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Continued fraction for alpha, the unique solution on [2,oo) of the equation alpha*log((2*e)/alpha)=1.
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%I #17 Jul 03 2024 10:54:32

%S 4,3,4,1,1,1,11,2,19,1,3,1,1,1,14,1,3,5,58,3,1,10,1,1,6,5,13,127,1,1,

%T 7,13,1,2,1,2,2,1,2,2,4,2,4,1,1,6,9,3,1,16,1,3,2,32,3,1,1,2,11,1,13,4,

%U 2,1,1,1,1,2,2,6,1,1,1,2,25,1,5,5,1,1,1,1,5,2,3,2,5,25,1,190,2,1,5,3,1,20,1,1,2,1,3

%N Continued fraction for alpha, the unique solution on [2,oo) of the equation alpha*log((2*e)/alpha)=1.

%C alpha is used to measure the expected height of random binary search trees.

%H B. Reed, <a href="http://doi.acm.org/10.1145/765568.765571">The height of a random binary search tree</a>, J. ACM, 50 (2003), 306-332.

%F alpha = -1/W(-exp(-1)/2), where W is the Lambert W function.

%F A195582(n)/A195583(n) = alpha*log(n) - beta*log(log(n)) + O(1), with beta = 1.953... (A195599).

%e 4.31107040700100503504707609644689027839156299804028805066937...

%p with(numtheory):

%p alpha:= solve(alpha*log((2*exp(1))/alpha)=1, alpha):

%p cfrac(evalf(alpha, 130), 100, 'quotients')[];

%t alpha = -1/ProductLog[-1/(2*E)]; ContinuedFraction[alpha, 101] (* _Jean-François Alcover_, Jun 20 2013 *)

%Y Cf. A195596 (decimal expansion), A195598 (Engel expansion), A195581, A195582, A195583, A195599, A195600, A195601.

%K nonn,cofr

%O 0,1

%A _Alois P. Heinz_, Sep 21 2011

%E Offset changed by _Andrew Howroyd_, Jul 03 2024