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A059929
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a(n) = Fibonacci(n)*Fibonacci(n+2).
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21
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0, 2, 3, 10, 24, 65, 168, 442, 1155, 3026, 7920, 20737, 54288, 142130, 372099, 974170, 2550408, 6677057, 17480760, 45765226, 119814915, 313679522, 821223648, 2149991425, 5628750624, 14736260450, 38580030723, 101003831722, 264431464440, 692290561601
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OFFSET
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0,2
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COMMENTS
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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
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
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
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
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
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
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REFERENCES
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Oskar Perron, Die Lehre von den Kettenbrüchen, Band I, 3. Auflage, B. G. Teubner, Stuttgart, 1954, pp. 61-61.
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LINKS
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FORMULA
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G.f.: (2*x-x^2) / ((1+x)*(1-3*x+x^2)).
Sum_{n>=1} 1/a(n) = 1.
Sum_{n>=1} (-1)^n/a(n) = 2 - sqrt(5).
Sum_{n>=1} 1/a(2n-1) = 1/phi = (sqrt(5) - 1)/2. - Franz Vrabec, Sep 15 2005
Sum_{n>=1} 1/a(2n) = (3 - sqrt(5))/2. - Franz Vrabec, Nov 30 2009
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
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
Sum_{n>=1} Fibonacci(n+1)/a(n) = 2. - Amiram Eldar, Jan 11 2022
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
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EXAMPLE
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G.f. = 2*x + 3*x^2 + 10*x^3 + 24*x^4 + 65*x^5 + 168*x^6 + ... - Michael Somos, Mar 18 2022
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MAPLE
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with(combinat): a:=n->fibonacci(n)*fibonacci(n+2): seq(a(n), n=0..26); # Zerinvary Lajos, Oct 07 2007
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MATHEMATICA
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PROG
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(PARI) for (n=0, 500, write("b059929.txt", n, " ", fibonacci(n)*fibonacci(n + 2))) \\ Harry J. Smith, Jun 30 2009
(Magma) [Fibonacci(n)*Fibonacci(n+2): n in [0..30]]; // Vincenzo Librandi, Jul 02 2014
(Sage) [fibonacci(n)*fibonacci(n+2) for n in range(30)] # G. C. Greubel, Nov 21 2018
(GAP) a:=List([0..30], n->Fibonacci(n)*Fibonacci(n+2));; Print(a); # Muniru A Asiru, Jan 05 2019
(Python)
from sympy import fibonacci
[fibonacci(n)*fibonacci(n+2) for n in range(30)] # Stefano Spezia, Jan 05 2019
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CROSSREFS
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Cf. A000045, A001519, A001654, A007598, A033999, A035513, A079472, A121801, A126358, A147973, A248161.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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