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A010215 Continued fraction for sqrt(167). 2

%I #26 Nov 22 2023 08:31:29

%S 12,1,11,1,24,1,11,1,24,1,11,1,24,1,11,1,24,1,11,1,24,1,11,1,24,1,11,

%T 1,24,1,11,1,24,1,11,1,24,1,11,1,24,1,11,1,24,1,11,1,24,1,11,1,24,1,

%U 11,1,24,1,11,1,24,1,11,1,24,1

%N Continued fraction for sqrt(167).

%H G. Xiao, <a href="http://wims.unice.fr/~wims/en_tool~number~contfrac.en.html">Contfrac</a>.

%H <a href="/index/Con#confC">Index entries for continued fractions for constants</a>.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,1).

%F From _Amiram Eldar_, Nov 22 2023: (Start)

%F Multiplicative with a(2) = 11, a(2^e) = 24 for e >= 2, and a(p^e) = 1 for an odd prime p.

%F Dirichlet g.f.: zeta(s) * (1 + 5/2^(s-1) + 13/4^s). (End)

%t ContinuedFraction[Sqrt[167],300] (* _Vladimir Joseph Stephan Orlovsky_, Mar 25 2011 *)

%K nonn,cofr,easy,mult

%O 0,1

%A _N. J. A. Sloane_

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