A269802 Decimal expansion of the number having (1,2,3,4,...) as its Fibonacci-nested interval sequence.

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

Offset: 0

Author

Clark Kimberling, Mar 05 2016

Keywords

• nonn
• cons
• easy

Comments

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.

Examples

			x = 0.68478826752267479338245582003...
		

Crossrefs

Cf. A000027, A000045, A269803.

Programs

Mathematica

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]]

Formula

x = sum{1/(P(k)F(k)), k >= 1}, where P(k) = F(1)*F(2)***F(k+2). F = A000045 (Fibonacci numbers).