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A044050 a(n) = "length" of the aliquot sequence for n.

%I A044050 #24 Feb 16 2025 08:32:38
%S A044050 1,2,2,3,2,1,2,3,4,4,2,7,2,5,5,6,2,4,2,7,3,6,2,5,1,7,3,1,2,15,2,3,6,8,
%T A044050 3,4,2,7,3,4,2,14,2,5,7,8,2,6,4,3,4,9,2,13,3,5,3,4,2,11,2,9,3,4,3,12,
%U A044050 2,5,4,6,2,9,2,5,5,5,3,11,2,7,5,6,2,6,3,9,7,7,2,10,4,6,4,4,2,9,2,3,4,5,2,18
%N A044050 a(n) = "length" of the aliquot sequence for n.
%C A044050 The aliquot sequence for n is the trajectory of n under repeated application of the map A001065 = x -> sigma(x) - x.
%C A044050 The trajectory will either have a transient part followed by a cyclic part, or have an infinite transient part and never cycle.
%C A044050 Sequence gives (length of transient part of trajectory) + (length of cycle if the trajectory did not reach 0). In other words, here we consider that the trajectory ends if we reach 1.
%C A044050 Given that A001065(n) is the sum of the divisors of n which are less than n, we have that the aliquot length A(n) = r-1 where r is the smallest integer such that A001065^r(n) = A001065^s(n) for some s<r. If this never happens (i.e., if r is infinite) then we set a(n) = 0. [Edited by _M. F. Hasler_, Nov 16 2013]
%C A044050 In the interval [1,1000] it is not known if the aliquot length is 0 for the numbers 276, 552, 564, 660 and 966.
%C A044050 The function sigma = A000203 (and thus A001065 = sigma - id) is defined only on the positive integers and not for 0, so the trajectory ends when 0 is reached. - _M. F. Hasler_, Nov 16 2013
%H A044050 T. D. Noe, <a href="/A044050/b044050.txt">Table of n, a(n) for n=1..275</a>
%H A044050 Wolfgang Creyaufmüller, <a href="http://www.aliquot.de/aliquote.htm">Prime Families</a>
%H A044050 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/AliquotSequence.html">Aliquot Sequence</a>
%H A044050 P. Zimmermann, <a href="http://www.loria.fr/~zimmerma/records/aliquot.html">Aliquot Sequences</a>
%e A044050 a(12) = 7:
%e A044050 12 is divisible by 1,2,3,4 and 6 so sigma(12)=16;
%e A044050 16 is divisible by 1,2,4 and 8 so sigma(16)=15;
%e A044050 15 is divisible by 1,3 and 5 so sigma(15)=9;
%e A044050 9 is divisible by 1 and 3 so sigma(9)=4;
%e A044050 4 is divisible by 1 and 2 so sigma(4)=3;
%e A044050 3 is divisible only by 1 so sigma(3)=1;
%e A044050 1 is not divisible by anything less than 1 so sigma(1)=0.
%e A044050 The aliquot sequence is therefore 16, 15, 9, 4, 3, 1, 0 which is 7 elements long. Therefore a(12) = 7.
%t A044050 f[n_]:=Plus@@Divisors[n]-n;lst2={};Do[lst={};a=k;Do[b=a;a=f[a];AppendTo[lst,a];If[a==0||a==b,Break[]],{n,7!}];AppendTo[lst2,Length[lst]],{k,5!}];lst2 (* _Vladimir Joseph Stephan Orlovsky_, Apr 24 2010 *)
%Y A044050 See A098007, A003023 for other versions. See A008886 for the aliquot sequence of 42.
%K A044050 nonn,nice
%O A044050 1,2
%A A044050 Aza Raskin (aza(AT)uchicago.edu), Jun 25 2003