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Search: id:A138742
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| A138742 |
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Let r_1 = 1. Let r_{m+1} = r_1 + 1/(r_2 + 1/(r_3 +...(r_{m-1} + 1/r_m)...)), a continued fraction of rational terms. Then row n of this irregular array contains the simple continued fraction terms of r_n. |
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+0
3
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| 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 4, 23
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OFFSET
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1,3
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COMMENT
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The number of terms in the continued fraction of r_n is A138743(n).
The continued fraction of the limit, as n->infinity, of r_n is sequence A053978.
r_n = A064845(n)/A064846(n).
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EXAMPLE
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{r_n}: 1, 1, 2, 5/3, 31/18, 1231/720,...
r_5 = 31/18, for instance, equals the simple continued fraction 1+ 1/(1 + 1/(2 + 1/(1 + 1/(1 +1/2)))). So, row 5 is (1,1,2,1,1,2).
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CROSSREFS
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Cf. A053978, A064845, A064846, A138743, A138744.
Sequence in context: A115570 A069288 A097795 this_sequence A082647 A029322 A112195
Adjacent sequences: A138739 A138740 A138741 this_sequence A138743 A138744 A138745
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KEYWORD
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more,nonn,tabf
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AUTHOR
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Leroy Quet (qq-quet(AT)mindspring.com), Mar 27 2008
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