A088167 -id:A088167 - cp's OEIS frontend

A124056 a(1)=1. a(n) = number of terms from among the first (n-1) terms of the sequence which divide a(n-1).

1, 1, 2, 3, 3, 4, 4, 5, 3, 5, 4, 6, 7, 3, 6, 9, 7, 4, 7, 5, 5, 6, 10, 8, 8, 9, 8, 10, 9, 9, 10, 10, 11, 3, 7, 6, 12, 17, 3, 8, 11, 4, 8, 13, 3, 9, 14, 8, 14, 9, 15, 14, 10, 12, 21, 14, 11, 5, 7, 7, 8, 15, 16, 16, 17, 4, 9, 16, 19, 3, 10, 14, 14, 15, 18, 23, 3, 11, 6, 17, 5, 8, 17, 6, 18, 27, 19

Offset: 1

Leroy Quet, Nov 03 2006

nonn

First occurrence of k: 1, 3, 4, 6, 8, 12, 13, 24, 16, 23, 33, 37, 44, 47, 51, 63, 38, 75, 69, 101, 55, 91, 76, 132, 102, ..., . - Robert G. Wilson v, Nov 05 2006

a(n+1) = number of preceding terms that are divisors of a(n); a(1) = 1. - Reinhard Zumkeller, May 23 2013

			a(12) is 6. a(1)=1, a(2)=1, a(3)=2, a(4)=3, a(5)=3, a(9)=3 and a(12)=6 are the seven terms that divide 6. So a(13)= 7.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A027750.

Cf. A088167 (counting divisors of n instead of those of a(n)).

import Data.List (isInfixOf)
a124056 n = a124056_list !! (n-1)
a124056_list = 1 : f [1] where
   f xs@(x:_) = y : f (y : xs) where
     y = length $ filter (flip elem $ a027750_row x) xs
-- Reinhard Zumkeller, May 23 2013

f[s_] := Append[s, Count[Mod[s[[ -1]], s], 0]]; Nest[f, {1, 1}, 86] (* Robert G. Wilson v *)
a[1]= 1; L[1]= {1}; a[n_]:=a[n]= Sum[If[Mod[a[n - 1], L[n - 1][[i]]]==0, 1, 0], {i,1,n-1}]; L[n_]:=L[n]= Table[a[i], {i, 1, n}]; L[87] (* Joel B. Lewis, Nov 05 2006 *)

More terms from Robert G. Wilson v and Joel B. Lewis, Nov 05 2006

A269347 With a(1) = 1, a(n) is the sum of all 0 < m < n for which a(m) divides n.

Original entry on oeis.org

1, 1, 3, 3, 3, 15, 3, 3, 30, 3, 3, 51, 3, 3, 84, 3, 3, 111, 3, 3, 150, 3, 3, 195, 3, 3, 246, 3, 3, 318, 3, 3, 366, 3, 3, 435, 3, 3, 510, 3, 3, 591, 3, 3, 684, 3, 3, 771, 3, 3, 882, 3, 3, 975, 3, 3, 1086, 3, 3, 1218, 3, 3, 1326, 3, 3, 1455

Offset: 1

Alec Jones, Feb 24 2016

For n > 2, I can prove that a(n) = 3 if 3 does not divide n, and in general, 3 divides a(n).
The base case is a(3) = 3. Suppose that the results hold for a(n) over 3 < n < k; we will show that the results hold for a(k) also. In the case that 3 does not divide k, then a(k) = 3, since a(1) and a(2) divide k but no other previous term can. This proves the first claim.
Otherwise, if 3 does divide k, then a(m) divides k for each 0 < m < k not divisible by 3; these numbers can be divided into k/3 pairs so that the sum of each pair is congruent to 0 modulo 3 (for instance, 1 + 2 == 4 + 5 == 7 + 8 == ... == 0 (mod 3)). If a(m) divides k for some 0 < m < k divisible by 3, this m does not change the congruence class of the sum that forms a(k). Thus, a(k) == 0 (mod 3) as required to prove the second claim.

			a(1) = 1;
a(2) = 1 because a(1) divides 2;
a(3) = 3 because a(1) and a(2) divide 3: 1+2=3;
a(4) = 3 because a(1) and a(2) divide 4: 1+2=3;
a(5) = 3 because a(1) and a(2) divide 5: 1+2=3;
a(6) = 15 because a(1), a(2), a(3), a(4), and a(5) divide 6: 1+2+3+4+5=15.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..1000 from Alec Jones)

Cf. A088167 which gives the number of m < n for which a(m) divides n.

Cf. A271326, A271328.

a269347 1 = 1
a269347 n = genericIndex a269347_list (n - 1)
a269347_list = map a [1..] where
  a n = sum $ filter ((==) 0 . mod n . a269347) [1..n-1]
-- Peter Kagey, Jun 17 2016

int[] terms = new int[1000];
terms[0] = 1;
for (int i = 1; i < 1000; i++) {
     int count = 0;
     for (int j = 0; j < i; j++) {
          if ((i + 1) % terms[j] == 0) {
               count = count + (j + 1);
          }
     }
     terms[i] = count;
}

a = {1}; Do[AppendTo[a, Total@ Select[Range[n - 1], Divisible[n, a[[#]]] &]], {n, 2, 66}]; a (* Michael De Vlieger, Mar 24 2016 *)

lista(nn) = {va = vector(nn); va[1] = 1; for (n=2, nn, va[n] = sum(k=1, n-1, k*((n % va[k])==0));); va;} \\ Michel Marcus, Feb 24 2016

from itertools import count, islice
from sympy import divisors
def A269347_gen(): # generator of terms
    A268347_dict = {1:1}
    yield 1
    for n in count(2):
        yield (s:=sum(A268347_dict.get(d,0) for d in divisors(n,generator=True)))
        A268347_dict[s] = A268347_dict.get(s,0) + n
A269347_list = list(islice(A269347_gen(),40)) # Chai Wah Wu, Nov 17 2022

def a(n)
  seq = [1]
  (2..Float::INFINITY).each do |i|
    return seq.last[0...n].last if seq.length > n
    indices = seq.each_index.select { |j| i % seq[j] == 0 }
    seq << indices.map(&:next).reduce(:+)
  end
end # Peter Kagey, Feb 25 2016

A351886 a(n) is the number of k < n such that a(k) AND n = 0 (where AND denotes the bitwise AND operator).

Original entry on oeis.org

0, 1, 2, 1, 4, 2, 3, 1, 8, 4, 6, 3, 8, 3, 4, 1, 16, 9, 11, 6, 14, 5, 8, 4, 17, 9, 10, 5, 10, 3, 5, 1, 32, 16, 21, 10, 24, 12, 15, 7, 26, 11, 17, 7, 16, 6, 11, 4, 39, 19, 20, 10, 24, 10, 11, 4, 26, 12, 15, 7, 12, 3, 6, 1, 64, 34, 34, 20, 41, 21, 21, 10, 45, 21

Offset: 0

Rémy Sigrist, Feb 23 2022

The definition is recursive: a(n) depends on prior terms (a(0), ..., a(n-1)); a(0) = 0 corresponds to an empty sum.

			The first terms, alongside the corresponding k's, are:
  n   a(n)  k's
  --  ----  -------------------------
   0     0  {}
   1     1  {0}
   2     2  {0, 1}
   3     1  {0}
   4     4  {0, 1, 2, 3}
   5     2  {0, 2}
   6     3  {0, 1, 3}
   7     1  {0}
   8     8  {0, 1, 2, 3, 4, 5, 6, 7}
   9     4  {0, 2, 4, 5}
  10     6  {0, 1, 3, 4, 7, 9}
  11     3  {0, 4, 9}
  12     8  {0, 1, 2, 3, 5, 6, 7, 11}

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Cf. A080100, A088167, A350677, A351887.

a:= proc(n) option remember; add(
     `if`(Bits[And](n, a(j))=0, 1, 0), j=0..n-1)
    end:
seq(a(n), n=0..80);  # Alois P. Heinz, Feb 28 2022

for (n=1, #a=vector(75), print1 (a[n]=sum(k=1, n-1, bitand(a[k], n-1)==0)", "))

a = []
[a.append(sum(a[k] & n == 0 for k in range(n))) for n in range(74)]
print(a) # Michael S. Branicky, Feb 24 2022

a(2^k) = 2^k.

A351887 a(n) is the number of k < n such that a(k) AND n = a(k) (where AND denotes the bitwise AND operator).

Original entry on oeis.org

0, 1, 1, 3, 1, 4, 2, 7, 1, 5, 2, 8, 3, 8, 6, 15, 1, 6, 3, 11, 2, 8, 7, 18, 4, 9, 8, 19, 7, 14, 14, 31, 1, 7, 4, 13, 4, 12, 10, 24, 5, 12, 9, 21, 11, 22, 19, 40, 1, 8, 5, 17, 5, 18, 13, 35, 8, 19, 15, 34, 15, 32, 28, 63, 1, 9, 4, 15, 6, 18, 12, 31, 7, 18, 11

Offset: 0

Rémy Sigrist, Feb 23 2022

The definition is recursive: a(n) depends on prior terms (a(0), ..., a(n-1)); a(0) = 0 corresponds to an empty sum.

			The first terms, alongside the corresponding k's, are:
  n   a(n)  k's
  --  ----  ------------------------
   0     0  {}
   1     1  {0}
   2     1  {0}
   3     3  {0, 1, 2}
   4     1  {0}
   5     4  {0, 1, 2, 4}
   6     2  {0, 5}
   7     7  {0, 1, 2, 3, 4, 5, 6}
   8     1  {0}
   9     5  {0, 1, 2, 4, 8}
  10     2  {0, 6}
  11     8  {0, 1, 2, 3, 4, 6, 8, 10}
  12     3  {0, 5, 11}

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Cf. A088167, A350802, A351886.

a:= proc(n) option remember; add(
     `if`(Bits[And](n, a(j))=a(j), 1, 0), j=0..n-1)
    end:
seq(a(n), n=0..80);  # Alois P. Heinz, Feb 28 2022

for (n=1, #a=vector(75), print1 (a[n]=sum(k=1, n-1, bitand(a[k], n-1)==a[k])", "))

a = []
[a.append(sum(a[k] & n == a[k] for k in range(n))) for n in range(75)]
print(a) # Michael S. Branicky, Feb 24 2022

A271774 a(1) = 1, then a(n) is the maximum of all 0 < m < n for which a(m) divides n.

Original entry on oeis.org

1, 1, 2, 3, 2, 5, 2, 7, 4, 7, 2, 11, 2, 13, 6, 13, 2, 17, 2, 19, 10, 19, 2, 23, 6, 23, 4, 27, 2, 29, 2, 31, 12, 31, 10, 33, 2, 37, 16, 37, 2, 41, 2, 43, 6, 43, 2, 47, 10, 49, 18, 47, 2, 53, 12, 53, 22, 53, 2, 59, 2, 61, 10, 61, 16, 61, 2, 67, 26, 67, 2, 71, 2

Offset: 1

Peter Kagey, Apr 14 2016

nonn

If n is an odd prime, then a(n) = 2 and a(n+1) = n. All n for which a(n) = 2 are odd primes. - Robert Israel, Apr 14 2016

			a(1) = 1 by definition.
a(2) = 1 because a(1) divides 2.
a(3) = 2 because a(2) divides 3.
a(4) = 3 because a(3) divides 4.
a(5) = 2 because a(2) divides 5.
a(6) = 5 because a(5) divides 6.
a(7) = 2 because a(2) divides 7.
a(8) = 7 because a(7) divides 8.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A088167, A269347, A271503, A271504, A271773.

A:= proc(n) option remember; local m;
    for m from n-1 by -1 do
      if n mod A(m) = 0 then return m fi
    od
end proc:
A(1):= 1:
seq(A(i),i=1..100); # Robert Israel, Apr 14 2016

a[1] = 1; a[n_] := a[n] = Block[{m = n - 1}, While[Mod[n, a[m]] > 0, m--]; m]; Array[a, 100] (* Giovanni Resta, Apr 14 2016 *)

A102128 a(1) = 1; a(n) = sum of previous terms which divide n.

Original entry on oeis.org

1, 1, 2, 4, 2, 6, 2, 12, 2, 10, 2, 34, 2, 14, 2, 20, 2, 24, 2, 54, 2, 22, 2, 70, 2, 26, 2, 46, 2, 46, 2, 36, 2, 68, 2, 94, 2, 38, 2, 74, 2, 62, 2, 70, 2, 138, 2, 94, 2, 60, 2, 82, 2, 114, 2, 74, 2, 58, 2, 172, 2, 124, 2, 68, 2, 94, 2, 242, 2, 234, 2, 154, 2, 222, 2, 118, 2, 110, 2, 114, 2

Offset: 1

Leroy Quet, Feb 14 2005

nonn

			Among the first 7 terms, the terms which divide 8 are 1, 1, 2, 4, 2 and 2.
So a(8) = 1 + 1 + 2 + 4 + 2 + 2 = 12.

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A088167.

Nest[Function[{a, n}, Append[a, Total@ Select[a, Mod[n, #] == 0 &]]] @@ {#, Length@ # + 1} &, {1}, 80] (* Michael De Vlieger, Nov 13 2018 *)

up_to = 20000;
A102128list(up_to) = { my(v=vector(up_to)); v[1] = 1; for(n=2,up_to,v[n] = sum(j=1,n-1,v[j]*!(n%v[j]))); (v); };
v102128 = A102128list(up_to);
A102128(n) = v102128[n]; \\ Antti Karttunen, Nov 10 2018

a(1) = 1; a(n) = [x^n] Sum_{k=1..n-1} a(k)*x^a(k)/(1 - x^a(k)). - Ilya Gutkovskiy, Dec 11 2017

More terms from John W. Layman, Mar 16 2005

A115751 a(1)=1. a(n) = number of positive divisors of n which are not among the first (n-1) terms of the sequence.

Original entry on oeis.org

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

Offset: 1

Leroy Quet, Mar 28 2006

nonn

There are only 40 distinct values among the first 10000 terms. The records occur at positions: 1, 4, 12, 30, 48, 72, 120, 180, 240, 360, 480, 720, 840, 1260, 1680, 2160, 2520, 4620, 5040, ... - Antti Karttunen, Oct 21 2017

			The divisors of 12 are 1, 2, 3, 4, 6 and 12. Of these, only the four divisors 3, 4, 6 and 12 do not occur among the first 11 terms of the sequence. So a(12) = 4.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A088167.

with(numtheory): a[1]:=1: for n from 2 to 120 do div:=divisors(n): M:=convert([seq(a[j],j=1..n-1)],set): a[n]:=nops(div minus M): od: seq(a[n],n=1..120); # Emeric Deutsch, Apr 01 2006

;; We define a mutual recurrence with the memoization-macro definec:
(definec (A115751 n) (if (= 1 n) n (length (remove (lambda (d) (zero? (modulo (Aauxseq_forA115751 (- n 1)) (A000040 d)))) (divisors n)))))
;; The other member of the mutual recurrence has not been submitted. Its n-th term keeps track in its prime factorization what distinct values has so far occurred in A115751(1) .. A115751(n). That is, iff value k has occurred in range a(1) .. a(n), then the n-th term of this auxiliary sequence is divisible by the k-th prime:
(definec (Aauxseq_forA115751 n) (if (= 1 n) 2 (lcm (A000040 (A115751 n)) (Aauxseq_forA115751 (- n 1)))))
(define (divisors n) (cons 1 (proper-divisors n)))
(define (proper-divisors n) (let loop ((k n) (divs (list))) (cond ((= 1 k) divs) ((zero? (modulo n k)) (loop (- k 1) (cons k divs))) (else (loop (- k 1) divs)))))
;; Antti Karttunen, Oct 21 2017

More terms from Emeric Deutsch, Apr 01 2006

A297003 a(n) = 2^(n-1), n=1,2,3; for n >= 4, a(n) is the number of the previous terms dividing n.

Original entry on oeis.org

1, 2, 4, 3, 1, 4, 2, 6, 3, 4, 2, 11, 2, 6, 4, 10, 2, 11, 2, 13, 4, 10, 2, 18, 2, 11, 4, 16, 2, 17, 2, 19, 7, 13, 3, 24, 2, 14, 7, 21, 2, 23, 2, 24, 5, 16, 2, 31, 4, 19, 6, 25, 2, 24, 6, 27, 7, 17, 2, 35, 2, 20, 9, 28, 5, 29, 2, 29, 6, 29, 2, 41, 2, 22, 8, 31

Offset: 1

Vladimir Shevelev, Dec 23 2017

			1-3) a(1)=1, a(2)=2 a(3)=4 by the definition;
4) Let n=4. From the previous terms {1,2,4} everyone divides 4, so a(4)=3;
5) Let n=5. From the previous terms {1,2,4,3} only 1 divides 5. So a(5)=1;
6) Let n=6. From the previous terms {1,2,4,3,1} exactly four divide 6. So a(6)=4; etc.

Peter J. C. Moses, Table of n, a(n) for n = 1..10000

Cf. A088167.

first[n_] := Fold[Append[#1, Count[#1, k_ /; Divisible[#2, k]]] &,
  2^Range[0, Min[n - 1, 2]], Range[4, n]] (* Michael De Vlieger, Dec 23 2017 *)

first(n) = my(res = vector(n), c = 0); for(x = 1, min(n, 3), res[x] = 1<<(x-1)); for(x=4, n, for(k=1, x-1, if(x%res[k]==0, c++)); res[x] = c; c = 0); res \\ Iain Fox, Dec 23 2017

def A297003_list(leng):
    L = [1, 2, 4]
    if leng < 4: return L[0:leng]
    for n in (4..leng) :
        count = 0
        for l in L: count += int(l.divides(n))
        L.append(count)
    return L
print(A297003_list(76)) # Peter Luschny, Dec 24 2017

a(p) = 2, where p is prime, other than 3 and 5.

More terms from Peter J. C. Moses, Dec 23 2017

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A088168 Smallest number m such that A088167(m)=n.

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A124056 a(1)=1. a(n) = number of terms from among the first (n-1) terms of the sequence which divide a(n-1).

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A269347 With a(1) = 1, a(n) is the sum of all 0 < m < n for which a(m) divides n.

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A351886 a(n) is the number of k < n such that a(k) AND n = 0 (where AND denotes the bitwise AND operator).

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A351887 a(n) is the number of k < n such that a(k) AND n = a(k) (where AND denotes the bitwise AND operator).

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A271774 a(1) = 1, then a(n) is the maximum of all 0 < m < n for which a(m) divides n.

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A102128 a(1) = 1; a(n) = sum of previous terms which divide n.

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