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%I #7 Mar 07 2016 12:40:02
%S 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,
%T 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,
%U 2,7,4,6,2,9,8,2,6,9,2,4,1,7,7,8,4,9
%N Decimal expansion of the number having (1,2,3,4,...) as its Fibonacci-nested interval sequence.
%C 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.)
%C 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.
%F x = sum{1/(P(k)F(k)), k >= 1}, where P(k) = F(1)*F(2)***F(k+2). F = A000045 (Fibonacci numbers).
%e x = 0.68478826752267479338245582003...
%t f[n_] := Fibonacci[n]; p[1] = f[3]; p[n_] := p[n - 1] f[n + 2]
%t Table[p[i]*f[i], {i, 1, 10}]
%t s = Sum[1/(p[i] f[i]), {i, 1, 200}]; RealDigits[N[s, 100]][[1]]
%Y Cf. A000027, A000045, A269803.
%K nonn,cons,easy
%O 0,1
%A _Clark Kimberling_, Mar 05 2016
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