A119996 Numerator of Sum_{k=1..n} 1/(Fibonacci(k)*Fibonacci(k+2)).
1, 5, 14, 39, 103, 272, 713, 1869, 4894, 12815, 33551, 87840, 229969, 602069, 1576238, 4126647, 10803703, 28284464, 74049689, 193864605, 507544126, 1328767775, 3478759199, 9107509824, 23843770273, 62423800997, 163427632718
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).
Programs
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GAP
F:=Fibonacci;; List([1..30], n-> F(n+1)*F(n+2)-1); # G. C. Greubel, Jul 23 2019
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Magma
[Fibonacci(n+1)* Fibonacci(n+2)-1: n in [1..30]]; // Vincenzo Librandi, Aug 14 2012
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Maple
with(combinat): seq(fibonacci(n+1)*fibonacci(n+2)-1, n=1..30); # Zerinvary Lajos, Jan 31 2008
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Mathematica
Numerator[Table[Sum[1/(Fibonacci[k]*Fibonacci[k+2]),{k,n}],{n,30}]] LinearRecurrence[{3,0,-3,1},{1,5,14,39},30] (* Harvey P. Dale, Aug 22 2011 *) -
PARI
vector(30, n, f=fibonacci; f(n+1)*f(n+2)-1) \\ G. C. Greubel, Jul 23 2019
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Sage
f=fibonacci; [f(n+1)*f(n+2)-1 for n in (1..30)] # G. C. Greubel, Jul 23 2019
Formula
a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4); a(0)=1, a(1)=5, a(2)=14, a(3)=39. - Harvey P. Dale, Aug 22 2011
G.f.: ((x-2)*x-1)/(x^4 - 3*x^3 + 3*x - 1). - Harvey P. Dale, Aug 22 2011
a(n) = Fibonacci(n+1)*Fibonacci(n+2) - 1. - Gary Detlefs, Mar 31 2012
a(n) = Sum_{k=1..n} Fibonacci(k+1)^2. Can be proved by induction from Gary Detlefs formula. - Joel Courtheyn, Mar 15 2021