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

0,5

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

Links

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

Example

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.

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 *)

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(str(an)))
        an = digsetcount[key]
        digsetcount[key] += 1
print(list(islice(agen(), 82))) # Michael S. Branicky, Mar 24 2025

Crossrefs

Cf. A309261, A326834, A364788, A380690.

Sequence in context: A257770 A027656 A142150 * A380690 A276457 A171181

Adjacent sequences: A380742 A380743 A380744 * A380746 A380747 A380748

Keyword

nonn,base

Author

Sergio Pimentel, Jan 31 2025

Status

approved