Consider a modified version of Collatz sequence:

$C(n)=\left\{ \begin{array}{ll} \frac{3n+1}{2} & n\ \mathrm{odd} \\ \frac{n}{2}& n\ \mathrm{even}\end{array} \right.$

Let $F_n$ be the smallest integer that satisfies $C^{F_n}(n)=1$, where $C^0(n)=n$, $C^1(n)=C(n)$, and $C^k(n) = C(C^{k-1}(n))$ for any integer $k>1$.

Now for any integer $n>0$ with a finite $F_n$ (can be infinite if Collatz conjecture is disproved) consider polynomial

$T_n(z)=n z^{F_n} + C(n) z^{F_n-1} + C^2(n) z^{F_n-2} + \cdots + C^{F_n-1} z + 1 = \sum_{k=0}^{F_n} C^k(n) z^{F_n-k}$

with roots $r^{(n)}_1, r^{(n)}_2, \ldots, r^{(n)}_{F_n}$.

Question: Can one prove that for any $n\neq9$, all roots are contained in an open disk of radius 1.5 around $(0,0)$, i.e.

$\forall 1\leq k \leq F_n, |r^{(n)}_k|<1.5$.

Note that $T_9(z)$ has a root at $r \approx -1.522093720599496 $. I have checked this for all $n<300000$, which obviously is not a large number, but it's striking to see how close some of the roots can get to the boundary of the disk but none pass it.

  • 1
    $\begingroup$ If you start at the other end, defining $T^*_n(z)=n+C(n)z+C^2(n)z^2+\cdots$, then you get a polynomial with roots the reciprocals of those of $T_n(z)$ when Collatz holds, and you get a power series should Collatz ever fail. $\endgroup$ Nov 21, 2012 at 23:24
  • $\begingroup$ Note that if you set out to find an $n$ yielding a large negative root, you'd want to make the coefficients of odd powers of $x$ small and those of even powers of $x$ large. In @Julian's answer, he finds that 393217 yields a large negative root. And if you follow that number's Collatz sequence, you are alternating between halving and the other operation (roughly multiplying by 1.5) which has the effect of making the even coefficients large and the odd ones small. $\endgroup$
    – 2'5 9'2
    Nov 21, 2012 at 23:25
  • $\begingroup$ @alex, it looks to me like Julian is reporting a large positive, not negative, root. $\endgroup$ Nov 21, 2012 at 23:27
  • $\begingroup$ @Gerry How is that possible with a polynomial whose coefficients are all positive? $\endgroup$
    – 2'5 9'2
    Nov 21, 2012 at 23:28
  • $\begingroup$ @alex, of course, you are right. Maybe Julian is just reporting the modulus of the root. $\endgroup$ Nov 21, 2012 at 23:38

1 Answer 1


$T_{393217}(z)$ has a root at $$z\approx-1.50000024999390139088569862099867130326048600135879415656217$$

I have done the computations up to $n=500000$. The only values I found such that the maximum of the modulus of the roots of $T_n$ is greater that $1.5$ are $n=9$ and $n=393217$. I also found (and you probably know already) that the maximum of the modulus of the roots of $T_n$ is strictly greater that $.5$ except for powers of $2$, in which case it is exactly $0.5$.

The following is an histogram of the maximum of the modulus of the roots of $T_n$ for $3\le n\le500000$.

enter image description here

  • 1
    $\begingroup$ Is your $z$ really $-1.5000\ldots$? These polynomials can't have positive real roots unless I misunderstand their construction. $\endgroup$
    – 2'5 9'2
    Nov 21, 2012 at 23:33
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    $\begingroup$ In the range $500000−1500000$ the only counterexample is for $n=524289$ with root $$z\approx−1.50000071910044097531042989367183380456409739542661084549869.$$ $\endgroup$
    – P..
    Nov 22, 2012 at 6:36
  • $\begingroup$ @alex.jordan I wrote the modulus. Thanks for pointing it out. $\endgroup$ Nov 22, 2012 at 7:13
  • $\begingroup$ @JuliánAguirre The difference from -1.5 in n=393217, could it be a numerical issue? What is the eps in the environment you did this calculation? $\endgroup$ Nov 22, 2012 at 8:32
  • 1
    $\begingroup$ @MohsenNosratinia In Mathematica using NSolve with 60 digits accuracy. $\endgroup$ Nov 22, 2012 at 11:12

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