

A130241


Maximal index k of a Lucas number such that Lucas(k) <= n (the 'lower' Lucas (A000032) Inverse).


24



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
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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

G. C. Greubel, Table of n, a(n) for n = 1..5000


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/(1x) * 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


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: A260998 A316846 * A130247 A209869 A087839 A106742
Adjacent sequences: A130238 A130239 A130240 * A130242 A130243 A130244


KEYWORD

nonn


AUTHOR

Hieronymus Fischer, May 19 2007, Jul 02 2007


STATUS

approved



