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%I #128 Sep 08 2022 08:45:03
%S 0,2,3,10,24,65,168,442,1155,3026,7920,20737,54288,142130,372099,
%T 974170,2550408,6677057,17480760,45765226,119814915,313679522,
%U 821223648,2149991425,5628750624,14736260450,38580030723,101003831722,264431464440,692290561601
%N a(n) = Fibonacci(n)*Fibonacci(n+2).
%C Expansion of golden ratio (1+sqrt(5))/2 as an infinite product: phi = Product_{i>=0} (1+1/(Fibonacci(2*i+1) * Fibonacci(2*i+3)-1)) * (1-1/(Fibonacci(2*i+2) * Fibonacci(2i+4)+1)). - _Thomas Baruchel_, Nov 11 2003
%C Each of these is one short of or one over the square of a Fibonacci number (A007598). This means that a rectangle sized F(n) by F(n + 2) units can't be converted into a square with sides of length F(n + 1) units unless one square unit of material is added or removed. - _Alonso del Arte_, May 03 2011
%C These are the integer parts of the numerators of the numbers with continued fraction representations [1, 2, 2, 2, ...], [1, 1, 2, 2, 2, ...], [1, 1, 1, 2, 2, 2, ...], etc., that is, sqrt(2), (2+sqrt(2))/2, 3-sqrt(2), (10+sqrt(2))/7, (24-sqrt(2))/14, etc. - _Geoffrey Caveney_, May 03 2014
%C a(n) appears also as the third component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, where F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), A079472(n+1), a(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - _Wolfdieter Lang_, Nov 01 2014
%C Numbers with a continued fraction expansion with the repeating sequence of length n [1, 1, ..., 1, 2], n-1 ones followed by a single two, for n > = 1, appear to be equal to (F(n) + sqrt(a(n)))/F(n+1), where F(n) = A000045(n). - _R. James Evans_, Nov 21 2018
%C The preceding conjecture is true. Proof: For n >= 1 let c(n) := confrac(repeat(1^{n-1}, 2)) where 1^{k} denotes 1 taken k times. This can be computed, e.g. from [Perron, third and fourth eq. on p. 62], as c(n) = (F(n) + sqrt(F(n+1)^2 - (-1)^n)) / F(n+1), which is the conjectured formula because F(n+1)^2 - (-1)^n = a(n). - _Wolfdieter Lang_, Jan 05 2019
%D Oskar Perron, Die Lehre von den Kettenbrüchen, Band I, 3. Auflage, B. G. Teubner, Stuttgart, 1954, pp. 61-61.
%H Muniru A Asiru, <a href="/A059929/b059929.txt">Table of n, a(n) for n = 0..2374</a> (first 501 terms from Harry J. Smith)
%H Tim Jones (producer), <a href="http://www.youtube.com/watch?v=TRNhnyGSVsA">Engineering Bits & Bytes: The Fibonacci Puzzle</a>, Wayne State University College of Engineering (2011).
%H Eric H. Kuo, <a href="https://arxiv.org/abs/math/0304090">Applications of graphical condensation for enumerating matchings and tilings</a>, arXiv:math/0304090 [math.CO], 2003.
%H Marc Renault, <a href="https://www.researchgate.net/publication/237102771_The_Fibonacci_Sequence_Under_Various_Moduli">Properties of the Fibonacci Sequence Under Various Moduli</a>, Master's Thesis, Wake Forest University, 1996.
%H Michel Waldschmidt, <a href="https://arxiv.org/abs/math/0312440">Open Diophantine problems</a>, arXiv:math/0312440 [math.NT], 2003-2004.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F a(n) = Fibonacci(n+1)^2 - (-1)^n = A007598(n+1) + A033999(n+1) = A000045(n+1)^2 - A033999(n).
%F G.f.: (2*x-x^2) / ((1+x)*(1-3*x+x^2)).
%F Sum_{n>=1} 1/a(n) = 1.
%F Sum_{n>=1} (-1)^n/a(n) = 2 - sqrt(5).
%F Sum_{n>=1} 1/a(2n-1) = 1/phi = (sqrt(5) - 1)/2. - _Franz Vrabec_, Sep 15 2005
%F Sum_{n>=1} 1/a(2n) = (3 - sqrt(5))/2. - _Franz Vrabec_, Nov 30 2009
%F a(n) = ((7+3*sqrt(5))/10)*((3+sqrt(5))/2)^(n-1) + ((7-3*sqrt(5))/10)*((3-sqrt(5))/2)^(n-1) + (3/5)*(-1)^(n-1). - _Tim Monahan_, Aug 09 2011
%F a(n) = (Lucas(n+1)^2 - Fibonacci(n+1)^2)/4. - _Vincenzo Librandi_, Aug 02 2014
%F a(n) = F(n-2)*F(n) + F(n-1)*F(n) + F(n-2)*F(n+1) + F(n-1)*F(n+1), where F=A000045, F(-2)=-1, F(-1)=1. - _Bruno Berselli_, Nov 03 2015
%F a(n) = A035513(1,n-1)*A035513(3,n-1)/2 = A035513(1,n-1)*A035513(4,n-1)/3. - _R. J. Mathar_, Sep 04 2016
%F a(n)+a(n+1) = A001519(n+2). - _R. J. Mathar_, Oct 19 2021
%F a(n) = 2*A001654(n) - A001654(n-1). - _R. J. Mathar_, Oct 19 2021
%F a(n)+a(n+3) = 2*F(2n+5) = A126358(n+2). - _Andrés Ventas_, Oct 25 2021
%F Sum_{n>=1} Fibonacci(n+1)/a(n) = 2. - _Amiram Eldar_, Jan 11 2022
%F a(n) = a(-2-n) and a(n) + a(n+3) = 2*(a(n+1) + a(n+2)) for all n in Z. - _Michael Somos_, Mar 18 2022
%e G.f. = 2*x + 3*x^2 + 10*x^3 + 24*x^4 + 65*x^5 + 168*x^6 + ... - _Michael Somos_, Mar 18 2022
%p with(combinat): a:=n->fibonacci(n)*fibonacci(n+2): seq(a(n), n=0..26); # _Zerinvary Lajos_, Oct 07 2007
%t Table[Fibonacci[n]*Fibonacci[n+2],{n,0,60}] (* _Vladimir Joseph Stephan Orlovsky_, Nov 17 2009 *)
%o (PARI) for (n=0, 500, write("b059929.txt", n, " ", fibonacci(n)*fibonacci(n + 2))) \\ _Harry J. Smith_, Jun 30 2009
%o (Magma) [Fibonacci(n)*Fibonacci(n+2): n in [0..30]]; // _Vincenzo Librandi_, Jul 02 2014
%o (Sage) [fibonacci(n)*fibonacci(n+2) for n in range(30)] # _G. C. Greubel_, Nov 21 2018
%o (GAP) a:=List([0..30],n->Fibonacci(n)*Fibonacci(n+2));; Print(a); # _Muniru A Asiru_, Jan 05 2019
%o (Python)
%o from sympy import fibonacci
%o [fibonacci(n)*fibonacci(n+2) for n in range(30)] # _Stefano Spezia_, Jan 05 2019
%Y Bisection of A070550.
%Y First differences of A059840.
%Y Cf. A000045, A001519, A001654, A007598, A033999, A035513, A079472, A121801, A126358, A147973, A248161.
%K nonn,easy
%O 0,2
%A _Henry Bottomley_, Feb 09 2001
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