5454,55. After discarding the last digit (here 5) and replacing in each batch the smaller number with +1 and larger number with 1, we get:
%e 1,1,+1,1,1,1,+1,+1,1,1,+1,+1,+1,+1,+1,1,1,+1,1,+1,1,+1,+1.
%e and summing these from RIGHT, we get the following partial sums:
%e 1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 5, 4, 3, 2, 1, 0, 1, 2, 1, 2, 1, 2, 1.
%e Retaining only the partial sums under the +1's (i.e. the rightmost one and all the partial sums that are larger than the preceding partial sum one step to the right) we obtain: 3,5,4,5,4,3,2,1,2,2,2 and 1. These, after appended to the number of 1's in the original term (2), yields the same term 2354543212221 from which we started from, which thus is a member of this sequence. Similarly, the term 2432211 belongs to this sequence, because the same procedure yields:
%e 22211,2322,43,4 and after discarding the last 4:
%e 1,1,1,+1,+1,+1,1,+1,+1,1,+1 and summing from the right:
%e 1, 2, 3, 4, 3, 2, 1, 2, 1, 0, 1.
%e collecting all the partial sums larger than their right neighbor (those under +1's), which appended after the number of 1's (2), results the same term 2432211.
%Y a(n) = A071158(A126300(n)) = A007623(A126311(n)). Subset of A126299. Cf. A126295.
%K nonn,hard,more,base
%O 0,3
%A _Antti Karttunen_, Jan 02 2007
