A367810 Lexicographically earliest sequence of distinct positive terms not ending in 0 such that the Levenshtein distance (Ld) between a(n) and a(n+1) is equal to the last digit of a(n).

1, 2, 11, 12, 3, 101, 102, 13, 201, 21, 22, 4, 1001, 1002, 103, 5, 10001, 10002, 1003, 14, 2001, 2002, 203, 6, 100001, 100002, 10003, 104, 2211, 211, 111, 112, 15, 20001, 20002, 202, 23, 105, 22211, 2221, 221, 121, 122, 16, 200001, 200002, 20003, 204, 1111, 1011, 1012, 106, 222211, 22221, 2222, 212, 17

Offset

1,2

Links

Table of n, a(n) for n=1..57.

Éric Angelini, More Levenshtein distances, Personal blog, December 2023.

Example

a(1) =  1 and a(2) =   2 are separated by an Ld of 1, and 1 is the last digit of a(1)
a(2) =  2 and a(3) =  11 are separated by an Ld of 2, and 2 is the last digit of a(2)
a(3) = 11 and a(4) =  12 are separated by an Ld of 1, and 1 is the last digit of a(3)
a(4) = 12 and a(5) =   3 are separated by an Ld of 2, and 2 is the last digit of a(4)
a(5) =  3 and a(6) = 101 are separated by an Ld of 3, and 3 is the last digit of a(5), etc.

Mathematica

a[1]=1; a[n_]:=a[n]=(k=1; While[MemberQ[Array[a, n-1], k] ||Mod[k, 10]==0|| EditDistance[ToString@a[n-1], ToString@k]!= Mod[a[n-1], 10], k++]; k); Array[a, 40]

Prog

(Python)
from itertools import islice
from Levenshtein import distance as Ld
def agen(): # generator of terms
    an, aset, mink = 1, {1}, 2
    while True:
        yield an
        s, k = str(an), mink
        target = int(s[-1])
        while k%10 == 0 or k in aset or Ld(s, str(k)) != target: k += 1
        an = k
        aset.add(k)
        while mink in aset or mink%10 == 0: mink += 1
print(list(islice(agen(), 57))) # Michael S. Branicky, Dec 01 2023

Crossrefs

Cf. A367638.

Sequence in context: A201187 A359207 A068225 * A043080 A108199 A137904

Adjacent sequences: A367807 A367808 A367809 * A367811 A367812 A367813

Keyword

base,nonn

Author

Eric Angelini and Giorgos Kalogeropoulos, Dec 01 2023

Status

approved