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Continued fraction for cube root of 58.
1

%I #17 Dec 27 2021 21:41:35

%S 3,1,6,1,2,1,10,1,2,7,1,4,15,37,1,6,1,2,2,1,2,8,10,39,75,1,2,2,1,1,2,

%T 1,10,18,1,4,3,2,1,2,1,2,1,5,1,4,1,1,3,181,1,2,1,5,11,2,16,1,2,1,2,3,

%U 1,1,1,3,3,1,1,15,3,8,3,3,2,2

%N Continued fraction for cube root of 58.

%H Robert Israel, <a href="/A010287/b010287.txt">Table of n, a(n) for n = 0..10000</a>

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

%p numtheory:-cfrac(58^(1/3),80,quotients)[1..-2]; # _Robert Israel_, Mar 10 2020

%t ContinuedFraction[Power[58, (3)^-1],100] (* _Harvey P. Dale_, Jan 15 2012 *)

%o (PARI) contfrac(sqrtn(58, 3)) \\ _Michel Marcus_, Mar 10 2020

%Y Cf. A010629 (cube root of 58).

%K nonn,cofr

%O 0,1

%A _N. J. A. Sloane_