

A130247


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


6



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
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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^24)/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/(1x))*(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 *)


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



