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

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

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: A339507 A305992 A082562 * A325840 A347764 A324740

Adjacent sequences: A159240 A159241 A159242 * A159244 A159245 A159246

Keyword

nonn,more

Author

Enrique Pérez Herrero, Apr 06 2009

Status

approved