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Continued fraction for cube root of 99.
1

%I #16 Sep 08 2022 08:44:37

%S 4,1,1,1,2,14,3,2,1,1,1,21,1,3,287,3,1,1,1,3,1,1,11,1,2,2,7,1,1,1,1,2,

%T 5,2,18,3,1,2,1,6,2,1,5,3,2,1,5,5,1,1,2,2,1,2,1,1,1,10,6,1,7,7,878,1,

%U 4,18,3,2,1,1,8,1,4,1,4,7,10

%N Continued fraction for cube root of 99.

%H Vincenzo Librandi, <a href="/A010327/b010327.txt">Table of n, a(n) for n = 0..3000</a>

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

%t ContinuedFraction[99^(1/3), 80] (* _Vincenzo Librandi_, Oct 08 2017 *)

%o (Magma) ContinuedFraction(99^(1/3)); // _Vincenzo Librandi_, Oct 08 2017

%o (PARI) contfrac(99^(1/3)) \\ _Felix Fröhlich_, Oct 08 2017

%Y Cf. A010669.

%K nonn,cofr

%O 0,1

%A _N. J. A. Sloane_