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The expression is the Binet formula for the Fibonacci numbers. Compute the two roots and round the result to the nearest integer:

The returned object is an integer. This floating-point arithmetic implementation correctly rounds the requested range through $n=50$; it is not an arbitrary-precision method for indefinitely large $n$. An integer recurrence is a useful independent numerical check.

Represent the doubly infinite ladder graph by $\mathbb{Z}\times \{0,1\}$, with horizontal edges between successive columns and a vertical rung at each column. An allowed self-avoiding walk is coded by a word in $R,V$. Two successive $V$ steps revisit the preceding graph vertex and are forbidden. Conversely, every word without consecutive $V$ steps is a self-avoiding walk: horizontal steps strictly increase the column, and a column is visited vertically at most once. This establishes a bijection, not merely an upper bound.

For the directed ladder self-avoiding walk count, ${\sigma }_0=1$, ${\sigma }_1=2$, and for $n\ge 2$ a word either ends in $R$, or ends in $RV$. Removing that final block givesLet $\phi =(1+\sqrt{5})/2$ be the golden ratio, and $\psi =(1-\sqrt{5})/2$. The Binet formula for the Fibonacci number yieldsSince $|\psi |<\phi$, taking $n$th roots gives ${\lim }_{n\to \infty }{\sigma }_n^{1/n}=\phi$. This use of the golden-ratio symbol is independent of the connection rate in Question 1.

LetSuppose $F_n=x^k$ is a perfect power with $x\ge 2$. The finitely many small $n$ can be absorbed into the final effective constant. The Binet formula gives $x^k\asymp {\varphi }^n$ andThus, forthe local Lipschitz equivalence of $u$ and $e^u-1$ at zero yieldsThe form cannot vanish: applying the nontrivial field automorphism of $\mathbb{Q}(\sqrt{5})$ to ${\varphi }^n=\sqrt{5}{x}^k$ would give ${\psi }^n=-\sqrt{5}{x}^k$, whose absolute values are incompatible.

Apply the Baker lower bound for a homogeneous linear form in logarithms with the variable-height number $x$ placed last. The parameters belonging to $\varphi$ and $\sqrt{5}$ are absolute constants, while $\log A_x\asymp \log x$. Moreover,so $n/\log A_x\ll k$ and therefore $B^{\ast }\ll k$. The refined lower bound becomesComparison with the exponential upper bound gives $k\le C_3\log k$. Since $k/\log k$ tends to infinity, this bounds $k$ by an effective absolute constant. Enlarging it to cover the discarded small indices proves the claim.

The diagonalization gives $A=SD{S}^{-1}$ and hence $A^{n-1}=S{D}^{n-1}{S}^{-1}$: the intervening factors $S^{-1}S$ cancel. The diagonal matrix power has entries ${\alpha }^{n-1},{\beta }^{n-1}$. Since $1-\beta =\alpha$ and $\alpha -1=-\beta$,Reading the second component gives the Binet formulaThe derivation includes $n=1$, and the expression gives $U_1=U_2=1$, consistent with the specified indexing.