A273376 Pick any pair of "1" digits in the sequence. Those two "1"s are separated by k digits. This is the lexicographically earliest sequence of distinct terms in which all the resulting values of k are distinct.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 20, 21, 22, 23, 24, 25, 26, 27, 12, 28, 29, 13, 30, 32, 33, 14, 34, 35, 36, 37, 38, 39, 40, 42, 43, 44, 15, 45, 46, 47, 48, 49, 50, 31, 52, 53, 54, 55, 41, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 51, 76, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87, 88, 89, 61, 90, 92, 93, 94, 95, 96, 97, 98, 99, 200, 202, 203, 204, 205, 206, 201

Offset

1,3

Comments

The sequence starts with a(1)=0. It is then always extended with the smallest integer not yet present and not leading to a contradiction (which would mean producing a value of k already seen).

Links

Eric Angelini, Table of n, a(n) for n = 1..1011

Example

The ten "k"s in the starting segment here are different [0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 20, 21,] and respectively equal to 8,10,11,15,1,2,6,0,4,3.
Indeed, there are k=8 digits between [1] and the "1" of [10] which are 2,3,4,5,6,7,8,9; there are k=10 digits between [1] and the first "1" of [11] which are 2,3,4,5,6,7,8,9,1,0;  there are k=11 digits between [1] and the second "1" of [11] which are 2,3,4,5,6,7,8,9,1,0,1; there are k=15 digits between [1] and the "1" of [21] which are 2,3,4,5,6,7,8,9,1,0,1,1,2,0,2.
There is k=1 digit between the "1" of [10] and the first "1" of [11] which is 0; there are k=2 digits between the "1" of [10] and the second "1" of [11] which are 0 and 1; there are k=6 digits between the "1" of [10] and the "1" of [21] which are 0,1,1,2,0,2.
There are k=0 digits between the first "1" of [11] and the second "1" of [11]; there are k=4 digits between the first "1" of [11] and the "1" of [21] which are 1,2,0,2.
There are k=3 digits between the second "1" of [11] and the "1" of [21] which are 2,0 and 2.

Prog

(Python)
from itertools import count, islice
def newdiffs(s, alen, locs1, diffs1):
    out = set(alen+i-j for i, c in enumerate(s, 1) if c == '1' for j in locs1)
    out |= set(i-j for j in range(len(s)-1) for i in range(j+1, len(s)) if '1' == s[i] == s[j])
    return out
def agen(): # generator of terms
    an, locs1, diffs1, mink, aset, alen = 0, [], set(), 1, {0}, 1
    while True:
        yield an
        an = next(k for k in count(mink) if k not in aset and ('1' not in (s:=str(k)) or not newdiffs(s, alen, locs1, diffs1)&diffs1))
        stran = str(an)
        if '1' in stran:
            diffs1 |= newdiffs(stran, alen, locs1, diffs1)
            locs1.extend([alen+i for i, c in enumerate(stran, 1) if c == '1'])
        alen += len(stran)
        aset.add(an)
        while mink in aset: mink += 1
print(list(islice(agen(), 100))) # Michael S. Branicky, Oct 01 2025

Crossrefs

Cf. A273880, A273881, A273882, A273883, A273884, A273885, A273886, A273887, A273888.

Sequence in context: A339018 A303948 A275413 * A247758 A247757 A247754

Adjacent sequences: A273373 A273374 A273375 * A273377 A273378 A273379

Keyword

nonn,base

Author

Eric Angelini and Jean-Marc Falcoz, May 30 2016

Status

approved