A130241 Maximal index k of a Lucas number such that Lucas(k) <= n (the 'lower' Lucas (A000032) Inverse).
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
Offset: 1
Keywords
Examples
a(10)=4, since Lucas(4)=7<=10 but Lucas(5)=11>10.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..5000
Crossrefs
Programs
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Magma
[Floor(Log((2*n+1)/2)/Log((1+Sqrt(5))/2)): n in [2..50]]; // G. C. Greubel, Sep 09 2018
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Mathematica
Join[{1}, Table[Floor[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* G. C. Greubel, Dec 24 2017 *)
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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
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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 A130241_list = list(islice(A130241_gen(),40)) # Chai Wah Wu, Jun 08 2022
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