A302173 Lexicographically first sequence of distinct terms such that any set of three successive digits can be reordered as {d, d+1, d+2}, d being the smallest of the three digits.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 78, 67, 56, 45, 34, 23, 12, 31, 20, 120, 123, 42, 312, 342, 345, 64, 53, 423, 453, 456, 75, 645, 675, 678, 97, 86, 756, 786, 789, 7867, 89, 7897, 867, 564, 534, 231, 201, 234, 567, 897, 8675, 6453, 4231, 2012, 3120, 1201, 2312, 3123, 1231, 2342, 3423, 1234, 2345

Offset

1,3

Comments

As the digit 0 has no predecessor and the digit 9 has no successor here, sets of successive digits like {1,0,9} and {8,9,0} are forbidden.

Links

Dominic McCarty, Table of n, a(n) for n = 1..10000 (first 624 terms from Jean-Marc Falcoz)

Example

Terms a(1) to a(10) are obvious;
a(11) is 78 because 78 is the smallest integer not yet in the sequence such that the elements of the sets {8,9,7} and {9,7,8} are three consecutive digits;
a(12) is 67 because 67 is the smallest integer not yet in the sequence such that the elements of the sets {7,8,6} and {8,6,7} are three consecutive digits;
a(13) is 56 because 56 is the smallest integer not yet in the sequence such that the elements of the sets {6,7,5} and {7,5,6} are three consecutive digits;
etc.

Prog

(Python)
a, runLength = [i for i in range(10)], 3
def helper(s, k, l, a):
    if k not in a: return k
    return min([helper(s[(2-l):]+str(i), int(str(k)+str(i)), l, a) for i in range(10) if (k!=0 or i!=0) and s.find(str(i))==-1 and (all(d[n]+1==d[n+1] for n in range(l-1)) if (d:=sorted([int((s+str(i))[n]) for n in range(l)])) else False)])
while len(a)<100: a.append(helper(("".join(map(str, a)))[(1-runLength):], 0, runLength, a))
print(a) # Dominic McCarty, Feb 02 2025

Crossrefs

Cf. A228326 for the same idea with sets of two digits.

Sequence in context: A153670 A229184 A331203 * A348799 A341909 A024663

Adjacent sequences: A302170 A302171 A302172 * A302174 A302175 A302176

Keyword

nonn,base,changed

Author

Eric Angelini and Jean-Marc Falcoz, Apr 03 2018

Status

approved