A064170 a(1) = 1; a(n+1) = product of numerator and denominator in Sum_{k=1..n} 1/a(k).

1, 1, 2, 10, 65, 442, 3026, 20737, 142130, 974170, 6677057, 45765226, 313679522, 2149991425, 14736260450, 101003831722, 692290561601, 4745030099482, 32522920134770, 222915410843905, 1527884955772562, 10472279279564026, 71778070001175617, 491974210728665290

Offset

1,3

Comments

The numerator and denominator in the definition have no common divisors >1.

Also denominators in a system of Egyptian fraction for ratios of consecutive Fibonacci numbers: 1/2 = 1/2, 3/5 = 1/2 + 1/10, 8/13 = 1/2 + 1/10 + 1/65, 21/34 = 1/2 + 1/10 + 1/65 + 1/442 etc. (Rossi and Tout). - Barry Cipra, Jun 06 2002

a(n)-1 is a square. - Sture Sjöstedt, Nov 04 2011

From Wolfdieter Lang, May 26 2020: (Start)

Partial sums of the reciprocals: Sum_{k=1..n} 1/a(k) equal 1 for n=1, and F(2*n - 1)/F(2*n - 3) for n >= 2, with F = A000045. Proof by induction. Hence a(n) = 1 for n=1, and F(2*n - 3)*F(2*n - 5) for n >= 2, with F(-1) = 1 (gcd(F(n), F(n+1) = 1). See the comment by Barry Cipra.

Thus a(n) = 1, for n = 1, and a(n) = 1 + F(2*(n-2))^2, for n >= 2 (by Cassini-Simson for even index, e.g., Vajda, p. 178 eq.(28)). See the Sture Sjöstedt comment.

The known G.f. of {F(2*n)^2} from A049684 leds then to the conjectured formula by R. J. Mathar below, and this proves also the recurrence given there..

From the partial sums the series Sum_{k>=1} 1/a(k) converges to 1 + phi, with phi = A001622. See the formulas by Gary W. Adamson and Diego Rattaggi below. (End)

References

S. Vajda, Fibonacci & Lucas Numbers, and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Links

Michael De Vlieger, Table of n, a(n) for n = 1..1199

Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.

Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.

Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.

C. Rossi and C. A. Tout, Were the Fibonacci Series and the Golden Section Known in Ancient Egypt?, Historia Mathematica, vol. 29 (2002), 101-113.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Formula

a(n) = Fibonacci(2*n-5)*Fibonacci(2*n-3), for n >= 3. - Barry Cipra, Jun 06 2002

Sum_{n>=3} 1/a(n) = 2/(1+sqrt(5)) = phi - 1, with phi = A001622. - Gary W. Adamson, Jun 07 2003

Conjecture: a(n) = 8*a(n-1)-8*a(n-2)+a(n-3), n>4. G.f.: -x*(2*x^2+x^3-7*x+1)/((x-1)*(x^2-7*x+1)). - R. J. Mathar, Jul 03 2009 [For a proof see the W. Lang comment above.]

a(n+1) = (A005248(n)^2 - A001906(n)^2)/4, for n => 0. - Richard R. Forberg, Sep 05 2013

From Diego Rattaggi, Apr 21 2020: (Start)

a(n) = 1 + A049684(n-2) for n>1.

Sum_{n>=2} 1/a(n) = phi = (1+sqrt(5))/2 = A001622.

Sum_{n>=1} 1/a(n) = phi^2 = 1 + phi. (End) [See a comment above for the proof]

a(n) = F(2*n - 3)*F(2*n - 5) = 1 + F(2*(n - 2))^2, for n >= 2, with F(-1) = 1. See the W. Lang comments above. - Wolfdieter Lang, May 26 2020

Example

1/a(1) + 1/a(2) + 1/a(3) + 1/a(4) = 1 + 1 + 1/2 + 1/10 = 13/5. So a(5) = 13 * 5 = 65.

Mathematica

A064170[1] := 1; A064170[n_] := A064170[n] = Module[{temp = Sum[1/A064170[i], {i, n - 1}]}, Numerator[temp] Denominator[temp]]; Table[A064170[n], {n, 20}](* Alonso del Arte, Sep 05 2013 *)
Join[{1}, LinearRecurrence[{8, -8, 1}, {1, 2, 10}, 23]] (* Jean-François Alcover, Sep 22 2017 *)

Crossrefs

Cf. A000045, A059929, A058038.

Cf. A033890 (first differences). - R. J. Mathar, Jul 03 2009

Cf. A001906, A001622.

Sequence in context: A371547 A130721 A167449 * A151410 A230050 A340467

Adjacent sequences: A064167 A064168 A064169 * A064171 A064172 A064173

Keyword

nonn,easy

Author

Leroy Quet, Sep 19 2001

Status

approved