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

Author

Leroy Quet, Sep 19 2001

Keywords

• nonn
• easy

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)

Examples

			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.
		

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).

Crossrefs

Cf. A000045, A059929, A058038.

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

Cf. A001906, A001622.

Programs

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 *)

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