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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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1
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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
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OFFSET
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1,1
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COMMENTS
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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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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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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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