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Search: id:A064170
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| A064170 |
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a(1) = 1; a(n+1) = product of numerator and denominator in sum{k = 1 to n} 1/a(k). |
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+0
3
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| 1, 1, 2, 10, 65, 442, 3026, 20737, 142130, 974170, 6677057, 45765226, 313679522, 2149991425, 14736260450, 101003831722, 692290561601, 4745030099482, 32522920134770, 222915410843905, 1527884955772562, 10472279279564026
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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The numerator and denominator in the definition have no common divisors >1.
For n >= 3, a(n) = Fibonacci(2*n-5)*Fibonacci(2*n-3). - Barry Cipra (bcipra(AT)rconnect.com), Jun 06 2002
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 (bcipra(AT)rconnect.com), Jun 06 2002
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REFERENCES
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Rossi and Tout, Historia Mathematica, vol. 29 (2002), 101-113.
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LINKS
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Leroy Quet, Home Page (listed in lieu of email address)
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FORMULA
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2/(1+sqrt5) = .6180339... = 1/2 + 1/10 + 1/65 + 1/442 + 1/3026...= Sum (3 through infinity) 1/a(n). - Gary W. Adamson (qntmpkt(AT)yahoo.com), 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)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 03 2009]
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EXAMPLE
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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.
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CROSSREFS
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Cf. A000045, A059929, A058038.
Cf. A033890 (first differences). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 03 2009]
Sequence in context: A078531 A130721 A167449 this_sequence A151410 A027307 A060206
Adjacent sequences: A064167 A064168 A064169 this_sequence A064171 A064172 A064173
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KEYWORD
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nonn
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AUTHOR
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Leroy Quet Sep 19 2001
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