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A255242
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.
4
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
OFFSET
1,4
COMMENTS
a(n) = 1 if n is prime.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Paolo P. Lava)
FORMULA
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:
a(3^k) = A001047(k-1);
a(5^k) = A016127(k-1);
a(7^k) = A016130(k-1);
a(11^k) = A016135(k-1).
EXAMPLE
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.
MAPLE
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);
MATHEMATICA
f[s_] := Flatten[Most[Divisors[#]] & /@ s]; a[n_] := Total@Flatten[FixedPointList[ f, {n}]] - n; Array[a, 100] (* Amiram Eldar, Apr 06 2019 *)
PROG
(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
KEYWORD
nonn,changed
AUTHOR
Paolo P. Lava, Feb 19 2015
STATUS
approved