A364788 -id:A364788 - cp's OEIS frontend

A248034 a(n+1) gives the number of occurrences of the last digit of a(n) so far, up to and including a(n), with a(0)=0.

0, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 2, 10, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 3, 10, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 4, 10, 5, 5, 6, 5, 7, 5, 8, 5, 9, 5, 10, 6, 6, 7, 6, 8, 6, 9, 6, 10, 7, 7, 8, 7, 9, 7, 10, 8, 8, 9, 8, 10, 9, 9, 10

Offset: 0

Eric Angelini and M. F. Hasler, Oct 11 2014

In other words, the number to the right of a comma gives the number of occurrences of the digit immediately to the left of the comma, counting from the beginning up to that digit or comma.

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Eric Angelini, Digit-counters updating themselves, SeqFan list, Oct 11 2014.
Zak Seidov, Graph of 10^6 terms

Cf. A249068 (analogous sequence in base 8).
Cf. A249009 (analogous sequence which uses the first, not the last digit).
Cf. A249069, A249070, A249144, A249148, A364788.

a:= proc(n) option remember; `if`(n=0, 0,
      coeff(b(n-1), x, irem(a(n-1), 10)))
    end:
b:= proc(n) option remember; `if`(n=0, 1, b(n-1)+
      add(x^i, i=convert(a(n), base, 10)))
    end:
seq(a(n), n=0..120);  # Alois P. Heinz, Oct 18 2014

nn = 120; a[0] = j = 0; c[] := 0; Do[Map[c[#]++ &, IntegerDigits[j]]; a[n] = j = c[Mod[j, 10]], {n, nn}]; Array[a, nn, 0] (* _Michael De Vlieger, Aug 07 2023 *)

c=vector(10);print1(a=0);for(n=1,99,apply(d->c[d+1]++,if(a,digits(a)));print1(","a=c[1+a%10]))
(MIT/GNU Scheme)
;; An implementation of memoization-macro definec can be found for example from: http://oeis.org/wiki/Memoization
(definec (A248034 n) (if (zero? n) n (vector-ref (A248034aux_digit_counts (- n 1)) (modulo (A248034 (- n 1)) 10))))
(definec (A248034aux_digit_counts n) (cond ((zero? n) (vector 1 0 0 0 0 0 0 0 0 0)) (else (let loop ((digcounts-for-n (vector-copy (A248034aux_digit_counts (- n 1)))) (n (A248034 n))) (cond ((zero? n) digcounts-for-n) (else (vector-set! digcounts-for-n (modulo n 10) (+ 1 (vector-ref digcounts-for-n (modulo n 10)))) (loop digcounts-for-n (floor->exact (/ n 10)))))))))
;; Antti Karttunen, Oct 22 2014

from itertools import islice
def A248034_gen(): # generator of terms
    c, clist = 0, [1]+[0]*9
    while True:
        yield c
        c = clist[c%10]
        for d in str(c):
            clist[int(d)] += 1
A248034_list = list(islice(A248034_gen(),30)) # Chai Wah Wu, Dec 13 2022

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

A248034 a(n+1) gives the number of occurrences of the last digit of a(n) so far, up to and including a(n), with a(0)=0.

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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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Author

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Programs

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python