A130242 Minimal index k of a Lucas number such that Lucas(k)>=n (the 'upper' Lucas (A000032) Inverse).

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

Offset: 0

Author

Hieronymus Fischer, May 19 2007, Jul 02 2007

Keywords

• nonn

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).

Examples

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

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000

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.

Programs

Mathematica

Join[{0, 0, 0}, Table[Ceiling[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* G. C. Greubel, Dec 24 2017 *)

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

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.