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A269802
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Decimal expansion of the number having (1,2,3,4,...) as its Fibonacci-nested interval sequence.
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1
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6, 8, 4, 7, 8, 8, 2, 6, 7, 5, 2, 2, 6, 7, 4, 7, 9, 3, 3, 8, 2, 4, 5, 5, 8, 2, 0, 0, 3, 7, 0, 5, 8, 3, 3, 1, 3, 2, 5, 4, 7, 8, 8, 5, 2, 8, 6, 2, 6, 3, 4, 2, 3, 9, 4, 6, 5, 2, 8, 6, 9, 2, 2, 1, 6, 4, 5, 1, 2, 7, 4, 6, 2, 9, 8, 2, 6, 9, 2, 4, 1, 7, 7, 8, 4, 9
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OFFSET
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0,1
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COMMENTS
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Suppose that r = (r(n)) is a sequence satisfying (i) 1 = r(1) > r(2) > r(3) > ... and (ii) r(n) -> 0. For x in (0,1], let n(1) be the index n such that r(n+1) , x <= r(n), and let L(1) = r(n(1))-r(n(1)+1). Let n(2) be the index n such that r(n(1)+1) < x <= r(n(1)+1) + L(1)r(n), and let L(2) = (r(n(2))-r(r(n)+1)L(1). Continue inductively to obtain the sequence (n(1), n(2), n(3), ... ), the r-nested interval sequence of x. Taking r = (1/F(n+1)) gives the Fibonacci-nested interval sequence of x. Here, F = A000045, the Fibonacci numbers.)
Conversely, given a sequence s= (n(1),n(2),n(3),...) of positive integers, the number x having satisfying NI(x) = s, is the sum of left-endpoints of nested intervals (r(n(k)+1), r(n(k))]; i.e., x = sum{L(k)r(n(k+1)+1), k >=1}, where L(0) = 1.
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LINKS
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FORMULA
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x = sum{1/(P(k)F(k)), k >= 1}, where P(k) = F(1)*F(2)***F(k+2). F = A000045 (Fibonacci numbers).
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EXAMPLE
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x = 0.68478826752267479338245582003...
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MATHEMATICA
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f[n_] := Fibonacci[n]; p[1] = f[3]; p[n_] := p[n - 1] f[n + 2]
Table[p[i]*f[i], {i, 1, 10}]
s = Sum[1/(p[i] f[i]), {i, 1, 200}]; RealDigits[N[s, 100]][[1]]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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