A130247 Inverse Lucas (A000032) numbers: index k of a Lucas number such that Lucas(k)=n; max(k|Lucas(k) < n), if there is no such index.

1, 0, 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), since a(Lucas(n))=n for n >= 0 (see A130241 and A130242 for other versions). Same as A130241 except for n=1.

Links

G. C. Greubel, Table of n, a(n) for n = 1..10000

Formula

a(n)=c(n), if (n^2-4)/5 is a square number, a(n)=s(n), if (n^2+4)/5 is a square number and a(n)=floor(log_phi(n)) otherwise, where s(n)=floor(arcsinh(n/2)/log(phi)), c(n)=floor(arccosh(n/2)/log(phi)) and phi=(1+sqrt(5))/2.

a(n) = A130241(n) except for n=2.

G.f.: g(x) = (1/(1-x))*(Sum_{k>=1} x^Lucas(k)) - x^2.

a(n) = floor(log_phi(n+1/2)) for n >= 3, where phi is the golden ratio.

Example

a(2)=0, since Lucas(0)=2; a(10)=4, since Lucas(4) = 7 < 10 but Lucas(5) = 11 > 10.

Mathematica

Join[{1, 0}, Table[Floor[Log[GoldenRatio, n + 1/2]], {n, 3, 50}]] (* G. C. Greubel, Dec 21 2017 *)

Prog

(Python)
from itertools import islice, count
def A130247_gen(): # generator of terms
    yield from (1, 0)
    a, b = 3, 4
    for i in count(2):
        yield from (i, )*(b-a)
        a, b = b, a+b
A130247_list = list(islice(A130247_gen(), 40)) # Chai Wah Wu, Jun 08 2022

Crossrefs

For partial sums see A130248. Other related sequences: A000032, A130241, A130242, A130245, A130249, A130255, A130259. Indicator sequence A102460. For Fibonacci inverse see A130233 - A130240, A104162.

Sequence in context: A327008 A316846 A130241 * A209869 A087839 A106742

Adjacent sequences: A130244 A130245 A130246 * A130248 A130249 A130250

Keyword

nonn

Author

Hieronymus Fischer, May 19 2007, Jul 02 2007

Status

approved