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Search: id:A081088
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| A081088 |
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Smallest partial quotients of a series of continued fractions that have a sum of unity, where the n-th continued fraction term is [0; a(n),a(n-1),...,a(1)], so that sum(n>0, [0; a(n),a(n-1),...,a(1)] ) = 1, with a(1)=2. |
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+0
5
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| 2, 2, 10, 260, 703300, 128651592765800, 11640481755119007104771565805489000, 17432395734015190050301181934013684788461125501100342391858949624062957005321114000
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OFFSET
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1,1
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COMMENT
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a(n+1) appears to be divisible by a(n) for n>0; a(n+1)/a(n) = A081089(n). Also log(a(n+1))/log(a(n)) -> 1+sqrt(2). The 8-th term has 79 digits, while the 9-th term has 199 digits.
The convergents of the continued fraction series equal the ratios of the terms of A081090: [0; a(n),a(n-1),...,a(1)] = A081090(n)/A080190(n+1) for n>1; thus a(n) = ( A081090(n+1) - A081090(n-1) )/A081090(n) for n>1. - Paul D. Hanna (pauldhanna(AT)juno.com), Mar 05 2003
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FORMULA
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a(n) = A081090(n)*A081090(n-1) for n>2, where A081090(n)^2 + 1 = a(n+1)/a(n).
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EXAMPLE
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1 = [0;2] + [0;2,2] + [0;10,2,2] + [0;260,10,2,2] + [0;703300,260,10,2,2] + [0;128651592765800,703300,260,10,2,2] +... = .5 + .4 + .0961538461 + .0038447319 + .0000014218 + ...
At n=4, [0;260,10,2,2] = A081090(4)/A081090(5) = 52/13525. - Paul D. Hanna (pauldhanna(AT)juno.com), Mar 05 2003
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CROSSREFS
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Cf. A081086, A081089, A081090.
Adjacent sequences: A081085 A081086 A081087 this_sequence A081089 A081090 A081091
Sequence in context: A011248 A069240 A000371 this_sequence A001038 A027623 A037234
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KEYWORD
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cofr,nonn
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AUTHOR
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Hans Havermann (pxp(AT)rogers.com) and Paul D. Hanna (pauldhanna(AT)juno.com), Mar 05 2003
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