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A269347
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With a(1) = 1, a(n) is the sum of all 0 < m < n for which a(m) divides n.
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8
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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
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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a = {1}; Do[AppendTo[a, Total@ Select[Range[n - 1], Divisible[n, a[[#]]] &]], {n, 2, 66}]; a (* Michael De Vlieger, Mar 24 2016 *)
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PROG
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(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
(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]
(Python)
from itertools import count, islice
from sympy import divisors
def A269347_gen(): # generator of terms
yield 1
for n in count(2):
yield (s:=sum(A268347_dict.get(d, 0) for d in divisors(n, generator=True)))
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CROSSREFS
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Cf. A088167 which gives the number of m < n for which a(m) divides n.
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KEYWORD
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easy,nonn,nice
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AUTHOR
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STATUS
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approved
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