[seqfan] Re: corollary to the Collatz conjecture?

[Bob Selcoe]

Hi Gottfried & Seqfans,

Yes, I think we are saying essentially the same thing.  I hope it was clear 
that "d = number of times 3 divides N" means d is the largest power of 3 
which divides N.

I simply eliminated even terms (i.e., used reduced Collatz sequences) in my 
analysis because the number of halving steps doesn't effect the "scaling up" 
by 3^k.  In other words, expanding on what I wrote previously:

Let N = n_0 be the first term in a reduced Collatz sequence, and let n_i {i 
= 1..z} be the i-th successor term.  Again, let d be the number of times 3 
divides n_0, and n_i -> 3 *n_i + 3^k when n_i is odd.  So:

n_0 = 3^d*a
n_1 = 3^(d+1)*b
n_2 = 3^(d+2)*c...
n_y = 3^k*m

where a ,b, c, ...m are not multiples of 3 when d <= k.  (Of course, n_0 = 
3^k*m when d > k, and m may be a multiple of 3).

Once n_y is reached, we can find n_z using the standard Collatz function 
starting with n_y/3^k = m_0 and m_i -> 3m + 1, then multiply each successor 
term by 3^k (i.e., "scaling up" 3^k).  So for example, m=7, k=2:

reduced Collatz:   7,  11,  17,   13,   5,   1, 1, 1,...
"scaled" Collatz:        99, 153, 117, 45, 9, 9, 9,...

So if the Collatz conjecture is true (or false), then its 3 + 3^k corollary 
necessarily is true (or false).

Cheers,
Bob

--------------------------------------------------
From: "Gottfried Helms" <helms at uni-kassel.de>
Sent: Tuesday, August 01, 2017 6:49 AM
To: <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: corollary to the Collatz conjecture?

> Am 31.07.2017 um 19:38 schrieb Bob Selcoe:
>> 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
>>
>
> let M=3^m with some m >=0
>
> then one step of transformation on the odd numbers (a->b) is
>
>        b = (a*3+M) / 2^A
>
> If we look at the question of cycles, then we should have
>
>          a = (a*3+M) / 2^A
>      a 2^A = 3a + M
>      a (2^A -3) =  M
>
> and of course, if we have a solution for the problem where M=1, so
>
>      a_0 (2^A_0 -3) = 1
>
> then of course we have the same set of solutions for the case M=3^m
> with a_0 * M instead of a_0
>
>      a_0*M (2^A_0 -3) = 1*M
>
> because the parenthese has no common part with M=3^m .
>
>   --------------
>
> Considering two instead of one transformation we have
>
>        b = (a*3+M) / 2^A
>        c = (b*3+M) / 2^B
>
>  and looking at cycles we must have c=a such that we can expand to
>
>        a = ((3a+M)/2^A*3+M) / 2^B
>          = ((3a+M)*3+M*2^A) / 2^(A+B)
>          = (3^2a +M*3+M*2^A) / 2^S     \\ I write S for A+B+... for longer 
> cycles
>    a(2^S - 3^2) = M( 3 + 2^A)
>
> and again, if we have a cycle for M=1 at a_0 then we have accordingly
> a cycle for M=3^m at a_0*3^M
>
> -----------------------------------------------------
>
> I think it is obvious how this generalizes, and I think that answers
> that question so that a full proof can be written easily.
>
>
> Gottfried
>
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