Dtyjlui

Generalize the $3x + 1$ problem as $cx pm 1$, where $c$ is a positive odd integer and $x$ is a positive integer iterated through the function as far as possible to discover a cycle. If $x$ is even, then you halve it. But if $x$ is odd, you do either $cx + 1$ or $cx - 1$ as the case may be. (If you prefer, $c$ may be negative and you disallow $cx - 1$ for the odd branch; then $|-3x + 1|$ and $3x - 1$ are kind of the same).

With $3x - 1$ and $5x + 1$ it is somewhat well-known that many $x$ don't lead to 1, while with $3x + 1$ the question is unresolved despite intense scrutiny by many professionals and amateurs. For which other $cx pm 1$ is the question of ultimate arrival at 1 still undetermined despite study by more than a few people? I would appreciate journal articles that look at several different $cx pm 1$.

upps:[update] I've just recently given the same answer in this question, maybe I'd delete this or the other one

• All $c$ of the form $c=2^C-1= {3,7,15,31,63,....}$ have a "trivial cycle" at $1 to 1$.


• All $c$ of the form $c=2^C+1= {3,5,9,17,33,....}$ have a "trivial cycle" at $-1 to -1$.


• $c=3$ has some extra cycles in the negative numbers, I've found none other than that known for instance in wikipedia


• $c=5$ has some extra cycles in the positive numbers, I've found only the two well known ones ($17 to 43 to 27 to 17 $ and $13 to 33 to 83 to 13 $ )


• $c=181$ ( $=lceil 2^{15/2} rceil-1$ ) has two cycles $27to 611to 27$ and $35 to 99 to 35$

I've searched in two modes : all odd numbers up to 9999 for $c$ checking possible cycle-lengthes up to 100 ; possible cycle lengthes up to 1000 (I think, have it not at hand at the moment) and optimal $c$ using the convergents of continued fractions excluding even extremely high $c$ requiring elements of a possible cycle $a_k$ below some upper bound taken by some formula depending on $c$ and projected cyclelength.

I've even taken negative values for $c$ and found some more cycles on small numbers but have it not at hand at the moment (but see some recent MSE-answer of mine)

Here are some references to the problem $qx+1$, for example:

• R.Steiner. On the "$QX+1$ problem", $Q$ odd.. Fibonacci Quarterly, 19(3), (1981), 285-288


• R.Steiner. On the "$QX+1$ problem", $Q$ odd. II. Fibonacci Quarterly, 19(4), (1981), 293-296

Author shows that for $q=5$ there is only one non-trivial cycle ($13to 208 to 13$), while for $q=7$ there are no non-trivial cycles.

• R. E. Crandall. On the "$3x+1$" Problem. Mathematics of Computation, 32, (1978), 1281-1292.

Here Crandall conjectured that for all $q geq 5$, there always exist some $n$ that never iterate to $1$.

• Z. Franco and C. Pomerance. On a Conjecture of Crandall concerning the $px+1$ Problem. Mathematics of Computation, 64(211), (1995), 1333-1336.

Here they show that Crandall conjecture is true if $q$ is a Wieferich number.

For exhaustive references on the problem I recommend to check The Ultimate Challenge: The $3x+1$ Problem by Jeffrey C. Lagarias. Point of the book is to summarize all results and references to the problem, so it seems a good place to start (the references for $qx+1$ are taken from there).

Here is one:

Consider the following operation on an arbitrary positive integer:

If the number is divisible by 12, divide it by 12.

If the number is divisible by 10, divide it by 10.

If the number is divisible by 8, divide it by 8.

If the number is divisible by 6, divide it by 6.

If the number is divisible by 4, divide it by 4.

If the number is divisible by 2, divide it by 2.

If the number is odd, multiply it by 5 and add 1.

The tests should be performed in this order.

The conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially.

This can be tested by running this simple python script:

#!/usr/bin/python



for i in range(2,1000):

    x = i

    while x != 1:        

        print "{0}, ".format(x),                       

        if x % 12 == 0:

            x = x / 12

        elif x % 10 == 0:

            x = x / 10

        elif x % 8 == 0:

            x = x / 8

        elif x % 6 == 0:

            x = x / 6

        elif x % 4 == 0:

            x = x / 4

        elif x % 2 == 0:

            x = x / 2

        else: 

            x = 5 * x + 1

    print "n"