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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. 8
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

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.

LINKS

Alec Jones, Table of n, a(n) for n = 1..1000

EXAMPLE

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.

MATHEMATICA

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

PROG

(Java)

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;

}

(PARI) 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

(Ruby)

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

(Haskell)

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

CROSSREFS

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

Cf. A271326, A271328.

Sequence in context: A147823 A341211 A335518 * A183554 A229847 A196529

Adjacent sequences:  A269344 A269345 A269346 * A269348 A269349 A269350

KEYWORD

easy,nonn,nice

AUTHOR

Alec Jones, Feb 24 2016

STATUS

approved

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Last modified January 16 19:48 EST 2022. Contains 350376 sequences. (Running on oeis4.)