a[m+2]=a[m+1]^2-a[m] & A Continued Fraction
Leroy Quet
qqquet at mindspring.com
Wed Jan 29 02:12:25 CET 2003
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"?
Thanks,
Leroy Quet
(Posted this to sci.math as well.)
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