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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 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 A324740 A262146 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 April 18 14:20 EDT 2021. Contains 343088 sequences. (Running on oeis4.)