A380745 a(0) = 0; a(n) = the number of times a(n-1) has the same digits in common with a previous term, in any permutation.

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

Offset: 0

Author

Sergio Pimentel, Jan 31 2025

Keywords

• nonn
• base

Comments

To find a(n), let L be the multiset of the digits of a(n-1). Then a(n) is the number of terms a(i), 0 <= i <= n-2, which also have L as the multiset of its digits. - N. J. A. Sloane, Mar 27 2025

Examples

			a(43) = 1 since a(42) = 21 and previously there is only one number in the sequence that contains both a 1 and a 2.
a(104) = 3 since a(103) = 11 and previously there are 3 numbers in the sequence that contain two 1's.
a(9942) = 14 since a(9941) = 155 and previously there are 14 numbers in the sequence that contain one 1 and two 5's.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A309261, A326834, A364788, A380690.

Programs

Mathematica

a[0] = 0; a[n_] := a[n] = Count[Array[a, n - 1, 0], ?(Sort[IntegerDigits[a[n - 1]]] == Sort[IntegerDigits[#]] &)]; Array[a, 100, 0] (* _Amiram Eldar, Jan 31 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(str(an)))
        an = digsetcount[key]
        digsetcount[key] += 1
print(list(islice(agen(), 82))) # Michael S. Branicky, Mar 24 2025