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%I #22 May 05 2024 08:56:07
%S 2,4,8,15,24,41,85,159,314,651,1267,2496,4977,9889,19731,38945,77356,
%T 154693,308051,615768,1229080,2456328,4908126,9815038,19620985,
%U 39237465,78466413,156910438,313788371,627528817
%N Number of elements in the continued fraction for Sum_{k=0..n} 1/(1+2^(2^k)).
%C Number of terms in the n-th partial sum of the Fermat number reciprocals.
%H Daniel Duverney, <a href="http://journal.ms.u-tokyo.ac.jp/pdf/jms080206.pdf">Irrationality of Fast Converging Series of Rational Numbers</a>, J. Math. Sci. Univ. Tokyo, 8 (2001), 275-316.
%H N. J. A. Sloane and James A. Sellers, <a href="https://doi.org/10.1016/j.disc.2004.11.014">On non-squashing partitions</a>, Discrete Mathematics, Vol. 294, No. 3 (2005), pp. 259-274; <a href="https://arxiv.org/abs/math/0312418">arXiv preprint</a>, arXiv:math/0312418 [math.CO], 2003.
%e The partial sum for n = 3 (four terms) is: 1/3 + 1/5 + 1/17 + 1/257 = 39062/65535 expressed in continued fraction gives: {0,1,1,2,9,1,2,1,1,2,2,1,2,1,5} that has 15 elements so: a(3) = 15.
%t Table[Length[ContinuedFraction[Sum[1/(1 + 2^2^k), {k, 0, v}]]], {v, 0, 20}]
%Y Cf. A056469, A051158.
%K nonn,more,changed
%O 0,1
%A _Enrique Pérez Herrero_, Apr 06 2009
%E Offset corrected and a(21)-a(29) added by _Amiram Eldar_, May 05 2024
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