A342383 -id:A342383 - cp's OEIS frontend

A336285 a(0) = 0; for n > 0, a(n) is the least positive integer not occurring earlier such that the digits in a(n-1)+a(n) are all distinct.

0, 1, 2, 3, 4, 5, 7, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 29, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 40, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 53, 50, 52, 51, 54, 55, 65, 58, 62, 61, 59, 64, 56, 67, 57, 63, 60, 66, 68, 69, 70, 72, 71, 74, 73, 75, 77

Offset: 0

Rémy Sigrist, Jul 22 2020

In other words, for any n > 0, a(n) + a(n+1) belongs to A010784.

The sequence is finite since there are only a finite number of positive integers with distinct digits, see A010784, although the exact number of terms is currently unknown.

			The first terms, alongside a(n) + a(n+1), are:
  n   a(n)  a(n)+a(n+1)
  --  ----  -----------
   0     0            1
   1     1            3
   2     2            5
   3     3            7
   4     4            9
   5     5           12
   6     7           13
   7     6           14
   8     8           17
   9     9           19
  10    10           21

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Image of the first 1000000 terms. The green line is a(n) = n.

Cf. A342383, A338466, A322845, A010784, A043537, A043096, A276633, A002378.

s=0; v=1; for (n=1, 67, print1 (v", "); s+=2^v; for (w=1, oo, if (!bittest(s, w) && #(d=digits(v+w))==#Set(d), v=w; break)))

def agen():
  alst, aset, min_unused = [0], {0}, 1
  yield 0
  while True:
    an = min_unused
    while True:
      while an in aset: an += 1
      t = str(alst[-1] + an)
      if len(t) == len(set(t)):
        alst.append(an); aset.add(an); yield an
        if an == min_unused: min_unused = min(set(range(max(aset)+2))-aset)
        break
      an += 1
g = agen()
print([next(g) for n in range(77)]) # Michael S. Branicky, Mar 11 2021

a(0)=0 added by N. J. A. Sloane, Mar 14 2021

A368181 a(1) = 1; for n > 1, a(n) is the smallest positive integer that has not yet appeared which shares no digit with the sum of all previous terms a(1)..a(n-1).

Original entry on oeis.org

1, 2, 4, 3, 5, 6, 7, 9, 8, 10, 11, 12, 13, 20, 22, 24, 23, 25, 14, 30, 15, 17, 33, 26, 16, 18, 19, 21, 27, 28, 31, 29, 34, 40, 41, 32, 35, 36, 38, 39, 37, 42, 44, 50, 43, 52, 45, 46, 47, 48, 49, 54, 55, 57, 56, 60, 53, 58, 59, 62, 63, 65, 70, 51, 61, 64, 66, 67, 69, 68, 71, 73, 74, 81, 90, 91, 77

Offset: 1

Scott R. Shannon, Dec 21 2023

The sequence is finite; after 14594 terms, where a(14594) = 20858, the sum of all terms is 173658294 which contains the digits 1..9, so the next term does not exist.

The largest term is a(12742) = 888888.

			a(14) = 20 as the sum of all terms a(1)..a(13) = 91, and 20 is the smallest unused number that does not contain the digits 1 or 9.

Scott R. Shannon, Table of n, a(n) for n = 1..14594

Cf. A362093, A362075, A342383, A342382.

from itertools import islice
def agen():
  s, aset, mink = 0, {0}, 1
  while True:
      k, dset = mink, set(str(s))
      if dset >= set("123456789"): break
      while k in aset or set(str(k)) & dset: k += 1
      an = k; aset.add(an); s += an; yield an
      while mink in aset: mink += 1
print(list(islice(agen(), 80))) # Michael S. Branicky, Dec 21 2023

A368347 a(1) = 1; for n > 1, a(n) is the smallest positive integer that has not yet appeared which contains all the distinct digits of the sum of all previous terms a(1)..a(n-1).

Original entry on oeis.org

1, 10, 11, 2, 24, 48, 69, 156, 123, 4, 84, 235, 67, 348, 128, 103, 134, 1457, 304, 308, 136, 2357, 1069, 178, 3567, 10239, 126, 182, 10247, 137, 13458, 12345, 567, 2458, 2068, 20567, 1378, 45689, 10348, 102347, 203479, 4568, 12456, 234568, 105689, 3089, 20689, 12678, 204589, 1048, 1023459

Offset: 1

Scott R. Shannon, Dec 22 2023

The sequence is infinite, although it is unknown if all positive numbers eventually appear. In the first 50000 terms the smallest number not to have appeared is 3. In the same range the largest value is a(49134) = 1023548967, with the sum of all previous terms at that point being 553402987165.

			a(3) = 11 as the sum of the first two terms is 1 + 10 = 11, which contains the distinct digit 1, and 11 is the smallest unused number to contain 1.
a(4) = 2 as the sum of the first three terms is 1 + 10 + 11 = 22, which contains the distinct digit 2, and 2 is the smallest unused number to contain 2.
a(5) = 24 as the sum of the first four terms is 1 + 10 + 11 + 2 = 24, which contains the distinct digits 2 and 4, and 24 is the smallest unused number to contain 2 and 4.

Scott R. Shannon, Table of n, a(n) for n = 1..10000

Cf. A368181, A362093, A362075, A342383, A342382.

cp's OEIS Frontend

A336285 a(0) = 0; for n > 0, a(n) is the least positive integer not occurring earlier such that the digits in a(n-1)+a(n) are all distinct.

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A368181 a(1) = 1; for n > 1, a(n) is the smallest positive integer that has not yet appeared which shares no digit with the sum of all previous terms a(1)..a(n-1).

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A368347 a(1) = 1; for n > 1, a(n) is the smallest positive integer that has not yet appeared which contains all the distinct digits of the sum of all previous terms a(1)..a(n-1).

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