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A159243 Number of elements in the continued fraction for sum(k=0..n, 1/(1+2^2^k)). 1

%I

%S 2,4,8,15,24,41,85,159,314,651,1267,2496,4977,9889,19731,38945,77356,

%T 154693,308051,615768,1229080

%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, James A. Sellers, <a href="http://arXiv.org/abs/math.CO/0312418">On non-squashing partitions</a>, Discrete Math., 294 (2005), 259-274.

%e The partial sum for k=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: f(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

%O 1,1

%A _Enrique PĂ©rez Herrero_, Apr 06 2009

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Last modified August 21 22:32 EDT 2019. Contains 326169 sequences. (Running on oeis4.)