Critical Period-Six Convergence for a Third-Order Rational Difference Equation

1. Introduction

Rational difference equations provide elementary-looking models whose global dynamics can exhibit sharp transitions between bounded convergence, periodic limiting behavior, and unbounded oscillation; see, for example, [1,2,3]. We consider

$$z_{n+1}={\displaystyle \frac{\alpha +z_n}{C{z}_{n-1}+z_{n-2}}},n=0,1,2,\dots ,$$

(1)

where $\alpha ,C>0$ and $z_{-2},z_{-1},z_0>0$. Positivity makes every denominator in (1) nonzero.

Camouzis and Ladas formulated a trichotomy for (1) in [4]; it was later recorded as Conjecture 6 by Ladas, Lugo, and Palladino [5]. Its proposed threshold is $\alpha C^2=1$: global equilibrium convergence above the threshold, convergence to a solution of period dividing six at the threshold, and the existence of unbounded positive solutions below it. Lugo and Palladino treated degenerate subcases in [6]. Huang and Knopf’s main boundedness theorem [7] applies to the full equation (1): for $\alpha ,C>0$, every positive solution is bounded when $\alpha C^2\ge 1$, including the boundary case, whereas positive unbounded solutions exist for suitable initial conditions when $\alpha C^2<1$.

Amleh and Ladas first posed the special equation

$$x_{n+1}={\displaystyle \frac{1+x_n}{x_{n-1}+x_{n-2}}}$$

as a period-six convergence problem and, in the same work, formulated the more general conjecture that for every $p>0$ each positive solution of

$$z_{n+1}={\displaystyle \frac{p^{-2}+z_n}{p{z}_{n-1}+z_{n-2}}}$$

(2)

converges to a period-six solution [8]. Equation (2) is precisely the critical family $\alpha C^2=1$ in (1), with $p=C$. Under the scaling $q=1/C$ and $x_n=C{z}_n$, it becomes

$$x_{n+1}={\displaystyle \frac{q+x_n}{x_{n-1}+q{x}_{n-2}}},q>0.$$

The normalized and original parametrizations of the periodic family are equivalent. More precisely, the classification proved in Corollary 2 shows that, after a cyclic choice of origin, a normalized cycle

$$\left(a,ac,c,a^{-1},{\left(ac\right)}^{-1},c^{-1}\right)$$

corresponds, upon writing $\varphi =z_0=a/C$ and $\psi =z_1=ac/C$, to the original-variable cycle

$$\left(\varphi ,\psi ,{\displaystyle \frac{\psi }{C\varphi }},{\displaystyle \frac{1}{C^2\varphi }},{\displaystyle \frac{1}{C^2\psi }},{\displaystyle \frac{\varphi }{C\psi }}\right),\varphi ,\psi >0.$$

(3)

Thus the two-parameter family obtained below is exactly the family associated with the Amleh–Ladas critical equation, expressed in normalized coordinates.

Spahn and Zeilberger later revisited the $q=1$ member by experimental mathematics and symbolic computation and described their treatment of the first four Amleh–Ladas conjectures, including the period-six case, as semi-rigorous [9]. The present paper supplies an exact proof for every $q>0$. Accordingly, the main theorem resolves the generalized Amleh–Ladas period-six conjecture and, equivalently, the critical branch of the Camouzis–Ladas trichotomy.

The critical proof has two ingredients. First, an elementary defect changes sign at every iterate. Second, an exact six-step factorization shows that each of the six residue subsequences is monotone. The positive numerator factor has 228 monomials and the positive denominator has 557 monomials, so printing either expansion in the article would obscure rather than clarify the argument. Both are supplied as machine-readable exact certificates. The archive contains distinct generation and verification programs. The certificate-reading verifier does not call the generator or factor the recurrence numerator; it reads the committed polynomials, checks their hashes and coefficient positivity, reconstructs the six iterates, and confirms the cross-multiplied identity over $\mathbb{Z}[a,b,c,q]$. Both stages use the same pinned SymPy environment, so this is an operationally separate exact verification stage rather than an implementation in a second computer-algebra system. No floating-point calculation enters the proof.

The paper is organized as follows. In Section 2 we determine the equilibrium and prove the exact local-stability threshold by an explicit Schur reduction. In Section 3 we normalize the critical equation and establish the certified six-step sign identity. The convergence theorem is proved in Section 4, together with the exact prime-period classification of the two-parameter periodic family. In Section 5 we normalize the general equation, state precisely the parameter range for the period-two and period-three exclusions, and derive a weighted reciprocal-mean entropy estimate. A numerical illustration of the proved critical theorem is given in Section 6; it is not used in any proof.

2. Equilibrium and the Local-Stability Threshold

Equation (1) has a unique positive equilibrium $\overline{z}$, determined by

$$(C+1){\overline{z}}^2-\overline{z}-\alpha =0.$$

(4)

Hence

$$\overline{z}={\displaystyle \frac{1+\sqrt{1+4\alpha (C+1)}}{2(C+1)}}.$$

(5)

Theorem 1

(Exact local-stability threshold). The positive equilibrium (5) is locally asymptotically stable if and only if

$$\alpha C^2>1.$$

It is nonhyperbolic and not locally asymptotically stable when $\alpha C^2=1$, and it is unstable when $\alpha C^2<1$.

Proof. 

Write $z_n=\overline{z}+u_n$. The linearization at $\overline{z}$ is

$$u_{n+1}=a{u}_n-{\displaystyle \frac{C}{C+1}}{u}_{n-1}-{\displaystyle \frac{1}{C+1}}{u}_{n-2},a={\displaystyle \frac{1}{(C+1)\overline{z}}}.$$

(6)

Since $(C+1){\overline{z}}^2=\overline{z}+\alpha >\overline{z}$, one has $0<a<1$. Put

$$t={\displaystyle \frac{1}{C+1}},r={\displaystyle \frac{C}{C+1}}=1-t.$$

The characteristic polynomial is

$$P\left(\lambda \right)={\lambda }^3-a{\lambda }^2+r\lambda +t.$$

(7)

We now display all Schur conditions rather than suppressing the reduction. For a real monic cubic

$$f\left(\lambda \right)={\lambda }^3+b{\lambda }^2+r\lambda +t,$$

with $\left|t\right|<1$, the Schur transform is

$${\displaystyle \frac{f\left(\lambda \right)-t{\lambda }^{3}f(1/\lambda )}{\lambda }}=(1-t^2){\lambda }^2+(b-tr)\lambda +(r-tb).$$

The cubic is Schur stable if and only if this quadratic is Schur stable. A real quadratic $A{\lambda }^2+B\lambda +D$, with $A>0$, is Schur stable exactly when

$$\left|D\right|<A,A+B+D>0,A-B+D>0$$

(see, for example, [10]). Here $b=-a$ and hence

$$A=1-t^2,B=-a-tr,D=r+at.$$

Because $D>0$, the first quadratic condition is $A-D>0$. Direct calculation gives

$$\begin{array}{cc}\hfill A-D& =t(r-a)={\displaystyle \frac{C\overline{z}-1}{\overline{z}{(C+1)}^2}},\hfill \\ \hfill A+B+D& =(1-t)(2-a)>0,\hfill \\ \hfill A-B+D& =(1+t)(a+2-2t)>0.\hfill \end{array}$$

Also $0<t<1$, so the initial condition $\left|t\right|<1$ is automatic. Consequently every characteristic root lies in the open unit disk exactly when $C\overline{z}>1$.

Using (4), one obtains

$$\alpha C^2-1=(C\overline{z}-1)\left(C(C+1)\overline{z}+1\right).$$

(8)

The second factor is positive, so $C\overline{z}>1$ is equivalent to $\alpha C^2>1$. In this case all characteristic roots lie strictly inside the unit disk, and the standard linearized-stability theorem for discrete dynamical systems yields local asymptotic stability of the equilibrium.

When $C\overline{z}=1$, necessarily $\alpha C^2=1$, and (7) factors as

$$P\left(\lambda \right)=\left(\lambda +{\displaystyle \frac{1}{C+1}}\right)({\lambda }^2-\lambda +1).$$

(9)

The quadratic factor has the primitive sixth roots $e^{\pm i\pi /3}$, so the equilibrium is nonhyperbolic. Nonhyperbolicity alone does not exclude nonlinear attraction, so the equality case requires an additional argument. At criticality introduce $x_n=C{z}_n$. For any $\epsilon >0$, the initial state

$$(x_{-2},x_{-1},x_0)=(1+\epsilon ,1+\epsilon ,1)$$

generates, by direct substitution in the critical recurrence, the nonconstant period-six orbit

$$\left(1+\epsilon ,1+\epsilon ,1,{\displaystyle \frac{1}{1+\epsilon }},{\displaystyle \frac{1}{1+\epsilon }},1\right).$$

As $\epsilon \downarrow 0$, these periodic initial states approach the equilibrium state $(1,1,1)$ in normalized coordinates, equivalently $(1/C,1/C,1/C)$ in the original variables, but their forward orbits do not converge to the equilibrium. Hence the critical equilibrium is not locally attractive and therefore is not locally asymptotically stable.

It remains to justify instability below the threshold. Recall that $r=1-t$. We first show that (7) has a root on the unit circle only when $a=r$. Suppose $P\left(e^{i\theta }\right)=0$. The values $P\left(1\right)=2-a>0$ and $P(-1)=-2-a+2t<0$ exclude $\theta =0,\pi$. Dividing the equation by $e^{i\theta }$ and taking imaginary parts gives

$$sin\theta \left(2cos\theta -a-t\right)=0,$$

so $2cos\theta =a+t$. Substitution in the real part yields

$$t(a+t-1)=0.$$

Since $t>0$, a unit-modulus root therefore forces $a=1-t=r$. Conversely, when $a=r$ the factorization (9) gives the two unit-modulus roots explicitly.

If $C\overline{z}<1$, then $a>r$, because $A-D=t(r-a)<0$. The Schur criterion fails, while the preceding calculation excludes roots on the unit circle. Hence at least one characteristic root has modulus strictly greater than one. The first-order companion map associated with (1) is continuously differentiable in a neighborhood of the positive equilibrium, and the eigenvalues of its derivative are the roots of (7). The standard linearized-instability theorem for discrete dynamical systems therefore implies that the equilibrium is unstable [10].    □

Remark 1.

The roots $e^{\pm i\pi /3}$ in (9) explain why the boundary dynamics is organized by six residue classes. The critical periodicity is therefore already visible in the linearization, although the global result requires a nonlinear argument.

3. Critical Normalization and an Exact Six-Step Identity

Assume throughout this section that

$$\alpha C^2=1.$$

(10)

Set

$$q={\displaystyle \frac{1}{C}},x_n=C{z}_n.$$

(11)

Then (1) becomes

$$x_{n+1}={\displaystyle \frac{q+x_n}{x_{n-1}+q{x}_{n-2}}},q>0.$$

(12)

Its positive equilibrium is $x=1$.

Define the defect

$$d_n=x_{n-2}{x}_n-x_{n-1}.$$

(13)

Lemma 1

(Alternating defect). Every positive solution of (12) satisfies

$$d_{n+1}=-{\displaystyle \frac{q{d}_n}{x_{n-1}+q{x}_{n-2}}}.$$

(14)

Consequently, either $d_n=0$ for every subsequent index, or the sign of $d_n$ alternates at each step. In particular,

$$sgn\left(d_{n+6}\right)=sgn\left(d_n\right).$$

Proof. 

Using (12),

$$\begin{array}{cc}\hfill d_{n+1}& =x_{n-1}{x}_{n+1}-x_n\hfill \\ & ={\displaystyle \frac{x_{n-1}(q+x_n)-x_n(x_{n-1}+q{x}_{n-2})}{x_{n-1}+q{x}_{n-2}}}\hfill \\ & =-{\displaystyle \frac{q(x_{n-2}{x}_n-x_{n-1})}{x_{n-1}+q{x}_{n-2}}}.\hfill \end{array}$$

The denominator is positive.    □

The factor appearing in the six-step identity below has a direct dynamical origin. For initial data $x_{-2}=a$, $x_{-1}=b$, and $x_0=c$, the equality $ac=b$ is exactly the condition $d_0=0$. By Lemma 1, it then follows that $d_n=0$ for all subsequent indices, so $x_n=x_{n-1}/x_{n-2}$ and direct iteration gives

$$\left(a,ac,c,a^{-1},{\left(ac\right)}^{-1},c^{-1}\right),$$

which repeats after six steps. Consequently, after denominators are cleared, the numerator of $x_6-x_0$ vanishes on the hypersurface $ac-b=0$ and is divisible by $ac-b$. The computer-assisted certificate is needed not to discover this structural factor, but to identify the remaining quotient exactly and prove that all of its coefficients are positive.

The following identity provides the global six-step monotonicity. Its verification is exact, but computer-assisted.

Proposition 1

(Certified six-step sign identity). Let $a,b,c,q>0$, and let (12) start from

$$x_{-2}=a,x_{-1}=b,x_0=c.$$

There exist polynomials $\mathcal{P},\mathcal{Q}\in \mathbb{Z}[a,b,c,q]$, every coefficient of which is strictly positive, such that

$$x_6-x_0=-{\displaystyle \frac{q^2(ac-b)\mathcal{P}(a,b,c,q)}{\mathcal{Q}(a,b,c,q)}}.$$

(15)

The polynomial $\mathcal{P}$ has 228 monomials and total degree 16; $\mathcal{Q}$ has 557 monomials and total degree 19.

Proof. 

Use the index correspondence

$$R_0=x_{-2}=a,R_1=x_{-1}=b,R_2=x_0=c,\dots ,R_8=x_6,$$

and define

$$R_{j+3}={\displaystyle \frac{q+R_{j+2}}{R_{j+1}+q{R}_j}},j=0,\dots ,5.$$

The certificate generator performs exact rational simplification over $\mathbb{Q}(a,b,c,q)$ and writes primitive expanded polynomials $\mathcal{P},\mathcal{Q}\in \mathbb{Z}[a,b,c,q]$, normalized to have positive leading coefficient and coefficient content one. A separate certificate-reading verifier then reads those stored files rather than regenerating them. It checks their SHA-256 hashes against hard-coded proof-critical values and cross-checks those values with both the certificate summary and the archive manifest. It also checks term counts, degrees, coefficient contents, and strict positivity of every coefficient. It reconstructs $R_3,\dots ,R_8$ from the recurrence and verifies the cross-multiplied identity

$$(R_8-c)\mathcal{Q}+q^2(ac-b)\mathcal{P}=0$$

by reducing its exact numerator to the zero polynomial over $\mathbb{Z}[a,b,c,q]$. This proves (15). The generator and this verifier are separate executable stages, but both use the same pinned SymPy stack; they are therefore described as operationally separate rather than as a second-CAS verification. The archive additionally contains a standard-library sparse-polynomial verifier that parses the committed polynomials, reconstructs the six rational iterates without SymPy, and checks the same cleared polynomial identity using integer dictionary arithmetic. This implementation-diverse check is not a formal proof-assistant verification, but it removes dependence on SymPy for the final identity test. Because all coefficients of $\mathcal{P}$ and $\mathcal{Q}$ are strictly positive, both polynomials are strictly positive on the positive orthant. All computations use exact integer or rational arithmetic; no numerical tolerance, interpolation, or random sampling is involved.    □

Corollary 1

(Monotonicity along residue classes). For every positive solution of (12),

$$sgn(x_{n+6}-x_n)=-sgn\left(d_n\right).$$

(16)

Therefore, for each $r\in \{0,1,\dots ,5\}$, the subsequence ${\left\{x_{6k+r}\right\}}_{k\ge 0}$ is monotone.

Proof. 

Apply Proposition 1 after shifting the initial index. All factors other than $d_n=x_{n-2}{x}_n-x_{n-1}$ are positive. By Lemma 1, the sign of $d_n$ is constant on each residue class modulo six.    □

4. Convergence to a Period Dividing Six

We now combine the preceding monotonicity with the boundedness theorem of Huang and Knopf [7].

Theorem 2

(Critical convergence). Assume $\alpha ,C>0$, $\alpha C^2=1$, and $z_{-2},z_{-1},z_0>0$. Then there are positive numbers $L_0,\dots ,L_5$ such that

$$\underset{k\to \infty }{lim}C{z}_{6k+r}=L_r,r=0,\dots ,5.$$

(17)

The bi-infinite repetition of $(L_0,\dots ,L_5)$ is a positive solution of (12). Hence every positive solution of (1) converges to a positive orbit whose period divides six.

Proof. 

The main theorem of Huang and Knopf [7] assumes $\alpha ,C>0$ and positive initial data and states that every positive solution of (1) is bounded whenever $\alpha C^2\ge 1$; hence it applies directly on the equality surface considered here. Thus the normalized solution of (12) satisfies $x_n\le M$ for some $M>0$. It is also bounded away from zero, because

$$x_{n+1}={\displaystyle \frac{q+x_n}{x_{n-1}+q{x}_{n-2}}}\ge {\displaystyle \frac{q}{(1+q)M}}>0.$$

(18)

By Corollary 1, each of the six residue subsequences is monotone; boundedness and (18) therefore imply the existence of finite positive limits $L_r$.

Fix r modulo six and let $n=6k+r$. Passing to the limit in (12) yields

$$L_{r+1}={\displaystyle \frac{q+L_r}{L_{r-1}+q{L}_{r-2}}},r(mod6).$$

(19)

Thus the six-periodic extension of $(L_0,\dots ,L_5)$ satisfies the recurrence. This proves convergence to a solution of period dividing six.    □

The critical equation also contains a large explicit family of exact period-six orbits.

Proposition 2

(Exact period-six family and prime-period classification). Let $a,c,q>0$, and initialize (12) by

$$x_{-2}=a,x_{-1}=ac,x_0=c.$$

Then

$$(x_{-2},x_{-1},x_0,x_1,x_2,x_3)=\left(a,ac,c,{\displaystyle \frac{1}{a}},{\displaystyle \frac{1}{ac}},{\displaystyle \frac{1}{c}}\right),$$

(20)

and $x_{n+6}=x_n$ for all n. The orbit is the equilibrium precisely when $(a,c)=(1,1)$; for every $(a,c)\ne (1,1)$ its prime period is six.

Proof. 

The initial defect is $d_0=ac-ac=0$. Equation (14) implies $d_n=0$ for every $n\ge 0$, so

$$x_n={\displaystyle \frac{x_{n-1}}{x_{n-2}}}.$$

Successive substitution gives (20) and then repeats after six steps. Directly, the first recurrence step is

$$x_1={\displaystyle \frac{q+c}{ac+qa}}={\displaystyle \frac{1}{a}},$$

and the remaining terms follow similarly.

It remains to determine the prime period. Any prime period dividing six is 1, 2, 3, or 6. If the displayed orbit has period dividing three, then $x_{-2}=x_1$, so $a=1/a$ and hence $a=1$; next $x_{-1}=x_2$ gives $ac=1/\left(ac\right)$, hence $c=1$. If it has period dividing two, then the even positions in (20) coincide, giving

$$a=c={\displaystyle \frac{1}{ac}}.$$

Thus $a=c$ and $a^3=1$, so again $a=c=1$. The period-one case is contained in either argument. Therefore every non-equilibrium member has prime period six.    □

Corollary 2

(Classification of all positive period-dividing-six orbits). Let $\left\{x_n\right\}$ be a positive periodic solution of (12) whose period divides six. Then there exist $a,c>0$ such that

$$(x_0,x_1,x_2,x_3,x_4,x_5)=\left(a,ac,c,{\displaystyle \frac{1}{a}},{\displaystyle \frac{1}{ac}},{\displaystyle \frac{1}{c}}\right).$$

(21)

Consequently, the equilibrium $x_n\equiv 1$ is the only solution with prime period less than six, and every other positive periodic solution whose period divides six has prime period six.

Proof. 

For a solution whose period divides six, $x_{n+6}-x_n=0$ for every n. Applying the shifted identity of Proposition 1, all factors except $d_n=x_{n-2}{x}_n-x_{n-1}$ are strictly positive; hence $d_n=0$ for every n. In particular,

$$x_{n-2}{x}_n=x_{n-1}\mathrm{for}\mathrm{every}n.$$

Set $a=x_0$ and $c=x_2$. The relation at $n=2$ gives $x_1=ac$, and the relations at $n=3,4,5$ successively give

$$x_3={\displaystyle \frac{1}{a}},x_4={\displaystyle \frac{1}{ac}},x_5={\displaystyle \frac{1}{c}}.$$

This proves (21). The prime-period assertion follows from Proposition 2.    □

Remark 2.

The exact family in Proposition 2 and its converse in Corollary 2 show that the critical limiting orbit cannot generally be replaced by the equilibrium. In particular, the conclusion of Theorem 2 can be sharpened: its limiting six-tuple is always of the form (21), and it is either the equilibrium or a prime period-six orbit.

5. Supercritical Normalization and Structural Reductions

We do not claim a proof of global equilibrium convergence for the entire supercritical region. Instead, this section records exact reductions that any such proof must exploit.

Let $\overline{z}$ be given by (5) and set

$$x_n={\displaystyle \frac{z_n}{\overline{z}}},p=C\overline{z},q=\overline{z},s=p+q-1.$$

(22)

Since $\alpha =(C+1){\overline{z}}^2-\overline{z}$, one has $s=\alpha /\overline{z}>0$, and the normalized equation is

$$x_{n+1}={\displaystyle \frac{s+x_n}{p{x}_{n-1}+q{x}_{n-2}}},s=p+q-1,$$

(23)

with equilibrium $x=1$. By (8),

$$\alpha C^2>1⟺p>1.$$

(24)

5.1. Exclusion of periods two and three

Proposition 3.

Assume $p>0$, $q>0$, and $p+q>1$. Then (23) has no positive prime period-two solution. Its only positive solution with period dividing two is $x_n\equiv 1$.

Proof. 

If $(a,b)$ is a positive period-two orbit, then

$$b(pb+qa)=s+a,a(pa+qb)=s+b.$$

Subtracting gives

$$(b-a)\left(p(a+b)+1\right)=0.$$

The second factor is positive, so $a=b$. The positive constant solutions satisfy $(p+q)a^2-a-s=0$. Since $s=p+q-1$ and $p+q>1$, the unique positive root is $a=1$.    □

Proposition 4.

Assume $p>0$, $q>0$, and $p+q>1$. Then (23) has no positive prime period-three solution. Its only positive solution with period dividing three is $x_n\equiv 1$.

Proof. 

Let $(a,b,c)$ be a positive period-three orbit. Then

$$\begin{array}{cc}\hfill b(pc+qb)& =s+a,\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill c(pa+qc)& =s+b,\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill a(pb+qa)& =s+c.\hfill \end{array}$$

(27)

Subtracting consecutive equations yields

$$\begin{array}{cc}\hfill (1+pc)(a-b)& =q(b+c)(b-c),\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill (1+pa)(b-c)& =q(c+a)(c-a),\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill (1+pb)(c-a)& =q(a+b)(a-b).\hfill \end{array}$$

(30)

Every coefficient multiplying a difference is positive. Thus the three differences $a-b$, $b-c$, and $c-a$ have the same sign unless they vanish. Since their sum is zero, all three vanish. The unique positive constant solution is 1.    □

5.2. A weighted reciprocal-mean reduction

Define

$$y_n={\displaystyle \frac{x_n{x}_{n-2}}{x_{n-1}}}>0.$$

(31)

Lemma 2.

Every positive solution of (23) satisfies

$$y_{n+1}={\displaystyle \frac{p+q-1+x_n}{p{x}_n+q{y}_n}},$$

(32)

that is,

$${\displaystyle \frac{1}{y_{n+1}}}={\displaystyle \frac{(p-1)x_n+x_n·1+q{y}_n}{(p-1)+x_n+q}}.$$

(33)

Moreover,

$$y_{n+1}-1=-{\displaystyle \frac{(p-1)(x_n-1)+q(y_n-1)}{p{x}_n+q{y}_n}}.$$

(34)

Proof. 

By (31) and (23),

$$y_{n+1}={\displaystyle \frac{x_{n+1}{x}_{n-1}}{x_n}}={\displaystyle \frac{s+x_n}{p{x}_n+q(x_n{x}_{n-2}/x_{n-1})}}.$$

Equations (33) and (34) follow by rearrangement and $s=p+q-1$.    □

When $p>1$, the right side of (33) is a weighted arithmetic mean of $x_n$, 1, and $y_n$, with positive weights $p-1$, $x_n$, and q, respectively. In particular,

$$min\{x_n,1,y_n\}\le {\displaystyle \frac{1}{y_{n+1}}}\le max\{x_n,1,y_n\},$$

with strict inequalities unless $x_n=y_n=1$ or the three entries coincide. The same representation yields the following exact Jensen-gap estimate.

Proposition 5

(Strict convex-entropy inequality). Let

$$\mathrm{\Phi }\left(t\right)=(t-1)logt,t>0.$$

(35)

If $p>1$, then every positive solution of (23) satisfies

$$(p{x}_n+q{y}_n)\mathrm{\Phi }\left(y_{n+1}\right)\le (p-1)\mathrm{\Phi }\left(x_n\right)+q\mathrm{\Phi }\left(y_n\right).$$

(36)

More precisely, with $D_n=p-1+x_n+q$ and weights

$${\omega }_{1,n}={\displaystyle \frac{p-1}{D_n}},{\omega }_{2,n}={\displaystyle \frac{x_n}{D_n}},{\omega }_{3,n}={\displaystyle \frac{q}{D_n}},$$

one has the exact identity

$$(p-1)\mathrm{\Phi }\left(x_n\right)+q\mathrm{\Phi }\left(y_n\right)-(p{x}_n+q{y}_n)\mathrm{\Phi }\left(y_{n+1}\right)=D_n{\mathcal{J}}_n,$$

(37)

where

$${\mathcal{J}}_n={\displaystyle \sum _{j=1}^3}{\omega }_{j,n}\mathrm{\Phi }\left({\xi }_{j,n}\right)-\mathrm{\Phi }\left({\displaystyle \sum _{j=1}^3}{\omega }_{j,n}{\xi }_{j,n}\right)\ge 0,({\xi }_{1,n},{\xi }_{2,n},{\xi }_{3,n})=(x_n,1,y_n).$$

Equality in (36) holds if and only if $x_n=y_n=1$.

Proof. 

The function $\mathrm{\Phi }$ is strictly convex because

$${\mathrm{\Phi }}^{″}\left(t\right)={\displaystyle \frac{t+1}{t^2}}>0,$$

and it satisfies $\mathrm{\Phi }(1/t)=\mathrm{\Phi }\left(t\right)/t$. By (33),

$${\displaystyle \frac{1}{y_{n+1}}}={\omega }_{1,n}{x}_n+{\omega }_{2,n}·1+{\omega }_{3,n}{y}_n.$$

The definition of the Jensen gap gives

$$\mathrm{\Phi }\left({\displaystyle \frac{1}{y_{n+1}}}\right)+{\mathcal{J}}_n={\displaystyle \frac{(p-1)\mathrm{\Phi }\left(x_n\right)+q\mathrm{\Phi }\left(y_n\right)}{D_n}},$$

because $\mathrm{\Phi }\left(1\right)=0$. Multiplication by $D_n$ and the identities

$$\mathrm{\Phi }(1/y_{n+1})={\displaystyle \frac{\mathrm{\Phi }\left(y_{n+1}\right)}{y_{n+1}}},{\displaystyle \frac{D_n}{y_{n+1}}}=p{x}_n+q{y}_n$$

yield (37), hence (36). Since all three weights are positive and $\mathrm{\Phi }$ is strictly convex, ${\mathcal{J}}_n=0$ exactly when $x_n=1=y_n$.    □

Remark 3

(What the entropy estimate does and does not prove). Proposition 5 is a local contraction statement for the reciprocal update $y_{n+1}$, not a Lyapunov theorem for the full third-order system. The quantity on the right of (36) is not the same state functional evaluated at time $n+1$, and the variable coefficient $p{x}_n+q{y}_n$ obstructs direct summation. Therefore the estimate alone does not imply convergence. Its precise value is that it supplies an explicit nonnegative dissipation term $D_n{\mathcal{J}}_n$ and identifies the unique zero-dissipation state. To complete the supercritical conjecture by this route one would need an additional comparison or a bounded-below finite-memory functional whose increment controls a sum of these Jensen gaps. The sixth-order organization at the critical boundary suggests examining functionals involving several consecutive phases, but no such functional is claimed here.

Problem 1.

For $p>1$ and $q>0$, prove or disprove that every positive solution of (23) converges to 1.

6. Numerical Illustration

This computation illustrates only the proved critical theorem and does not enter its logical proof. For (12), take $q=0.7$ and

$$(x_{-2},x_{-1},x_0)=(0.3,2,1.1).$$

As guaranteed by Theorem 2, each residue subsequence is monotone and converges; the six limiting values need not coincide. A double-precision run of $\mathrm{20,000}$ steps gives

$$(L_0,\dots ,L_5)\approx (1.22815622,0.76883005,0.62600347,0.81422867,1.30067757,1.59743524).$$

The maximum residual in the classification relations

$$L_1=L_0{L}_2,L_3=L_0^{-1},L_4=L_1^{-1},L_5=L_2^{-1}$$

is below $3\times {10}^{-15}$. These numerical values are only a reproducible consistency check; Figure 1 displays the six subsequences separately.

7. Conclusion

The threshold $\alpha C^2=1$ has two complementary meanings for (1). It is exactly the local-stability boundary of the positive equilibrium, and it is the parameter surface on which nonlinear period-six limiting behavior survives. Using the boundedness theorem of Huang and Knopf together with the exact factorization (15), we proved that every positive critical solution converges along its six residue classes to a positive orbit of period dividing six. We then classified all such positive periodic orbits: after a cyclic choice of origin they are exactly the family (21), with the constant member equal to the equilibrium and every nonconstant member of prime period six. This establishes the generalized Amleh–Ladas conjecture and the critical convergence assertion of the trichotomy.

Above the threshold, the equilibrium is locally asymptotically stable. More generally, throughout the parameter domain $p>0$, $q>0$, $p+q>1$, the normalized equation has no positive prime two- or three-cycles. The weighted reciprocal-mean representation (33) and the exact Jensen-gap identity (37) provide a strict local contraction mechanism when $p>1$. They do not by themselves define a Lyapunov function for the third-order recurrence; constructing the additional comparison or finite-memory functional needed for global equilibrium convergence remains open.