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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, 2456328, 4908126, 9815038, 19620985, 39237465, 78466413, 156910438, 313788371, 627528817
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OFFSET
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0,1
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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 n = 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: a(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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EXTENSIONS
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Offset corrected and a(21)-a(29) added by Amiram Eldar, May 05 2024
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STATUS
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approved
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