The n-th continuant ${\displaystyle K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})}$  is defined recursively by

${\displaystyle K_{0}=1;\,}$ 

${\displaystyle K_{1}(x_{1})=x_{1};\,}$ 

${\displaystyle K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})=x_{n}K_{n-1}(x_{1},\;x_{2},\;\ldots ,\;x_{n-1})+K_{n-2}(x_{1},\;x_{2},\;\ldots ,\;x_{n-2}).\,}$ 

• The continuant ${\displaystyle K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})}$  can be computed by taking the sum of all possible products of x₁,...,x_n, in which any number of disjoint pairs of consecutive terms are deleted (Euler's rule). For example,

  ${\displaystyle K_{5}(x_{1},\;x_{2},\;x_{3},\;x_{4},\;x_{5})=x_{1}x_{2}x_{3}x_{4}x_{5}\;+\;x_{3}x_{4}x_{5}\;+\;x_{1}x_{4}x_{5}\;+\;x_{1}x_{2}x_{5}\;+\;x_{1}x_{2}x_{3}\;+\;x_{1}\;+\;x_{3}\;+\;x_{5}.}$ 

It follows that continuants are invariant with respect to reversing the order of indeterminates: ${\displaystyle K_{n}(x_{1},\;\ldots ,\;x_{n})=K_{n}(x_{n},\;\ldots ,\;x_{1}).}$ 

• The continuant can be computed as the determinant of a tridiagonal matrix:

  ${\displaystyle K_{n}(x_{1},\;x_{2},\;\ldots ,\;x_{n})=\det {\begin{pmatrix}x_{1}&1&0&\cdots &0\\-1&x_{2}&1&\ddots &\vdots \\0&-1&\ddots &\ddots &0\\\vdots &\ddots &\ddots &\ddots &1\\0&\cdots &0&-1&x_{n}\end{pmatrix}}.}$ 

• ${\displaystyle K_{n}(1,\;\ldots ,\;1)=F_{n+1}}$ , the (n+1)-st Fibonacci number.
• ${\displaystyle {\frac {K_{n}(x_{1},\;\ldots ,\;x_{n})}{K_{n-1}(x_{2},\;\ldots ,\;x_{n})}}=x_{1}+{\frac {K_{n-2}(x_{3},\;\ldots ,\;x_{n})}{K_{n-1}(x_{2},\;\ldots ,\;x_{n})}}.}$ 
• Ratios of continuants represent (convergents to) continued fractions as follows:

  ${\displaystyle {\frac {K_{n}(x_{1},\;\ldots ,x_{n})}{K_{n-1}(x_{2},\;\ldots ,\;x_{n})}}=[x_{1};\;x_{2},\;\ldots ,\;x_{n}]=x_{1}+{\frac {1}{\displaystyle {x_{2}+{\frac {1}{x_{3}+\ldots }}}}}.}$ 

• The following matrix identity holds:

  ${\displaystyle {\begin{pmatrix}K_{n}(x_{1},\;\ldots ,\;x_{n})&K_{n-1}(x_{1},\;\ldots ,\;x_{n-1})\\K_{n-1}(x_{2},\;\ldots ,\;x_{n})&K_{n-2}(x_{2},\;\ldots ,\;x_{n-1})\end{pmatrix}}={\begin{pmatrix}x_{1}&1\\1&0\end{pmatrix}}\times \ldots \times {\begin{pmatrix}x_{n}&1\\1&0\end{pmatrix}}}$ .

  • For determinants, it implies that

    ${\displaystyle K_{n}(x_{1},\;\ldots ,\;x_{n})\cdot K_{n-2}(x_{2},\;\ldots ,\;x_{n-1})-K_{n-1}(x_{1},\;\ldots ,\;x_{n-1})\cdot K_{n-1}(x_{2},\;\ldots ,\;x_{n})=(-1)^{n}.}$ 

  • and also

    ${\displaystyle K_{n-1}(x_{2},\;\ldots ,\;x_{n})\cdot K_{n+2}(x_{1},\;\ldots ,\;x_{n+2})-K_{n}(x_{1},\;\ldots ,\;x_{n})\cdot K_{n+1}(x_{2},\;\ldots ,\;x_{n+2})=(-1)^{n+1}x_{n+2}.}$ 

A generalized definition takes the continuant with respect to three sequences a, b and c, so that K(n) is a polynomial of a₁,...,a_n, b₁,...,b_n−1 and c₁,...,c_n−1. In this case the recurrence relation becomes

${\displaystyle K_{0}=1;\,}$ 

${\displaystyle K_{1}=a_{1};\,}$ 

${\displaystyle K_{n}=a_{n}K_{n-1}-b_{n-1}c_{n-1}K_{n-2}.\,}$ 

Since b_r and c_r enter into K only as a product b_rc_r there is no loss of generality in assuming that the b_r are all equal to 1.

The generalized continuant is precisely the determinant of the tridiagonal matrix

${\displaystyle {\begin{pmatrix}a_{1}&b_{1}&0&\ldots &0&0\\c_{1}&a_{2}&b_{2}&\ldots &0&0\\0&c_{2}&a_{3}&\ldots &0&0\\\vdots &\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&0&\ldots &a_{n-1}&b_{n-1}\\0&0&0&\ldots &c_{n-1}&a_{n}\end{pmatrix}}.}$ 

In Muir's book the generalized continuant is simply called continuant.

• Thomas Muir (1960). A treatise on the theory of determinants. Dover Publications. pp. 516–525.
• Cusick, Thomas W.; Flahive, Mary E. (1989). The Markoff and Lagrange Spectra. Mathematical Surveys and Monographs. Vol. 30. Providence, RI: American Mathematical Society. p. 89. ISBN 0-8218-1531-8. Zbl 0685.10023.
• George Chrystal (1999). Algebra, an Elementary Text-book for the Higher Classes of Secondary Schools and for Colleges: Pt. 1. American Mathematical Society. p. 500. ISBN 0-8218-1649-7.