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%I #21 Jun 09 2022 02:27:03
%S 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,
%T 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,
%U 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
%N Inverse Lucas (A000032) numbers: index k of a Lucas number such that Lucas(k)=n; max(k|Lucas(k) < n), if there is no such index.
%C 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.
%H G. C. Greubel, <a href="/A130247/b130247.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n)=c(n), if (n^2-4)/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.
%F a(n) = A130241(n) except for n=2.
%F G.f.: g(x) = (1/(1-x))*(Sum_{k>=1} x^Lucas(k)) - x^2.
%F a(n) = floor(log_phi(n+1/2)) for n >= 3, where phi is the golden ratio.
%e a(2)=0, since Lucas(0)=2; a(10)=4, since Lucas(4) = 7 < 10 but Lucas(5) = 11 > 10.
%t Join[{1, 0}, Table[Floor[Log[GoldenRatio, n + 1/2]], {n, 3, 50}]] (* _G. C. Greubel_, Dec 21 2017 *)
%o (Python)
%o from itertools import islice, count
%o def A130247_gen(): # generator of terms
%o yield from (1,0)
%o a, b = 3, 4
%o for i in count(2):
%o yield from (i,)*(b-a)
%o a, b = b, a+b
%o A130247_list = list(islice(A130247_gen(),40)) # _Chai Wah Wu_, Jun 08 2022
%Y 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.
%K nonn
%O 1,3
%A _Hieronymus Fischer_, May 19 2007, Jul 02 2007
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