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A071851
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The limit of a continued fraction.
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1
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1, 4, 6, 8, 4, 4, 8, 8, 4, 2, 5, 0, 8, 9, 4, 8, 5, 9, 8, 7, 8, 9, 8, 1, 6, 4, 8, 8, 4, 8, 4, 6, 6, 8, 8, 8, 6, 9, 2, 0, 0, 0, 5, 7, 3, 7, 3, 4, 0, 5, 2, 9, 7, 2, 0, 9, 1, 6, 4, 4, 6, 7, 1, 2, 1, 4, 2, 9, 5, 1
(list; constant; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Consider the following Quet-like continued fraction recursion: x_0 = 1, x_1 = 1 + 1/x_1, x_2 = 1 + 1/(x_1 + 1/x_2), x_3 = 1 + 1/(x_1 + 1/(x_2 + 1/x_3)), ... x_n = 1 + 1/(x_1 + 1/(x_2 + 1/(x_3 + ... + 1/x_n))).
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REFERENCES
| Paul D. Hanna (pauldhanna(AT)juno.com), in a posting to sci.math news group entitled 'Limit of Continued Fraction Recursion' dated Jul 29 2002, 12:22.
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EXAMPLE
| Initial convergents are: x_0 = 1, x_1 = 1.618033988749895, x_2 = 1.431683416590579, x_3 = 1.477931798482607, x_4 = 1.466017958390778, x_5 = 1.469072006453889, x_6 = 1.468289031006081, x_7 = 1.468489818230137, x_8 = 1.468438335506229, x_9 = 1.468451536645148, x_10 = 1.46844815169046, ...
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MAPLE
| f := proc(m, n) local u, x, expr, k; expr := x; u := 1; for k to n do expr := simplify(subs(x = u + 1/x, expr)); u := fsolve(x = expr, x = 0 .. 2); if m < k then print(k, u) end if end do end proc
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CROSSREFS
| The partial quotients are in A072981.
Sequence in context: A201520 A179022 A021685 * A083256 A083257 A201336
Adjacent sequences: A071848 A071849 A071850 * A071852 A071853 A071854
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KEYWORD
| nonn,cons
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AUTHOR
| Paul D. Hanna (pauldhanna(AT)juno.com), Robert G. Wilson v (rgwv(AT)rgwv.com) and Denis Feldmann (denis.feldmann(AT)wanadoo.fr), Jul 29 2002
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