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Continued fraction for sqrt(325).
5

%I #25 Feb 15 2024 13:01:48

%S 18,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,

%T 36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,

%U 36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36,36

%N Continued fraction for sqrt(325).

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

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (1).

%F From _Elmo R. Oliveira_, Feb 14 2024: (Start)

%F a(n) = 36 for n >= 1.

%F G.f.: 18*(1+x)/(1-x).

%F E.g.f.: 36*exp(x) - 18.

%F a(n) = 18*A040000(n) = 9*A040002(n) = 6*A040006(n). (End)

%e 18 + 1/(36 + 1/(36 + 1/(36 + 1/(36 + ...)))) = sqrt(325).

%p with(numtheory): Digits := 300: convert(evalf(sqrt(325)),confrac);

%t Block[{$MaxExtraPrecision=1000}, ContinuedFraction[Sqrt[325],100]] (* or *) PadRight[{18},100,{36}] (* _Harvey P. Dale_, Dec 01 2020 *)

%Y Cf. A041612/A041613 (convergents).

%Y Cf. A040000, A040002, A040006.

%K nonn,cofr,easy

%O 0,1

%A _N. J. A. Sloane_