%I #9 Jul 11 2021 09:49:01
%S 1,2,3,4,5,6,7,8,9,10,11,16,13,14,15,16,17,21,19,22,21,22,23,55,25,26,
%T 27,28,29,259,31,32,33,34,35,55,37,38,39,50,41,259,43,50,45,46,47,76,
%U 49,50,51,52,53,259,55,64,57,58,59,172,61,62,63,64,65,259
%N Maximum peak of aliquot sequence starting at n.
%C According to Catalan's conjecture all aliquot sequences end in a prime followed by 1, a perfect number, a friendly pair or an aliquot cycle. Some sequences seem to be open ended and keep growing forever i.e. 276. Most sequences only go down (i.e. 10 - 8 - 7 - 1), so for most cases in this sequence, a(n) = n. The first number to achieve a significantly high peak is 138
%H Jinyuan Wang, <a href="/A115060/b115060.txt">Table of n, a(n) for n = 1..275</a>
%H W. Creyaufmueller, <a href="http://www.aliquot.de/aliquote.htm">Aliquot Sequences</a>.
%H Paul Zimmerman, <a href="http://www.loria.fr/~zimmerma/records/aliquot.html">Aliquot Sequences</a>.
%e a(24)=55 because the aliquot sequence starting at 24 is: 24 - 36 - 55 - 17 - 1 so the maximum peak of this sequence is 55.
%o (Python)
%o from sympy import divisor_sigma as sigma
%o def aliquot(n):
%o alst = []; seen = set(); i = n
%o while i and i not in seen: alst.append(i); seen.add(i); i = sigma(i) - i
%o return alst
%o def aupton(terms): return [max(aliquot(n)) for n in range(1, terms+1)]
%o print(aupton(66)) # _Michael S. Branicky_, Jul 11 2021
%Y Cf. A003023, A005114, A007906, A037020, A044050, A063769, A098007, A098008, A098009, A098010.
%K nonn
%O 1,2
%A _Sergio Pimentel_, Mar 06 2006
%E More terms from _Jinyuan Wang_, Jul 11 2021