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A130245
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Number of Lucas numbers (A000032) <= n.
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16
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0, 1, 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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0,3
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COMMENTS
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Partial sums of the Lucas indicator sequence A102460.
For n>=2, we have a(A000032(n)) = n + 1.
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LINKS
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FORMULA
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a(n) = 1 +floor(log_phi((n+sqrt(n^2+4))/2)) = 1 +floor(arcsinh(n/2)/log(phi)) for n>=2, where phi = (1+sqrt(5))/2.
G.f.: g(x) = 1/(1-x)*sum{k>=0, x^Lucas(k)}.
a(n) = 1 +floor(log_phi(n+1/2)) for n>=1, where phi is the golden ratio.
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EXAMPLE
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a(9)=5 because there are 5 Lucas numbers <=9 (2,1,3,4 and 7).
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MATHEMATICA
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Join[{0}, Table[1+Floor[Log[GoldenRatio, (2*n+1)/2]], {n, 1, 100}]] (* G. C. Greubel, Sep 09 2018 *)
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PROG
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(PARI)
A102460(n) = { my(u1=1, u2=3, old_u1); if(n<=2, sign(n), while(n>u2, old_u1=u1; u1=u2; u2=old_u1+u2); (u2==n)); };
\\ Or just as:
(Magma) [0] cat [1+Floor(Log((2*n+1)/2)/Log((1+Sqrt(5))/2)): n in [1..100]]; // G. C. Greubel, Sep 09 2018
(Python)
from itertools import count, islice
def A130245_gen(): # generator of terms
yield from (0, 1, 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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