A249144 -id:A249144 - cp's OEIS frontend

A249143 Length of the rightmost run in the binary representation of A241944: a(n) = A136480(A249144(n)).

1, 1, 2, 1, 1, 3, 1, 3, 1, 2, 2, 2, 1, 1, 2, 3, 2, 3, 1, 1, 1, 1, 5, 1, 5, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 6, 1, 6, 1, 2, 1, 1, 1, 2, 5, 2, 1, 4, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 1, 2, 4, 1, 1, 2, 1, 3, 3, 3, 1, 2, 1, 1, 7, 1, 7, 1, 2, 1, 1, 1, 2, 1, 4, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 6, 2, 2, 4, 2, 1, 1, 1, 4, 2, 1, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 3, 3, 3, 1, 1, 2, 2, 1

Offset: 1

Antti Karttunen, Oct 24 2014

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

Cf. A136480, A249144, A249145.

(define (A249143 n) (A136480 (A249144 n)))

a(n) = A136480(A249144(n)).

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.

Original entry on oeis.org

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

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)))))))))

A249146 a(0) = 0, after which a(n) gives the total number of runs of the same length as the maximal run in the binary representation of a(n-1) [i.e., A043276(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

Original entry on oeis.org

0, 1, 2, 4, 1, 6, 2, 9, 3, 4, 5, 15, 1, 16, 2, 19, 7, 1, 21, 26, 8, 2, 32, 1, 34, 3, 9, 10, 43, 11, 12, 14, 4, 15, 3, 16, 4, 17, 5, 58, 6, 18, 19, 21, 71, 8, 9, 22, 23, 10, 84, 24, 11, 26, 27, 29, 12, 31, 2, 99, 13, 34, 14, 15, 5, 108, 37, 38, 40, 16, 6, 41, 42, 130, 3, 43, 44, 46, 17, 18, 47, 7, 19, 49

Offset: 0

Antti Karttunen, Oct 22 2014

nonn

			a(0) = 0 (by definition), and 0 is also '0' in binary we consider it to contain a single run of length one.
For n = 1, we see that in a(0) there is one run of length 1, which is total number of runs of length 1 so far in terms a(0) .. a(n-1), thus a(1) = 1.
For n = 2, we see that the only and thus also the longest run of a(1) = 1 ('1' also in binary) has occurred two times in total (once in a(0) and a(1)), thus a(2) = 2.
For n = 3, we see that there are two runs in a(2) = 2 ('10' in binary), both one bit long, and so far there has occurred four such runs in total (namely once in a(0) and a(1), twice in a(2)), thus a(3) = 4.
For n = 4, we see that the longest run of a(3) = 4 ('100' in binary) is two bits long, and it is so far the first and only two-bit run in the sequence, thus a(4) = 1.
For n = 5, we see that the longest run of a(4) = 1 ('1' in binary) is one bit long, and so far there has occurred 6 such one-bit runs in terms a(0) .. a(4), thus a(5) = 6.
For n = 6, we see that the longest run of a(5) = 6 ('110' in binary) is two bits long, and so far there has occurred 2 such two bit runs (once in terms a(3) and a(5)), thus a(6) = 2.

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A043276, A249145, A249144, A248034.

cp's OEIS Frontend

A249143 Length of the rightmost run in the binary representation of A241944: a(n) = A136480(A249144(n)).

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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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A249146 a(0) = 0, after which a(n) gives the total number of runs of the same length as the maximal run in the binary representation of a(n-1) [i.e., A043276(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

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