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A130241
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Maximal index k of a Lucas number such that Lucas(k)<=n (the 'lower' Lucas (A000032) Inverse).
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23
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1, 1, 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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Inverse of the Lucas sequence (A000032), nearly, since a(Lucas(n))=n for n>=1 (see A130242 and A130247 for other versions). For n>=2, a(n)+1 is equal to the partial sum of the Lucas indicator sequence (see A102460). Identical to A130247 except for n=2.
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FORMULA
| a(n)=floor(log_phi((n+sqr(n^2+4))/2))=floor(arsinh(n/2)/ln(phi)) where phi=(1+sqr(5))/2.
a(n)=A130242(n+1)-1 for n>=2. a(n)=A130247(n) except for n=2.
G.f.: g(x)=1/(1-x)*sum{k>=1, x^Lucas(k)}.
a(n)=floor(log_phi(n+1/2)) for n>=2, where phi is the golden ratio.
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EXAMPLE
| a(10)=4, since Lucas(4)=7<=10 but Lucas(5)=11>10.
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CROSSREFS
| For partial sums see A130243. Other related sequences: A000032, A130242, A130245, A130247, A130249, A130255, A130259. Indicator sequence A102460. Fibonacci inverse see A130233 - A130240, A104162.
Sequence in context: A117806 A085423 * A130247 A087839 A106742 A106733
Adjacent sequences: A130238 A130239 A130240 * A130242 A130243 A130244
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KEYWORD
| nonn
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AUTHOR
| Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), May 19 2007, Jul 02 2007
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