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

0,1

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.

Links

Table of n, a(n) for n=0..85.

Formula

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

Example

x = 0.68478826752267479338245582003...

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]]

Crossrefs

Cf. A000027, A000045, A269803.

Sequence in context: A195701 A247038 A195492 * A269991 A171784 A200133

Adjacent sequences: A269799 A269800 A269801 * A269803 A269804 A269805

Keyword

nonn,cons,easy

Author

Clark Kimberling, Mar 05 2016

Status

approved