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A130242 Minimal index k of a Lucas number such that Lucas(k)>=n (the 'upper' Lucas (A000032) Inverse). 10
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,4
COMMENTS
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).
LINKS
FORMULA
a(n) = ceiling(log_phi((n+sqrt(n^2-4))/2))=ceiling(arccosh(n/2)/log(phi)) where phi=(1+sqrt(5))/2.
a(n) = A130241(n-1) + 1 = A130245(n-1) for n>=3.
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.
EXAMPLE
a(10)=5, since Lucas(5)=11>=10 but Lucas(4)=7<10.
MATHEMATICA
Join[{0, 0, 0}, Table[Ceiling[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* G. C. Greubel, Dec 24 2017 *)
PROG
(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
A130242_list = list(islice(A130242_gen(), 40)) # Chai Wah Wu, Jun 08 2022
CROSSREFS
For partial sums see A130244.
Other related sequences: A000032, A130241, A130245, A130247, A130250, A130256, A130260.
Indicator sequence A102460.
Fibonacci inverse see A130233 - A130240, A104162.
Sequence in context: A225486 A325282 A305233 * A130245 A087793 A030411
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, May 19 2007, Jul 02 2007
STATUS
approved

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Last modified March 28 18:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)