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A130241
Maximal index k of a Lucas number such that Lucas(k) <= n (the 'lower' Lucas (A000032) Inverse).
25
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
1,3
COMMENTS
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.
LINKS
FORMULA
a(n) = floor(log_phi((n+sqrt(n^2+4))/2)) = floor(arcsinh((n+1)/2)/log(phi)) where phi=(1+sqrt(5))/2.
a(n) = A130242(n+1) - 1 for n>=2.
a(n) = A130247(n) except for n=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.
EXAMPLE
a(10)=4, since Lucas(4)=7<=10 but Lucas(5)=11>10.
MATHEMATICA
Join[{1}, Table[Floor[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* G. C. Greubel, Dec 24 2017 *)
PROG
(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
A130241_list = list(islice(A130241_gen(), 40)) # Chai Wah Wu, Jun 08 2022
CROSSREFS
For partial sums see A130243. Other related sequences: A000032, A130242, A130245, A130247, A130249, A130255, A130259. Indicator sequence A102460. Fibonacci inverse see A130233 - A130240, A104162.
Sequence in context: A324728 A327008 A316846 * A130247 A209869 A087839
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, May 19 2007, Jul 02 2007
STATUS
approved