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A130241
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Maximal index k of a Lucas number such that Lucas(k) <= n (the 'lower' Lucas (A000032) Inverse).
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25
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1, 1, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9
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OFFSET
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1,3
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COMMENTS
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Inverse of the Lucas sequence (A000032), nearly, since a(Lucas(n))=n for n>=1 (see A130242 and A130247 for other versions). For n>=2, a(n)+1 is equal to the partial sum of the Lucas indicator sequence (see A102460). Identical to A130247 except for n=2.
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LINKS
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FORMULA
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a(n) = floor(log_phi((n+sqrt(n^2+4))/2)) = floor(arcsinh((n+1)/2)/log(phi)) where phi=(1+sqrt(5))/2.
G.f.: g(x) = 1/(1-x) * Sum{k>=1, x^Lucas(k)}.
a(n) = floor(log_phi(n+1/2)) for n>=2, where phi is the golden ratio.
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EXAMPLE
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a(10)=4, since Lucas(4)=7<=10 but Lucas(5)=11>10.
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MATHEMATICA
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Join[{1}, Table[Floor[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* G. C. Greubel, Dec 24 2017 *)
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PROG
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(PARI) for(n=1, 50, print1(floor(log((2*n+1)/2)/log((1+sqrt(5))/2)), ", ")) \\ G. C. Greubel, Sep 09 2018
(Magma) [Floor(Log((2*n+1)/2)/Log((1+Sqrt(5))/2)): n in [2..50]]; // G. C. Greubel, Sep 09 2018
(Python)
from itertools import count, islice
def A130241_gen(): # generator of terms
a, b = 1, 3
for i in count(1):
yield from (i, )*(b-a)
a, b = b, a+b
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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