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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

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

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 *)

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

Adjacent sequences:  A130239 A130240 A130241 * A130243 A130244 A130245

KEYWORD

nonn

AUTHOR

Hieronymus Fischer, May 19 2007, Jul 02 2007

STATUS

approved

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Last modified July 24 06:04 EDT 2021. Contains 346273 sequences. (Running on oeis4.)