A326834 -id:A326834 - cp's OEIS frontend

A343102 a(n) is the sum of the number of times the digits in n (without repetition) have appeared in the sequence.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 1, 2, 2, 2, 2, 2, 2, 2, 2, 19, 11, 8, 8, 8, 8, 8, 8, 14, 9, 11, 8, 8, 0, 1, 0, 0, 0, 8, 2, 16, 11, 10, 1, 1, 1, 2, 1, 10, 3, 17, 19, 10, 1, 1, 0, 1, 1, 9, 4, 20, 26, 14, 3, 5, 3, 2, 3, 11, 6, 21, 30, 15, 6, 4, 3, 5, 1, 10, 5, 31, 42, 24, 16, 15, 14, 14, 10

Offset: 0

Scott R. Shannon, Apr 05 2021

The fixed points > 0 in the first one million terms are 10, 310, 341, 351, 514. It is likely no more exist.

			a(0) to a(9) = 0 as the digits 0 to 9 have not appeared in the sequence.
a(10) = 10 as 1 has not appeared while 0 has appeared ten times, thus a(10) = 0 + 10 = 10.
a(11) = 1 as the repetitions of 1 in 11 are ignored, and 1 has appeared once in the sequence.
a(12) = 2 as 1 has appeared twice while 2 has not appeared, thus a(12) = 2 + 0 = 2.
a(20) = 19 as 2 has appeared eight times while 0 has appeared eleven times, thus a(20) = 8 + 11 = 19.
a(22) = 8 as the repetitions of 2 in 22 are ignored, and 2 has appeared eight times in the sequence.

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

Cf. A343103 (count all digits in n), A326834, A004207, A309261, A331162.

Block[{a = {}, d = ConstantArray[0, 10]}, Do[AppendTo[a, Total@ Map[d[[If[# == 0, 10, #] ]] &, Union@ IntegerDigits[i]]]; Set[d, d + DigitCount[a[[i + 1]] ]], {i, 0, 87}]; a] (* Michael De Vlieger, Apr 05 2021 *)

A343103 a(n) is the sum of the number of times each digit in n (taken with repetition) has appeared in the sequence.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 2, 2, 1, 2, 2, 2, 2, 2, 2, 19, 11, 16, 8, 8, 8, 9, 8, 12, 11, 11, 11, 9, 0, 0, 0, 1, 0, 4, 3, 16, 16, 10, 2, 2, 1, 4, 2, 6, 5, 17, 20, 14, 2, 4, 2, 5, 3, 6, 5, 22, 25, 23, 8, 9, 9, 10, 6, 11, 11, 19, 27, 22, 5, 6, 7, 10, 6, 8, 9, 25, 33, 29, 11, 10, 12, 14, 9

Offset: 0

Scott R. Shannon, Apr 05 2021

The fixed points > 0 in the first one million terms are 10, 101, 194. It is likely no more exist.

			a(0) to a(9) = 0 as the digits 0 to 9 have not appeared in the sequence.
a(10) = 10 as 1 has appeared zero times and 0 has appeared ten times, thus a(10) = 0 + 10 = 10.
a(11) = 2 as 1 has appeared once in the sequence, and as 1 appears twice in 11, a(11) = 1 + 1 = 2.
a(12) = 2 as 1 has appeared twice and 2 has appeared zero times, thus a(12) = 2 + 0 = 2.
a(20) = 19 as 2 has appeared eight times and 0 has appeared eleven times, thus a(20) = 8 + 11 = 19.
a(22) = 16 as 2 has appeared eight times in the sequence, and as 2 appears twice in 22, a(22) = 8 + 8 = 16.

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

Cf. A343102 (count only unique digits in n), A326834, A004207, A309261, A331162.

Block[{a = {}, d = ConstantArray[0, 10]}, Do[AppendTo[a, Total@ Map[d[[If[# == 0, 10, #] ]] &, IntegerDigits[i]]]; Set[d, d + DigitCount[a[[i + 1]] ]], {i, 0, 87}]; a] (* Michael De Vlieger, Apr 05 2021 *)

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

Original entry on oeis.org

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

Sergio Pimentel, Jan 30 2025

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

			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.

Cf. A309261, A326834, A364788.

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 *)

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

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.

Original entry on oeis.org

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

Sergio Pimentel, Jan 31 2025

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

			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.

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

Cf. A309261, A326834, A364788, A380690.

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 *)

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

cp's OEIS Frontend

A343102 a(n) is the sum of the number of times the digits in n (without repetition) have appeared in the sequence.

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A343103 a(n) is the sum of the number of times each digit in n (taken with repetition) has appeared in the sequence.

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Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

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

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Mathematica

Python

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.

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Author

Keywords

Comments

Examples

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Crossrefs

Programs

Mathematica

Python