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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 *) PROG (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 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 October 7 07:33 EDT 2022. Contains 357270 sequences. (Running on oeis4.)