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

%I A130242 #17 Jun 09 2022 02:26:36
%S A130242 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,
%T A130242 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,
%U A130242 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
%N A130242 Minimal index k of a Lucas number such that Lucas(k)>=n (the 'upper' Lucas (A000032) Inverse).
%C A130242 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).
%H A130242 G. C. Greubel, <a href="/A130242/b130242.txt">Table of n, a(n) for n = 0..5000</a>
%F A130242 a(n) = ceiling(log_phi((n+sqrt(n^2-4))/2))=ceiling(arccosh(n/2)/log(phi)) where phi=(1+sqrt(5))/2.
%F A130242 a(n) = A130241(n-1) + 1 = A130245(n-1) for n>=3.
%F A130242 G.f.: x/(1-x)*(2x^2+sum{k>=2, x^Lucas(k)}).
%F A130242 a(n) = ceiling(log_phi(n-1/2)) for n>=3, where phi is the golden ratio.
%e A130242 a(10)=5, since Lucas(5)=11>=10 but Lucas(4)=7<10.
%t A130242 Join[{0, 0, 0}, Table[Ceiling[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* _G. C. Greubel_, Dec 24 2017 *)
%o A130242 (Python)
%o A130242 from itertools import islice, count
%o A130242 def A130242_gen(): # generator of terms
%o A130242     yield from (0,0,0,2)
%o A130242     a, b = 3, 4
%o A130242     for i in count(3):
%o A130242         yield from (i,)*(b-a)
%o A130242         a, b = b, a+b
%o A130242 A130242_list = list(islice(A130242_gen(),40)) # _Chai Wah Wu_, Jun 08 2022
%Y A130242 For partial sums see A130244.
%Y A130242 Other related sequences: A000032, A130241, A130245, A130247, A130250, A130256, A130260.
%Y A130242 Indicator sequence A102460.
%Y A130242 Fibonacci inverse see A130233 - A130240, A104162.
%K A130242 nonn
%O A130242 0,4
%A A130242 _Hieronymus Fischer_, May 19 2007, Jul 02 2007