A195615 Numerators b(n) of Pythagorean approximations b(n)/a(n) to 2.
15, 273, 4895, 87841, 1576239, 28284465, 507544127, 9107509825, 163427632719, 2932589879121, 52623190191455, 944284833567073, 16944503814015855, 304056783818718321, 5456077604922913919, 97905340104793732225, 1756840044281364266127, 31525215456959763058065
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..797
- Index entries for linear recurrences with constant coefficients, signature (17,17,-1).
Programs
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Magma
[Fibonacci(3*n+1)*Fibonacci(3*n+2): n in [1..40]]; // G. C. Greubel, Feb 13 2023
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Mathematica
LinearRecurrence[{17,17,-1}, {15,273,4895}, 40] (* G. C. Greubel, Feb 13 2023 *)
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PARI
Vec(x*(15+18*x-x^2)/((1+x)*(1-18*x+x^2)) + O(x^50)) \\ Colin Barker, Jun 04 2015
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SageMath
[fibonacci(3*n+1)*fibonacci(3*n+2) for n in range(1,41)] # G. C. Greubel, Feb 13 2023
Formula
From Colin Barker, Jun 04 2015: (Start)
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3).
G.f.: x*(15 + 18*x - x^2)/((1+x)*(1-18*x+x^2)). (End)
a(n) = ((-1)^n + (2+sqrt(5))*(9+4*sqrt(5))^n + (2-sqrt(5))*(9+4*sqrt(5))^(-n))/5. - Colin Barker, Mar 04 2016
From G. C. Greubel, Feb 13 2023: (Start)
a(n) = Fibonacci(3*n+1)*Fibonacci(3*n+2).
a(n) = (1/5)*(4*A049629(n) + (-1)^n - 5*[n=0]). (End)
Comments