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.
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
Keywords
Examples
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.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Paolo P. Lava)
- Jon Maiga, Computer-generated formulas for A255242, Sequence Machine.
Crossrefs
Programs
-
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 *)
-
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
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).
From Antti Karttunen, Nov 22 2024: (Start)
a(n) = A330575(n) - n.
Also, following formulas were conjectured by Sequence Machine:
a(n) = (A191161(n)-n)/2.
a(n) = Sum_{d|n} A001065(d)*A074206(n/d). [Compare to David A. Corneth's Apr 13 2020 formula for A330575]
(End)
Comments