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Search: id:A159243
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| A159243 |
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Number of elements in the continued fraction for sum(k=0,n,1/1+2^2^k) |
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2
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| 2, 4, 8, 15, 24, 41, 85, 159, 314, 651, 1267, 2496, 4977, 9889, 19731, 38945, 77356, 154693, 308051, 615768, 1229080
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Number of terms in the n-th partial sum of the Fermat number reciprocals.
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LINKS
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Daniel Duverney, Irrationality of Fast Converging Series of Rational Numbers, Discrete Math., 294 (2005), 259-274. , On non-squashing partitions, Discrete Math., 294 (2005), 259-274.
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EXAMPLE
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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
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MATHEMATICA
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Table[Length[ContinuedFraction[Sum[1/(1 + 2^2^k), {k, 0, v}]]], {v, 0, 20}]
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CROSSREFS
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Cf. A056469
Cf. A051158 [From Barbarel Tres Mil (barbarel3000(AT)yahoo.es), Nov 17 2009]
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KEYWORD
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nonn
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AUTHOR
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Barbarel Tres Mil (barbarel3000(AT)yahoo.es), Apr 06 2009
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