0,3

Partial sums of the Lucas indicator sequence A102460. For n>=2, we have a(A000032(n))=n+1.

a(n)=1+floor(log_phi((n+sqr(n^2+4))/2))=1+floor(arsinh(n/2)/ln(phi)) for n>=2, where phi=(1+sqr(5))/2.

a(n)=A130241(n)+1=A130242(n+1) for n>=2.

G.f.: g(x)=1/(1-x)*sum{k>=0, x^Lucas(k)}.

a(n)=1+floor(log_phi(n+1/2)) for n>=1, where phi is the golden ratio.

a(9)=5 because there are 5 Lucas numbers <=9 (2,1,3,4 and 7).

For partial sums see A130246. Other related sequences: A000032, A130241, A130242, A130247, A130249, A130253, A130255, A130259. Indicator sequence A102460. Fibonacci inverse see A130233 - A130240, A104162, A108852.

Sequence in context: A087827 A136528 A130242 this_sequence A087793 A030411 A132913

Adjacent sequences: A130242 A130243 A130244 this_sequence A130246 A130247 A130248

nonn

Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), May 19 2007, Jul 02 2007