[seqfan] Recurrences for continued fractions of sqrt(k)?

Georg.Fischer georg.fischer at t-online.de
Fri Feb 19 18:48:17 CET 2021


Dear Seqfans,

the OEIS has a large collection of the - periodic - continued
fractions of sqrt(n): periods, numerators/denominators of convergents,
period lengths etc. (cf. 
<https://oeis.org/wiki/Index_to_continued_fractions_for_sqrt(n)>)

For most numerator/denominator sequences the signature of
a linear recurrence with constant coefficients is given, but
in <https://oeis.org/A041555> for example it is only conjectured.

In general, A006702 states:
"*Apparently* the generating function of the sequence of the 
denominators of continued fraction convergents to sqrt(n) is always 
rational and of form p(x)/[1 - C*x^m + x^(2m)], or equivalently, the 
denominators satisfy the linear recurrence b(n+2m) = C*b(n+m) - b(n). If 
so, then *it seems* that a(n) is half the value of C for each nonsquare 
n, or 1. See A003285 for the conjecture regarding m. The same 
conjectures apply to the sequences of the numerators of continued 
fraction convergents to sqrt(n)."

My question is this conjecture - I'm rather convinced that it
should be true.

A shorter example shows the pattern:

A041552	Numerators of continued fraction convergents to sqrt(294).
17, 103, 120, 583, 703, *4801*, 163937, 988423 ...
Period of CF: 17;6,1,4,1,6,34 (cf. my table in A308778)
Recurrence signature: (0,0,0,0,0,*9602*,0,0,0,0,0,-1)

I checked a set of cases; the central element C of the signature
is always 2*numerator(m), where m is the period length, as
highlighted in the example.

I suppose that there is somebody in our group how says
immediately: Oh, trivial, this is known because ...?

Then we could remove these uncertainities in A006702,
A003285 etc., and maybe the central terms C could even
end up in a new (derived) sequence.

Best regards
Georg




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