A380690 a(0) = 0; a(n) = the number of times a(n-1) has all digits in common with a previous term.

0, 0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, 0, 8, 0, 9, 0, 10, 0, 11, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 12, 0, 12, 1, 13, 0, 13, 1, 14, 0, 14, 1, 15, 0, 15, 1, 16, 0, 16, 1, 17, 0, 17, 1, 18, 0, 18, 1, 19, 0, 19, 1, 20, 0, 20, 1, 21

Offset: 0

Author

Sergio Pimentel, Jan 30 2025

Keywords

• nonn
• base

Comments

Every number should appear a limited number of times in the sequence as opposed as in A326834.

Examples

			a(24) = 2 since a(23) = 1 and previously there are two numbers that only have the digit '1': a(22) = 11 and a(2) = 1.
a(4062) = 113 since a(4061) = 112 and previously there are 113 occurrences of numbers that only have the digits '1' and '2' such as 12,21,112,121,122.
		

Crossrefs

Cf. A309261, A326834, A364788.

Programs

Mathematica

a[0] = 0; a[n_] := a[n] = Count[Array[a, n - 1, 0], ?(Union[IntegerDigits[a[n - 1]]] == Union[IntegerDigits[#]] &)]; Array[a, 100, 0] (* _Amiram Eldar, Jan 30 2025 *)

Python

from collections import Counter
from itertools import count, islice
def agen(): # generator of terms
    an, digsetcount = 0, Counter()
    while True:
        yield an
        key = "".join(sorted(set(str(an))))
        an = digsetcount[key]
        digsetcount[key] += 1
print(list(islice(agen(), 80))) # Michael S. Branicky, Jan 30 2025