A269803 -id:A269803 - cp's OEIS frontend

A305412 a(n) = F(n)*F(n+1) + F(n+2), where F = A000045 (Fibonacci numbers).

1, 3, 5, 11, 23, 53, 125, 307, 769, 1959, 5039, 13049, 33929, 88451, 230957, 603667, 1578823, 4130829, 10810469, 28295411, 74067401, 193893263, 507590495, 1328842801, 3478880593, 9107706243, 23844088085, 62424315227, 163428464759, 427860443429, 1120151837069

Offset: 0

Vincenzo Librandi, Jun 05 2018

Index entries for linear recurrences with constant coefficients, signature (3,1,-5,-1,1).

Cf. A001654, A269803.
Cf. A059769: F(n)*F(n+1) - F(n+2), with offset 3.
Equals A000045 + A286983.
First differences are listed in A059727 (after 0).

List([0..35], n -> Fibonacci(n)*Fibonacci(n+1)+Fibonacci(n+2)); # Muniru A Asiru, Jun 06 2018

[Fibonacci(n)*Fibonacci(n+1)+Fibonacci(n+2): n in [0..30]];

Table[Fibonacci[n] Fibonacci[n+1] + Fibonacci[n+2], {n, 0, 30}]

G.f.: (1 - 5*x^2 - 2*x^3 + x^4)/((x + 1)*(1 - 3*x + x^2)*(1 - x - x^2)).
a(n) = 3*a(n-1) + a(n-2) - 5*a(n-3) - a(n-4) + a(n-5).
5*a(n) = (-1)^(n+1) +5*F(n+2) + A002878(n). - R. J. Mathar, Nov 14 2019

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

Original entry on oeis.org

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

Clark Kimberling, Mar 05 2016

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.

			x = 0.68478826752267479338245582003...

Cf. A000027, A000045, A269803.

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]]

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

cp's OEIS Frontend

A305412 a(n) = F(n)*F(n+1) + F(n+2), where F = A000045 (Fibonacci numbers).

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