0
$\begingroup$
step[n_] := 
  step[n] = Plus @@ Delete[Divisors[n], Length[Divisors[n]]];

maxLength = 0;
For[n = 220, n < 223, n++,
 chain = NestWhileList[step, n, # > 1 && # < 1000000 &];
 chain = Delete[chain, Length[chain]];
 If[Length[chain] > maxLength,
  maxLength = Length[chain];
  bestN = Min[chain];,
  Nothing[]
  ]
 ]

does not work, I guess because n=220 indeed produces an infinite List (am1c4ble cha1n) which ist precisely what I am looking for. But how do I test for that, so how do I add && # is not part of chain already?

Or am I assuming what the problem is incorrectly? Working test for example with n=221, n<223 terminating and having the correct chain

$\endgroup$
5
  • 3
    $\begingroup$ Have you considered using the fourth argument of NestWhileList, All? This would supply you with the information you need to check if # has appeared previously. $\endgroup$
    – C. E.
    Jul 15, 2016 at 15:03
  • $\begingroup$ No, missed that, will try. $\endgroup$
    – Erik Itter
    Jul 15, 2016 at 15:06
  • $\begingroup$ obviously, just did not mention it in order to not index it in search engines - also I beblieve the p0j3ct 3u13r is to sensitive about that. Who wants to cheat ... have at it... $\endgroup$
    – Erik Itter
    Jul 15, 2016 at 19:31
  • $\begingroup$ no, do not get how to use it with All NestWhileList[step, 222, # > 1 && # < 1000000 &, All] already seems to be an infinite loop $\endgroup$
    – Erik Itter
    Jul 15, 2016 at 19:38
  • 1
    $\begingroup$ A single piece of my solution from six years ago: NestWhileList[ds, n, Signature@{1, n, ##2} =!= 0 &, All] where ds is similar to your step. I used Signature to check for duplicates. One could also make use of UnsameQ or the new-in-10 DuplicateFreeQ with different performance profiles. (6745) I'll leave the rest for you to figure out. $\endgroup$
    – Mr.Wizard
    Jul 16, 2016 at 2:39

1 Answer 1

2
$\begingroup$

C.E.:

using the fourth argument of NestWhileList, All

which solves the example with Mr.Wizard:

NestWhileList[step, n, Signature@{1, n, ##2} =!= 0 &, All]
$\endgroup$
3
  • $\begingroup$ if this should still not be appropriate how do I need to proceed to mark the question as answered? $\endgroup$
    – Erik Itter
    Jul 16, 2016 at 15:29
  • $\begingroup$ you deleted my answere, that I put there to mark it as answered refering to the answers provided as comments. $\endgroup$
    – Erik Itter
    Jul 16, 2016 at 15:31
  • $\begingroup$ As I said, your previous post was not an answer in any way. This one is much better. (In any event, see this meta thread.) $\endgroup$ Jul 16, 2016 at 15:32

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