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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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Number of terms in the n-th 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), 275-316.

N. J. A. Sloane, James A. Sellers, On non-squashing partitions, Discrete Math., 294 (2005), 259-274.

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: A301629 A305992 A082562 * A325840 A262146 A089140

Adjacent sequences:  A159240 A159241 A159242 * A159244 A159245 A159246

KEYWORD

nonn,more

AUTHOR

Enrique Pérez Herrero, Apr 06 2009

STATUS

approved

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Last modified July 21 00:39 EDT 2019. Contains 325189 sequences. (Running on oeis4.)