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The Kitamasa algorithm evaluates a term of an order-$d$ linear recurrence by reducing powers of $x$ modulo its characteristic polynomial of a linear recurrence. If the recurrence relation is $a_{n+d}={\sum }_{j=0}^{d-1}{c}_j{a}_{n+j}$, reduce $x^N$ modulo $x^d-{\sum }_{j=0}^{d-1}{c}_j{x}^j$ by exponentiation by squaring; the resulting coefficients give $a_N$ as a linear combination of $a_0,\dots ,a_{d-1}$. Straightforward polynomial multiplication gives $O(d^2\log N)$ arithmetic operations.

A linear recurrence relation expresses each term of a sequence as a linear combination of earlier terms, possibly with an added forcing term. A constant-coefficient homogeneous recurrence of order $r$ has the formTogether with $r$ consecutive initial conditions, it determines the following terms uniquely. For example, the Fibonacci sequence satisfies $F_{n+2}=F_{n+1}+F_n$, $F_0=0$, $F_1=1$.

Substituting $u_n={\lambda }^n$ gives the characteristic polynomial of a linear recurrence. Distinct characteristic roots give independent geometric solutions; repeated roots introduce factors polynomial in $n$. This converts repeated term-by-term computation into polynomial or matrix arithmetic, useful for obtaining a distant term efficiently.

The Lucas U-sequence has $U_0(P,Q)=0$, $U_1(P,Q)=1$ and $U_{n+2}=P{U}_{n+1}-Q{U}_n$. Its first terms are $0,1,P,P^2-Q,P^3-2PQ$. At $(P,Q)=(1,-1)$ it is the Fibonacci sequence.

If the roots $\alpha ,\beta$ of its characteristic polynomial of a linear recurrence are distinct, $U_n=({\alpha }^n-{\beta }^n)/(\alpha -\beta )$. This identity follows by checking the two initial conditions and the linear recurrence relation. At a repeated root $\alpha =\beta$, the corresponding formula is $U_n=n{\alpha }^{n-1}$ for $n\ge 1$.

There is no forbidden pair in a sequence of length zero or one, so $p_0=p_1=1$. For $n⩾2$, divide acceptable sequences into two disjoint cases. If the first digit differs from $3$, its probability is $2/3$, and the remaining $n-1$ independent digits must avoid the pair. If the first digit is $3$, the next must differ from $3$, which has combined probability $2/9$, and the remaining $n-2$ digits must avoid the pair. The boundary between this prefix and the remaining digits cannot create a forbidden pair, because the prefix ends in a digit other than $3$. ConsequentlyThe characteristic polynomial of a linear recurrence is $r^2-\frac{2}{3}r-\frac{2}{9}$, with distinct rootsWriting $p_n=A{r}_{+}^n+B{r}_{-}^n$, the initial conditions give $A+B=1$ and $A{r}_{+}+B{r}_{-}=1$. ThereforeFor example, $p_2=8/9$, since only one of the nine equally likely pairs is forbidden. This is the case with success probability $1/3$ of the avoidance recurrence for two consecutive successes.

For a constant-coefficient linear recurrence relation over the complex numbers, a nonzero root $\mu$ of multiplicity $m$ in its characteristic polynomial of a linear recurrence supplies the $m$ linearly independent solutionsSubstitute $y_n={\mu }^n{z}_n$. Each factor $E-\mu$ of the shift linear operator becomes $\mu$ times the forward difference operator on $z_n$, so its $m$th power kills every polynomial in $n$ of degree less than $m$. Their independence follows from the Lagrange root bound over a field. The same argument works over a field of characteristic zero; the displayed monomials need modification in positive characteristic.

A zero root needs separate initial-data treatment: for characteristic polynomial ${\lambda }^m$, the recurrence relation is $y_{n+m}=0$, leaving the first $m$ values arbitrary.