Probability distribution of Fibonacci sequence minima

From papers

Consider the Fibonacci recurrence $a_n=a_{n-1}+a_{n-2}$ and its chiral recurrence $a_n=-a_{n-1}+a_{n-2}$ with random integer initialization $(a_0,a_1)\in\mathbb Z^2$ . Let $F_n$ and $L_n$ denote the Fibonacci and Lucas numbers, and let $P(n)$ be the probability that the sequence has its absolute minimum at position $n$ . Probability distribution of Fibonacci minima.

P(n)=\begin{cases}\dfrac14,&n=0,\\[4pt]\dfrac1\pi\left(\arctan\left(\dfrac{L_{n-2}}{L_{n-1}}\right)-\arctan\left(\dfrac{L_n}{L_{n+1}}\right)\right),&n>1\text{ and }n\text{ even},\\[6pt]\dfrac1\pi\left(\arctan\left(\dfrac{L_n}{L_{n+1}}\right)-\arctan\left(\dfrac{L_{n-2}}{L_{n-1}}\right)\right),&n>1\text{ and }n\text{ odd}. \end{cases}

Moreover, $P(n)=P(1-n)$ . The supplied text gives no resolution status.

Progress summary

Open

The formula is recorded as a conjecture, and no verified proof or counterexample has been found.

The problem asserts an explicit probability law for where the absolute minimum occurs in two related Fibonacci recurrences, together with a symmetry under reversing the index. It is presented as a conjecture rather than an established theorem.

October 2025 conjecture formulation

The paper Modular Periodicity of Random Initialized Recurrences states this distribution as Conjecture 20 and derives the proposed expression from intervals involving Lucas-number ratios. The scan found no proof, counterexample, independent verification, withdrawal, or retraction for this specific conjecture.

Current status (as of August 2026): The distribution remains an unproved conjecture, with no confirmed proof or counterexample recorded in the retrieved sources.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Marc T. Pudelko, “Modular Periodicity of Random Initialized Recurrences”, arXiv:2510.24882 (2026).

1 week ago

For the square-uniform integer initialization explicitly specified in the source, the limiting probability at position zero is $5/24$ , not $1/4$ .

Choose

(a_0,a_1)=(x,y)

uniformly from

[-N,N]^2\cap\mathbb Z^2,

and extend the Fibonacci recurrence in both directions. Its neighboring terms are

a_{-2}=2x-y,\qquad a_{-1}=y-x,\qquad a_0=x,\qquad a_1=y,\qquad a_2=x+y.

If $x\ne0$ , put $r=y/x$ . A unique global absolute minimum at index zero requires

|r|>1, \qquad |r-1|>1, \qquad |r+1|>1, \qquad |2-r|>1.

Their intersection is exactly

r<-2\quad\text{or}\quad r>3.

These conditions are also sufficient. After simultaneously negating $x,y$ if necessary, assume $x>0$ . For $r<-2$ , the forward terms have a common negative sign and strictly increasing absolute value, while the backward terms alternate signs with strictly increasing absolute value. For $r>3$ , the forward terms are positive and increasing, and the backward terms again alternate with increasing absolute value. Therefore no later index attains a smaller absolute value. The cases $x=0$ , $y\ne0$ also have their unique minimum at index zero.

Hence the exact number of square-lattice initializations having unique minimum at zero is

C_N= 2N + 2\sum_{j=1}^{\lfloor(N-1)/2\rfloor}(N-2j) + 2\sum_{j=1}^{\lfloor(N-1)/3\rfloor}(N-3j).

Consequently

\lim_{N\to\infty}P_N(0) = \lim_{N\to\infty}\frac{C_N}{(2N+1)^2} = \frac5{24}.

Since

\frac5{24}\ne\frac14,

the conjectured distribution contradicts its own stated square-uniform sampling rule.

The discrepancy comes from treating a uniform distribution on an expanding square as though it induced a uniform polar angle. It does not; the arctangent expression corresponds to a different rotationally invariant sampling scheme.

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Shivam Patel · 22 hours ago