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A255242
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Calculate the aliquot parts of a number n and take their sum. Then repeat the process calculating the aliquot parts of all the previous aliquot parts and add their sum to the previous one. Repeat the process until the sum to be added is zero. Sequence lists these sums.
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4
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0, 1, 1, 4, 1, 8, 1, 12, 5, 10, 1, 30, 1, 12, 11, 32, 1, 36, 1, 38, 13, 16, 1, 92, 7, 18, 19, 46, 1, 74, 1, 80, 17, 22, 15, 140, 1, 24, 19, 116, 1, 90, 1, 62, 51, 28, 1, 256, 9, 62, 23, 70, 1, 136, 19, 140, 25, 34, 1, 286, 1, 36, 61, 192, 21, 122, 1, 86, 29, 114
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OFFSET
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1,4
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COMMENTS
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a(n) = 1 if n is prime.
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LINKS
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FORMULA
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a(1) = 0.
a(2^k) = k*2^(k-1) = A001787(k), for k>=1.
a(n^k) = (n^k-2^k))/(n-2), for n odd prime and k>=1.
In particular:
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EXAMPLE
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The aliquot parts of 8 are 1, 2, 4 and their sum is 7.
Now, let us calculate the aliquot parts of 1, 2 and 4:
1 => 0; 2 => 1; 4 => 1, 2. Their sum is 0 + 1 + 1 + 2 = 4.
Let us calculate the aliquot parts of 1, 1, 2:
1 => 0; 1 = > 0; 2 => 1. Their sum is 1.
We have left 1: 1 => 0.
Finally, 7 + 4 + 1 = 12. Therefore a(8) = 12.
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MAPLE
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with(numtheory): P:=proc(q) local a, b, c, k, n, t, v;
for n from 1 to q do b:=0; a:=sort([op(divisors(n))]); t:=nops(a)-1;
while add(a[k], k=1..t)>0 do b:=b+add(a[k], k=1..t); v:=[];
for k from 2 to t do c:=sort([op(divisors(a[k]))]); v:=[op(v), op(c[1..nops(c)-1])]; od;
a:=v; t:=nops(a); od; print(b); od; end: P(10^3);
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MATHEMATICA
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f[s_] := Flatten[Most[Divisors[#]] & /@ s]; a[n_] := Total@Flatten[FixedPointList[ f, {n}]] - n; Array[a, 100] (* Amiram Eldar, Apr 06 2019 *)
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PROG
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(PARI) ali(n) = setminus(divisors(n), Set(n));
a(n) = my(list = List(), v = [n]); while (#v, my(w = []); for (i=1, #v, my(s=ali(v[i])); for (j=1, #s, w = concat(w, s[j]); listput(list, s[j])); ); v = w; ); vecsum(Vec(list)); \\ Michel Marcus, Jul 15 2023
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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