Collatz conjecture but with $n^2-1$ instead of $3n+1.$ Does the sequence starting with $13$ go to infinity?

Question

Let's consider the following variant of Collatz $(3n+1) : $

If $n$ is odd then $n \to n^2-1.$

$1\to 0.$

$3\to 8\to 1\to 0.$

$5\to 24\to 3\to 0.$

$7\to 48\to 3\to 0.$

$9\to 80\to 5\to 0.$

$11\to 120\to 15\to 224\to 7\to 0.$

$\color{red}{13\to 168\to 21\to 440\to 55\to 3024\to 189\to 35720\to 4465\to 19936224\to 623007\to\ldots\ ?}\ $

Does the sequence starting with $13$ go off to infinity? If yes, what is a proof? If no, is there a starting number whose sequence does go off to infinity, and how do we prove either that such a number must exist, or even better that a specific starting number goes off to infinity?

Here is some Python code I ran, which suggests that the numbers in the sequence starting at $13$ quickly become large:

n=13
num_loops=0
print(n)
while n!=0:
    if n%2==0:
        n//=2
    else: n=n**2-1
    print('\n', n)
    num_loops+=1
    if num_loops==70:
        print("too many loops")
        break

Answer 1

The set of solutions can be written as table, segmented by e'th powers of $2$ :

  t |     e=0                  |       e=1                  |       e=2                  |    e=3                      |    e=4                       |   ...
----+--------------------------+----------------------------+----------------------------+-----------------------------+------------------------------+---- 
    |         ...              |           ...              |           ...              |           ...               |            ...               |
  -4|  56*2^2  =  (-15)^2-1    |   105*2^3  =  (-29)^2-1    |   203*2^4  =  (-57)^2-1    |   399*2^5  =  (-113)^2-1    |    791*2^6  =  (-225)^2-1    | 
  -3|  30*2^2  =  (-11)^2-1    |    55*2^3  =  (-21)^2-1    |   105*2^4  =  (-41)^2-1    |   205*2^5  =   (-81)^2-1    |    405*2^6  =  (-161)^2-1    | 
  -2|  12*2^2  =   (-7)^2-1    |    21*2^3  =  (-13)^2-1    |    39*2^4  =  (-25)^2-1    |    75*2^5  =   (-49)^2-1    |    147*2^6  =   (-97)^2-1    | 
    |
  -1|   2*2^2  =   (-3)^2-1    |     3*2^3  =   (-5)^2-1    |     5*2^4  =   (-9)^2-1    |     9*2^5  =   (-17)^2-1    |     17*2^6  =   (-33)^2-1    | 
   0|   0*2^2  =    (1)^2-1    |     1*2^3  =    (3)^2-1    |     3*2^4  =    (7)^2-1    |     7*2^5  =    (15)^2-1    |     15*2^6  =    (31)^2-1    | 
    |
   1|   6*2^2  =    (5)^2-1    |    15*2^3  =   (11)^2-1    |    33*2^4  =   (23)^2-1    |    69*2^5  =    (47)^2-1    |    141*2^6  =    (95)^2-1    | 
   2|  20*2^2  =    (9)^2-1    |    45*2^3  =   (19)^2-1    |    95*2^4  =   (39)^2-1    |   195*2^5  =    (79)^2-1    |    395*2^6  =   (159)^2-1    | 
   3|  42*2^2  =   (13)^2-1    |    91*2^3  =   (27)^2-1    |   189*2^4  =   (55)^2-1    |   385*2^5  =   (111)^2-1    |    777*2^6  =   (223)^2-1    | 
   4|  72*2^2  =   (17)^2-1    |   153*2^3  =   (35)^2-1    |   315*2^4  =   (71)^2-1    |   639*2^5  =   (143)^2-1    |   1287*2^6  =   (287)^2-1    | 
    |         ...              |           ...              |           ...              |           ...               |            ...               |
----+--------------------------+----------------------------+----------------------------+-----------------------------+------------------------------+---- 

meaning always $$n_2 \cdot 4\cdot 2^e = n_1 ^2 -1 \tag 1 $$

Here $n_2=f_e(t)$ can be written as a quadratic and $n_1=g_e(t)$ as a linear polynomial in $t \in \mathbb Z$. The shown data would have rowindices $t$ in the interval $t=-4..+4$.

We can construct the polynomials in $t$ from the values in a column with the rowindex taken as $t$ in the interval $-4...4$ and can then write $$ \begin{array} {} f_e(t) &= 4 \cdot 2^e \cdot t \cdot (t+1)-2 \cdot t+2^e-1 \\ g_e(t) &= 4 \cdot 2^e \cdot t+2 \cdot 2^e-1 \end{array} \tag 2$$

A better memorizable version is this:
$$ \begin{array} {} f_e(t) = 2^e \cdot (2 t +1)^2-(2 t+1) \\ g_e(t) = 2^e \cdot 2 \cdot (2 \cdot t+1 )-1 \end{array} \tag 3$$ and replacing the odd values $2t+1$ by $u \in \mathbb Z $ \ $ 2 \mathbb Z $ gives the even shorter: $$ \begin{array} {} f_e(u) = u \cdot (2^e \cdot u-1) \\ g_e(u) = 2 \cdot (2^e \cdot u-1)+1 \end{array} \tag 4$$

To prove equality according to (1) it suffices to expand and insert this into the equation (1): $$ 4 \cdot 2^e \cdot f_e(u) = g_e(u)^2 - 1 \tag 5$$ We find the equality in a few steps.

---

Relation to Sil's and EricSnyder's comments and @WillJagy's analysis: in the table the $4$'th and $5$'th rows contain the cases, where $n_2 \lt \mid n_1 \mid $, which has been problematized by Will Jagy's answer; his notation $(2^k,x,y)$ there is my notation $(4 \cdot 2^e,n_1,n_2)$ here.

Update: This idea seems to allow to proceed to a proof, that there are only finitely many other orbits towards zero than that of $n_1=2^k \pm 1$ with $k>2$ (or $4 \cdot 2^e$ with $e>0$), and that there are no cycles and that thus all other orbits diverge infinitely.

The critical two types of $n_1$ occur regularly in row $t=-1$ and $t=0$, so we need only find $n_1$ where the related $n_2$ has the form $2^k \pm 1$ in the other rows $t<-1 $ or $t>0$, which would iterate then to the center rows and thus to zero.

(a): Searching the rectangle $(t,e)=(-800...800,1..31)$ we find only three solutions: $$ (t,e,n_1,n_2) \in \{ (1,2,23,33), (-23,1,-181,4095), (1,1,11,15) \} \tag 6$$ so $n_1=11$, $n_1=23$ and $n_1=181$ iterate towards zero.

(b): Now it could happen, that the $n_1$ values $(23),(181),(11)$ occur as $n_2$ values elsewhere, with three other values $n_1$ and are thus themselves iterates.
But this does not happen, and fortunately we need only a finite searchspace to prove this: using the ordering of the $n_2$ values in a column and along the rows.

So if we could prove, that indeed the three cases (a) are the only ones, we had solved the problem in the OP. It seems to me, that an "A. Baker"-style proof might be possible to limit the search-space to a finite rectangle, but I've not so much expertise to do that on my own, at least at the moment. But there are results about the distance of squares and powers of $2$ available, so maybe we need only refer to one of that results.
See for instance this recent question in MSE and especially Mike Bennet's answer.

Update2: Proof for (eq.6)
The case is indeed solvable; there is an article of Laszlo Szalai, 2001 who deals with a slightly generalized form of the question, from which by

• Theorem1 (Szalay) "if the positive integers n,m, and x with $n \gt m$ satisfy $$2^{n-m} + 1 = {x^2-1\over 2^m} $$ then (...uninteresting cases omitted...) (iii) $(n,m,x) \in \{(5,4,7),(9,4,23) \}$." The case $x=23$ is exactly that cases where $x$ is not of the form $2^k - 1$ and yet $f_e(x)$ is of that form and thus $x$ can iterate towards zero.
• Theorem2 (Szalay) "if the positive integers n,m, and x with $n \gt m$ satisfy $$2^{n-m} - 1 = {x^2-1\over 2^m} $$ then (...uninteresting cases omitted...) (iii) $(n,m,x) \in \{(5,3,5),(7,3,11),(15,3,181) \}$." The cases $x=11$ and $x=181$ are exactly that cases where $x$ is not of the form $2^k + 1$ and yet $f_e(x)$ is of that form and thus both $x$ can iterate towards zero.

Thus: we have the proof, that

• the three cases found in (eq 6) $$ (t,e,n_1,n_2) \in \{ (1,2,23,33), (-23,1,-181,4095), (1,1,11,15) \} $$ are the only solutions not in the rows $4$ and $5$ ($t=-1$ or $t=0$) in the above table, which iterate towards zero.
• Since all other cases not in the rows $4$ and $5$ ($t=-1$ or $t=0$) increase by one iteration, all that other cases must diverge to infinity.

Szalay, László: The equations $2^N \pm 2^M \pm 2^L = z^2 $, Indag. Mathem. N.S., 131-142, 2002. See the downloadable links at the comments/answer of my earlier question in MO

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Update3: Curious fact on the iteration of $a_1=13$. The fractional part of the double-log to base 2 seems to converge to some limit. See the protocol of the first 30 iterations (odd steps only):

 k   log_2(log_2(a_k))
-----------------------
[1, 1.88769671439]
[2, 2.13498231840]
[3, 2.53140883896]
[4, 2.91881409881]
[5, 3.59984673966]
[6, 4.26670326276]
[7, 5.02226916018]
[8, 5.95409288384]
[9, 6.88253436364]
[10, 7.85787043544]
[11, 8.84851166124]
[12, 9.84223857933]
[13, 10.8390917800]
[14, 11.8379099584]
[15, 12.8371215392]
[16, 13.8367271679]
[17, 14.8365299418]
[18, 15.8364559751]
[19, 16.8364066619]
[20, 17.8363820046]
[21, 18.8363727580]
[22, 19.8363681347]
[23, 20.8363658230]
[24, 21.8363635114]
[25, 22.8363627408]
[26, 23.8363623555]
[27, 24.8363622111]
[28, 25.8363621147]
[29, 26.8363620786]
[30, 27.8363620606]

Couldn't proceed further due to digit-length of occuring integers $a_k$ with finite memory... Defining $$b=2^{2^{0.836362060564583494096400006341-3}} \approx 1.1673140482117283628568828285728 ...$$ then it seems, we get approximation $a_k \to b^{2^k}$ depending on increasing precision of $b$ in the

Answer 2

About Eric Snyder's comment: suppose $ x^2 - 1 = 2^k y$ with $x,y$ odd. We wish to investigate what happens when $y \leq x \; . \; \;$ Note that $\gcd(x+1, x-1) = 2$ because $x \equiv 1,3 \pmod 4.$ One of the $x \pm 1 $ is $\equiv 2 \pmod 4. $

Let us make the name $ \delta = \pm 1.$ Then we may demand $$ x + \delta \equiv 2 \pmod 4$$ This tells us that the integer $ \frac{x + \delta}{2} $ is odd. We also have $$ \frac{x + \delta}{2} \frac{x - \delta}{2} = 2^{k-2}y$$ By repeated division by $2$ it follows that $$ \frac{x - \delta}{2^{k-1}} = w $$ is an odd positive integer, with $$\frac{x + \delta}{2} \; \frac{x - \delta}{2^{k-1}} = y $$ IF WE ASSUME $$ w = \frac{x - \delta}{2^{k-1}} \geq 3, $$ we find $$ y = w \frac{x + \delta}{2} \geq 3 \frac{x + \delta}{2} \geq \frac{3x - 3}{2}$$ The assumption $ w \neq 1$ has led us to $y \geq \frac{3x - 3}{2}$ The hypothesis that $x \geq y$ now says $x \geq \frac{3x - 3}{2},$ or $ 2x \geq 3x - 3,$ or $3 \geq x$

If $ x \geq y$ and $ x \geq 5$ in $ x^2 - 1 = 2^k y,$ we find that $w=1$ in $ \frac{x - \delta}{2^{k-1}} = w .$ So that $ x - \delta = 2^{k-1} .$ or $$ x = \pm 1 + 2^{k-1}$$