a[m+2]=a[m+1]^2-a[m] & A Continued Fraction

[cloitre]

le 29/01/03 2:12, Leroy Quet à qqquet at mindspring.com a écrit :

> This is probably well known.
> 
> Let a[1] = 1, a[2] = 2; and for all positive integers m:
> 
> a[m+2] = a[m+1]^2 - a[m].
> 
> (1, 2, 3, 7, 46,...)
> 
> This is sequence A058181 of the EIS, with a shift in the indexing.
> ( 
> http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?A
> num=A058181
> )
> 
> So:
> 
> If c[m] = [a[1];-a[2],a[3],-a[4],a[5],...,+-a[m]],
> 
> where c[m] is the m_th convergent of a continued fraction (where the
> terms are the a[m]'s alternating in sign); then:
> 
> a[m+2] a[m+1] = 
> 
> (c[m] + 1)(c[m+1] + 1) /(c[m] - c[m+1]),
> 
> for all positive integer m's.
> 
> --
> 
> Related: Sequence A058182:
> ( 
> http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?A
> num=A058182
> )
> I don't quite understand what is meant on this page of the EIS under
> "comments".
> What is a "continuant"?
> 

The continuant polynomial C(n,x(1),x(2),....,x(n)) is defined as follows :

C(0,)=1

C(1,x(1))=x(1)

C(n,x(1),x(2),..,x(n))= C(n-1,x(1),..,x(n-1))+C(n-2,x(1),..,x(n-2))

And they are connected to continued fraction many ways :

for example :

a(0)+1a(1)/1+a(2)/1+..../1+a(n)=C(a(0),a(1),...,a(n))/C(a(1),a(2),...,a(n))

Benoit Cloitre