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%I #63 Jan 10 2021 11:16:23
%S 1,1,2,10,65,442,3026,20737,142130,974170,6677057,45765226,313679522,
%T 2149991425,14736260450,101003831722,692290561601,4745030099482,
%U 32522920134770,222915410843905,1527884955772562,10472279279564026,71778070001175617,491974210728665290
%N a(1) = 1; a(n+1) = product of numerator and denominator in Sum_{k=1..n} 1/a(k).
%C The numerator and denominator in the definition have no common divisors >1.
%C 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
%C a(n)-1 is a square. - _Sture Sjöstedt_, Nov 04 2011
%C From _Wolfdieter Lang_, May 26 2020: (Start)
%C 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_.
%C 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.
%C 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..
%C 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)
%D S. Vajda, Fibonacci & Lucas Numbers, and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
%H Michael De Vlieger, <a href="/A064170/b064170.txt">Table of n, a(n) for n = 1..1199</a>
%H Christian Aebi and Grant Cairns, <a href="https://arxiv.org/abs/2006.07566">Lattice Equable Parallelograms</a>, arXiv:2006.07566 [math.NT], 2020.
%H Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Nemeth/nemeth7.html">Ellipse Chains and Associated Sequences</a>, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2016volume16/FG2016volume16.pdf#page=423">Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences</a>, Forum Geometricorum, Volume 16 (2016) 419-427.
%H C. Rossi and C. A. Tout, <a href="http://dx.doi.org/10.1006/hmat.2001.2334">Were the Fibonacci Series and the Golden Section Known in Ancient Egypt?</a>, Historia Mathematica, vol. 29 (2002), 101-113.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).
%F a(n) = Fibonacci(2*n-5)*Fibonacci(2*n-3), for n >= 3. - _Barry Cipra_, Jun 06 2002
%F Sum_{n>=3} 1/a(n) = 2/(1+sqrt(5)) = phi - 1, with phi = A001622. - _Gary W. Adamson_, Jun 07 2003
%F 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.]
%F a(n+1) = (A005248(n)^2 - A001906(n)^2)/4, for n => 0. - _Richard R. Forberg_, Sep 05 2013
%F From _Diego Rattaggi_, Apr 21 2020: (Start)
%F a(n) = 1 + A049684(n-2) for n>1.
%F Sum_{n>=2} 1/a(n) = phi = (1+sqrt(5))/2 = A001622.
%F Sum_{n>=1} 1/a(n) = phi^2 = 1 + phi. (End) [See a comment above for the proof]
%F 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
%e 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.
%t 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 *)
%t Join[{1}, LinearRecurrence[{8, -8, 1}, {1, 2, 10}, 23]] (* _Jean-François Alcover_, Sep 22 2017 *)
%Y Cf. A000045, A059929, A058038.
%Y Cf. A033890 (first differences). - _R. J. Mathar_, Jul 03 2009
%Y Cf. A001906, A001622.
%K nonn,easy
%O 1,3
%A _Leroy Quet_, Sep 19 2001
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