A380690 - OEIS

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

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OFFSET

0,5

COMMENTS

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

LINKS

Table of n, a(n) for n=0..79.

EXAMPLE

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.

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 *)

PROG

(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