# [seqfan] Re: corollary to the Collatz conjecture?

Bob Selcoe rselcoe at entouchonline.net
Mon Jul 31 19:38:08 CEST 2017

```Hi Gottfried & Seqfans,

Thanks.  Here's a link to the Crandall paper.  http://www.ams.org/journals/mcom/1978-32-144/S0025-5718-1978-0480321-3/S0025-5718-1978-0480321-3.pdf.  It doesn't appear to address this issue in particular, but it may be buried somewhere in the text.

Anyway, it turns out my conjecture is equivalent to the *standard* 3n+1 Collatz conjecture. The reasoning is straightforward:

Dealing only with odd terms N in the sequences, let d be the number of times 3 divides N. It takes k-d steps for a successor term to be a multiple of 3^k (obviously if d > k, the number of steps is 0).  Let 1 <= j < k-d. The sequences cannot loop prior to k-d steps, since 3^(j-k+d) is a non-integer.  Once a term reaches a multiple of 3^k, the sequences behave like a standard Collatz sequence "scaled up" 3^k times, so the conjectures are equivalent.

(Thanks to Jack Brennan for pointing this out and to Jean-Paul Allouche for verifying my analysis).

Cheers,
Bob

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From: "Gottfried Helms" <helms at uni-kassel.de>
Sent: Saturday, July 29, 2017 3:42 AM
To: <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: corollary to the Collatz conjecture?

> Am 28.07.2017 um 17:47 schrieb Bob Selcoe:
>>
>> Hi Seqfans,
>>
>> I'm wondering if there is a corollary to the Collatz conjecture for k >= 0, that 3x + 3^k eventually terminate with a 4-term loop [3^k => 4*3^k => 2*3^k => 3^k].
>>
>> So 3x+1 is simply k=0 with terminal loop [1,4,2,1];  k=1 is 3x+3 with loop [3,12,6,3];  k=2 is 3x+9 with loop [9,36,18,9]; etc.
>>
>> There do not appear to be any OEIS entries pertaining to k > 0.
>>