A249070 -id:A249070 - 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

A246359 Maximum digit in the factorial base expansion of n (A007623).

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4

Offset: 0

Antti Karttunen, Oct 20 2014

Maximum entry in n-th row of A108731.

			Factorial base representation of 46 is "1320" as 46 = 1*4! + 3*3! + 2*2! + 0*1!, and the largest of these digits is 3, thus a(46) = 3.

Antti Karttunen, Table of n, a(n) for n = 0..10080
Index entries for sequences related to factorial base representation

Cf. A007623, A034968, A051903, A060130, A084558, A099563, A257684, A257687, A276076.

Cf. also A249070.

nn = 96; m = 1; While[Factorial@ m < nn, m++]; m; Table[Max@ IntegerDigits[n, MixedRadix[Reverse@ Range[2, m]]], {n, 0, nn}] (* Version 10.2, or *)
f[n_] := Block[{a = {{0, n}}}, Do[AppendTo[a, {First@ #, Last@ #} &@ QuotientRemainder[a[[-1, -1]], Times @@ Range[# - i]]], {i, 0, #}] &@ NestWhile[# + 1 &, 0, Times @@ Range[# + 1] <= n &]; Most@ Rest[a][[All, 1]] /. {} -> {0}]; Table[Max@ f@ n, {n, 0, 96}] (* Michael De Vlieger, Aug 29 2016 *)

def a007623(n, p=2): return n if nIndranil Ghosh, Jun 21 2017

From Antti Karttunen, Aug 29 2016: (Start)
a(0) = 0; for n >= 1, a(n) = 1 + a(A257684(n)).
a(0) = 0; for n >= 1, a(n) = max(A099563(n), a(A257687(n))).
a(n) = A051903(A276076(n)).
(End)

A249069 a(n+1) gives the number of occurrences of the first digit of a(n) in factorial base (i.e., A099563(a(n))) so far amongst the factorial base representations of all the terms up to and including a(n), with a(0)=0.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 1, 7, 9, 12, 2, 13, 3, 16, 5, 6, 18, 1, 19, 2, 21, 3, 25, 27, 30, 32, 35, 38, 40, 41, 43, 45, 48, 13, 14, 15, 16, 18, 6, 53, 20, 7, 57, 21, 8, 64, 24, 65, 27, 69, 28, 72, 10, 73, 11, 76, 12, 33, 80, 13, 34, 85, 14, 37, 89, 15, 41, 94, 17, 46, 96, 1, 97, 2, 99, 3, 103, 4, 48, 49, 50

Offset: 0

Antti Karttunen, Oct 20 2014

			   a(0) =  0 (by definition)
   a(1) =  1 ('1' in factorial base), as 0 has occurred once in all the preceding terms.
   a(2) =  1 as 1 has occurred once in all the preceding terms.
   a(3) =  2 ('10' in factorial base), as digit '1' has occurred two times in total in all the preceding terms.
   a(4) =  3 ('11' in factorial base), as '1' occurs once in each a(1) and a(2) and a(3).
   a(5) =  5 ('21' in factorial base), as '1' occurs once in each of a(1), a(2) and a(3) and twice at a(4).
   a(6) =  1 as '2' so far occurs only once at a(5)
   a(7) =  7 = '101'
   a(8) =  9 = '111'
   a(9) = 12 = '200'
  a(10) =  2 = '2'
  a(11) = 13 = '201'
  a(12) =  3 = '11'
  a(12) =  3 = '11'
  a(13) = 16 = '220'
  a(14) =  5 = '21'
  a(15) =  6 = '100'
  a(16) = 18 = '300'
  a(17) =  1 = '1'
  a(18) = 19 = '301'
  a(19) =  2 = '10'
  a(20) = 21 = '311'
  a(21) =  3 = '11'
  a(22) = 25 = '1001'
  a(23) = 27 = '1011'
  a(24) = 30 = '1100'
  a(25) = 32 = '1110'
  a(26) = 35 = '1121'
  a(27) = 38 = '1210' as the leftmost digit '1' has occurred 38 times in total in the factorial base expansions of the preceding terms a(0) - a(26).
etc.

Antti Karttunen, Table of n, a(n) for n = 0..10080

Cf. A249009 (analogous sequence in base-10).
Differs from a variant A249070 for the first time at n=27, where a(27) = 38, while A249070(27) = 7.
Cf. also A007623, A099563, A246359.

A249148 a(1) = 1, after which, if a(n-1) = 1, a(n) = 1 + the total number of 1's that have occurred in the sequence so far, otherwise a(n) = the total number of times the least prime dividing a(n-1) [i.e., A020639(a(n-1))] occurs as a divisor (counted with multiplicity for each term) in the previous terms from a(1) up to and including a(n-1).

Original entry on oeis.org

1, 2, 1, 3, 1, 4, 3, 2, 4, 6, 7, 1, 5, 1, 6, 8, 11, 1, 7, 2, 12, 14, 15, 6, 16, 20, 22, 23, 1, 8, 26, 27, 10, 28, 30, 31, 1, 9, 13, 2, 32, 37, 1, 10, 38, 39, 14, 40, 43, 1, 11, 3, 15, 16, 47, 1, 12, 49, 7, 8, 52, 54, 55, 9, 22, 56, 59, 1, 13, 5, 10, 60, 62, 63, 25, 14, 64, 70, 71, 1, 14, 72, 75, 28, 77, 15, 29, 1, 15, 30, 78, 79, 1, 16, 83

Offset: 1

Antti Karttunen, Oct 24 2014

nonn

Inspired by A248034.

After a(1), it is very likely that 1's occur only just after primes, although they do not necessarily occur after every prime. For example, 13 is the first prime whose initial occurrence is not followed by 1.

			a(1) = 1 by definition.
For n = 2, we see that a(1) = 1, which is the only 1 that has occurred in the sequence so far, and thus a(2) = 1+1 = 2.
For n = 3, we see that a(2) = 2, with the least prime dividing it being 2, which has occurred so far only once (namely in a(2)), thus a(3) = 1.
For n = 4, we see that a(3) = 1, and there has occurred two 1's so far (as a(1) and a(3)), thus a(4) = 2+1 = 3.
For n = 5, we see that a(4) = 3, with the least prime dividing it being 3, which has occurred now just once, thus a(5) = 1.
For n = 6, we see that a(5) = 1, and there has occurred three 1's so far (as a(1), a(3) and a(5)), thus a(6) = 3+1 = 4.
For n = 7, we see that a(6) = 4 = 2*2, with its least prime 2 dividing it two times, and also occurring once at a(2), thus a(7) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A020639, A049084, A055396, A061395, A249147, A248034, A249144, A249069, A249070.

A049084(n) = if(isprime(n), primepi(n), 0); \\ This function from Charles R Greathouse IV
A249148_write_bfile(up_to_n) = { my(pfcounts, n, a_n, f, k); pfcounts = vector(up_to_n); a_n = 1; for(n = 1, up_to_n, if((1 == a_n), pfcounts[1]++; a_n = pfcounts[1], f=factor(a_n); for(i=1,#f~,k = A049084(f[i,1])+1; pfcounts[k] += f[i,2]); a_n = pfcounts[A049084(f[1,1])+1]); write("b249148.txt", n, " ", a_n)); };
A249148_write_bfile(10000);
(MIT/GNU Scheme) ;; With memoizing definec-macro from Antti Karttunen's IntSeq-library and factor function from Aubrey Jaffer's SLIB-library.
(definec (A249148 n) (if (= 1 n) 1 (vector-ref (A249148aux_primefactor_counts (- n 1)) (A055396 (A249148 (- n 1))))))
(definec (A249148aux_primefactor_counts n) (cond ((= 1 n) (vector 2)) (else (let* ((a_n (A249148 n)) (copy-of-prevec (vector-copy (A249148aux_primefactor_counts (- n 1)))) (newsize (max (vector-length copy-of-prevec) (+ 1 (A061395 a_n)))) (pf_counts_vec (vector-grow copy-of-prevec newsize))) (let loop ((pf_indices (map A049084 (factor a_n)))) (cond ((null? pf_indices) pf_counts_vec) (else (vector-set! pf_counts_vec (car pf_indices) (+ 1 (or (vector-ref pf_counts_vec (car pf_indices)) 0))) (loop (cdr pf_indices)))))))))

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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A246359 Maximum digit in the factorial base expansion of n (A007623).

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A249069 a(n+1) gives the number of occurrences of the first digit of a(n) in factorial base (i.e., A099563(a(n))) so far amongst the factorial base representations of all the terms up to and including a(n), with a(0)=0.

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