

A159243


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


1



2, 4, 8, 15, 24, 41, 85, 159, 314, 651, 1267, 2496, 4977, 9889, 19731, 38945, 77356, 154693, 308051, 615768, 1229080
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OFFSET

1,1


COMMENTS

Number of terms in the nth partial sum of the Fermat number reciprocals.


LINKS

Table of n, a(n) for n=1..21.
Daniel Duverney, Irrationality of Fast Converging Series of Rational Numbers, J. Math. Sci. Univ. Tokyo, 8 (2001), 275316.
N. J. A. Sloane, James A. Sellers, On nonsquashing partitions, Discrete Math., 294 (2005), 259274.


EXAMPLE

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


MATHEMATICA

Table[Length[ContinuedFraction[Sum[1/(1 + 2^2^k), {k, 0, v}]]], {v, 0, 20}]


CROSSREFS

Cf. A056469, A051158.
Sequence in context: A339507 A305992 A082562 * A325840 A324740 A262146
Adjacent sequences: A159240 A159241 A159242 * A159244 A159245 A159246


KEYWORD

nonn,more


AUTHOR

Enrique Pérez Herrero, Apr 06 2009


STATUS

approved



