A367638 -id:A367638 - cp's OEIS frontend

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

Eric Angelini and Giorgos Kalogeropoulos, Dec 01 2023

			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.

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

Cf. A367638.

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]

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

A367797 The successive digits of the number k are the successive "inside Levenshtein distances" of k (except for the last digit of k). See the Comment section for the definition of an "inside Levenshtein distance".

Original entry on oeis.org

10, 12, 13, 14, 15, 16, 17, 18, 19, 111, 211, 2020, 2122, 2230, 2231, 2234, 2235, 2236, 2237, 2238, 2239, 3121, 31131, 32131, 32233, 32340, 32341, 32345, 32346, 32347, 32348, 32349, 42232, 422242, 432242, 432450, 432451, 432456, 432457, 432458, 432459, 433242, 433344, 532342, 5433353, 5433455, 5433560

Offset: 1

Eric Angelini and Giorgos Kalogeropoulos, Nov 30 2023

Let's consider 2023 and compute the successive traditional Levenshtein distances between 2 and 023, 20 and 23, 202 and 3 (the so-called inside Lds).
We have:
  Ld 2<>023 = 2,
  Ld 20<>23 = 1,
  Ld 202<>3 = 3.
The successive Lds of 2023 are 2, 1 and 3.

			a(1) = 10 has an iLd of 1 (the Levenshtein distance between 1 and 0) and this iLd of 1 is the first digit of a(1);
a(47) = 5433560 is in the sequence because its successive Lds are:
  Ld 5<>433560 = 5
  Ld 54<>33560 = 4
  Ld 543<>3560 = 3
  Ld 5433<>560 = 3
  Ld 54335<>60 = 5
  Ld 543356<>0 = 6.
We see that the rightmost column above reproduces a(47), except for the last digit.

Éric Angelini, Inside Levenshtein distances, Personal blog, November 2023.

Cf. A367638.

from Levenshtein import distance as Ld
def ok(n):
    s = str(n)
    if n < 10: return False # convention, though condition is vacuously True
    return all(Ld(s[:i+1], s[i+1:]) == int(s[i]) for i in range(len(s)-1))
print([k for k in range(10**7) if ok(k)]) # Michael S. Branicky, Dec 01 2023

cp's OEIS Frontend

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).

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