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Search: id:A126336
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| A126336 |
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Irregular table where the first row is (1). Row n is the continued fraction terms of the rational equal to the sum of the reciprocals of all the terms in the previous rows. |
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+0
4
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| 1, 1, 2, 2, 2, 3, 2, 4, 3, 4, 1, 11, 6, 3, 1, 7, 2, 8, 2, 2, 92, 9, 1, 1, 6, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 2, 22, 3, 5, 2, 3, 1, 1, 15, 3, 4, 2, 26, 1, 7, 12, 1, 7, 2, 1, 26, 1, 1, 4, 33, 13, 1, 5, 1, 13, 1, 8, 1, 13, 1, 18, 2, 39, 4, 1, 2, 10, 6, 1, 4, 1, 20, 43, 1, 1, 3, 1, 21, 1, 1, 2, 2, 49, 1
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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The continued fractions, for rows 3 and up, each have a final term >=2. The number of terms in the n-th row is A126337(n). The sum of the reciprocals of the terms in rows 1 through n is A126338(n)/A126339(n).
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LINKS
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Leroy Quet, Home Page (listed in lieu of email address)
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EXAMPLE
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The sum of the reciprocals of the terms of the first 6 rows is 1 +1 +1/2 +1/2 +1/ 2 +1/3 +1/2 +1/4 +1/3 = 59/12. 59/12 equals the continued fraction 4 +1/(1 +1/11). So row 7 is (4,1,11).
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MATHEMATICA
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f[l_List] := Append[l, ContinuedFraction[Plus @@ (1/# &) /@ Flatten[l]]]Flatten@Nest[f, {{1}}, 15] (*Chandler*)
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CROSSREFS
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Cf. A126337, A126338, A126339.
Sequence in context: A076709 A110021 A036013 this_sequence A134446 A125749 A014085
Adjacent sequences: A126333 A126334 A126335 this_sequence A126337 A126338 A126339
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KEYWORD
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easy,nonn,tabf
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AUTHOR
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Leroy Quet Dec 25 2006
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EXTENSIONS
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Extended by Ray Chandler (rayjchandler(AT)sbcglobal.net), Dec 26 2006
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