

A135472


Shortest and lexicographically earliest string of decimal digits with property that when made into cycle every pair of digits from 0,0 to 9,9 can be seen exactly once.


1



0, 0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, 0, 8, 0, 9, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 5, 5, 6, 5, 7, 5, 8, 5, 9, 6, 6, 7, 6, 8, 6, 9, 7, 7, 8, 7, 9, 8, 8, 9, 9
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OFFSET

1,5


COMMENTS

Comments from Max Alekseyev, Feb 14 2008: (Start) It is easy to prove that such a string exists and, moreover, it can be closed into a circle of length 100.
Namely, let us construct a (directed) de Bruijn graph G on vertices V={0,1,2,3,4,5,6,7,8,9}, where every vertex is connected to every other vertex (including itself  so there is a selfloop at every vertex) by a directed arc. The arcs in G "encode" all possible 2digit strings.
Any string over the alphabet V can be viewed as a path in G. If the string contains all 2digit strings as substrings, then the corresponding path passes through every arc in G. The shortest such path is an Eulerian one (if it exists) and it indeed exists in G.
The indegree of every vertex in G equals its outdegree, implying that there exists an Eulerian cycle. Such a cycle has length 100 and visit every vertex 10 times.
So we want to find an Eulerian cycle resulting in the lexicographically earliest string. Such a cycle can be easily found by traversing G in a greedy manner. (End)


LINKS

Table of n, a(n) for n=1..100.


CROSSREFS

Cf. A102167.
Sequence in context: A276457 A171181 A034948 * A008723 A263397 A008803
Adjacent sequences: A135469 A135470 A135471 * A135473 A135474 A135475


KEYWORD

nonn,fini,full,base,nice


AUTHOR

Patrick A. Kirol (sunwukong(AT)povn.com), Feb 08 2008


EXTENSIONS

Confirmed by Max Alekseyev, Joshua Zucker and Joerg Arndt, Feb 14 2008
Edited by N. J. A. Sloane, Feb 18 2008


STATUS

approved



