On the margin of continued fraction blogging I would like to present simple result as regards to continuant polynomials. Continued fraction {[a_0,a_1,...,a_n]} may be expressed as quotient of two polynomials of {(a_0,a_1,...,a_n)}, named continuant (see continuants on Wikipedia)

{[a_0,a_1,...,a_n]} = K(a_0,a_1,...,a_n)/K(a_1,...,a_n)

For example {K(a_0,a_1,a_2,a_3) = a_{0} a_{1} a_{2} a_{3} + a_{0} a_{1} + a_{0} a_{3} + a_{2} a_{3} + 1}

Continuants has many interesting recurrence relations some of which You may find in Graham, Knuth, Patashnik book „Concrete mathematics”. The most important of this recurrences is as follows:

\displaystyle  K(a_0,a_1,...,a_n,x) = xK(a_0,a_1,...,a_n)+K(a_0,a_1,...,a_n) \ \ \ \ \

During my plays with matrix representation of continued fractions I found interesting relation:

\displaystyle  K( a_0,a_1,\ldots ,a_k,a_{(k+1)},\ldots ,a_n ) = \ \ \ \ \

\displaystyle  K(a_0,a_1,\ldots ,a_k,1,1,a_{(k+1)}, \ldots ,a_n) - K(a_0,a_1,\ldots ,a_k,1,a_{(k+1)}, \ldots ,a_n) \ \ \ \ \

that is – between variables {a_k} and {a_{(k+1)}} You put two of „1” in the first and one „1” in the second term. You may consider this as generalisation of Fibonacci recurrence – because if You put all {a_i =1} You obtain Fibonacci numbers.

For example:

\displaystyle  K(a_0,a_1,a_2) = (a_0 a_ 1+1)a_2+a_0 \ \ \ \ \

\displaystyle  K(a_0,a_1,1,1,a_2) = (2 a_0 a_1+ a_0+2)a_2+a_0 a_1+a_0+1 \ \ \ \ \

\displaystyle  K(a_0,a_1,1,a_2) = (a_2+1)(a_0 a_1+1)+a_0 a_2 \ \ \ \ \

And it is true that:

\displaystyle  K(a_0,a_1,a_2) = K(a_0,a_1,1,1,a_2) - K(a_0,a_1,1,a_2) \ \ \ \ \

How to prove it? W have:

{ x = [a_{0},a_{1},a_{2},...a_{n}] =\frac{K(a_0,a_1,\ldots ,a_n)}{K(a_1,\ldots ,a_n)}= - \frac{1}{2} \frac{Tr \left( M(S- L) \right) }{Tr \left( M(S-T) \right) }}

where {tr} is trace of a matrix, and maybe You know that I define M as follows (see here ):

\displaystyle  M = S \prod_{i=0}^{n} S(ST)^{a_{i}} \ \ \ \ \

So we have the following formulas true:

\displaystyle  K(a_{0},a_{1},a_{2} \ldots a_{n}) =\frac{1}{2} Tr \left( M(S- L) \right) \ \ \ \ \

\displaystyle  K(a_{1},a_{2} \ldots a_{n}) = - Tr \left( M(S-T) \right) \ \ \ \ \

Please note that upper formula has different arguments than the lower so they agree.

And there is true that {S^2 =I} and {T^2 = I+T} which we may use to produce „unity decomposition” as below:

\displaystyle  I = T^2 -T = S(ST)S(ST) - S(ST) \ \ \ \ \

Then You may insert it in any place between {S(ST)^{a_{i}}S(ST)^{a_{i+1}}} which gives expression above and many more if You consider {T^p} for {p \in Z} etc.

Remark 1: proved here formula should be possible also to prove from recursion relation directly – so if You would like to try this way – please let me know.

Remark 2: I post this on mathoverflow, and someone provide the sketch of proof by induction – but it seems to be only a sketch – and rather incomplete.