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A130242
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Minimal index k of a Lucas number such that Lucas(k)>=n (the 'upper' Lucas (A000032) Inverse).
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10
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0, 0, 0, 2, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 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, 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, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10
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OFFSET
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0,4
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COMMENTS
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Inverse of the Lucas sequence (A000032), nearly, since a(Lucas(n))=n except for n=1 (see A130241 and A130247 for other versions). For n>=2, a(n+1) is equal to the partial sum of the Lucas indicator sequence (see A102460).
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LINKS
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FORMULA
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a(n) = ceiling(log_phi((n+sqrt(n^2-4))/2))=ceiling(arccosh(n/2)/log(phi)) where phi=(1+sqrt(5))/2.
G.f.: x/(1-x)*(2x^2+sum{k>=2, x^Lucas(k)}).
a(n) = ceiling(log_phi(n-1/2)) for n>=3, where phi is the golden ratio.
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EXAMPLE
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a(10)=5, since Lucas(5)=11>=10 but Lucas(4)=7<10.
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MATHEMATICA
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Join[{0, 0, 0}, Table[Ceiling[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* G. C. Greubel, Dec 24 2017 *)
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PROG
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(Python)
from itertools import islice, count
def A130242_gen(): # generator of terms
yield from (0, 0, 0, 2)
a, b = 3, 4
for i in count(3):
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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