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a(1)=1, a(2)=3, a(n) = a(n-1) + n if the minimal positive integer not yet in the sequencer is greater than a(n-1), else a(n) = a(n-1)-1.
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%I #14 Aug 09 2017 23:07:39

%S 1,3,2,6,5,4,11,10,9,8,7,19,18,17,16,15,14,13,12,32,31,30,29,28,27,26,

%T 25,24,23,22,21,20,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37,

%U 36,35,34,33,87,86,85,84,83,82,81,80,79,78,77,76,75,74,73,72,71,70,69

%N a(1)=1, a(2)=3, a(n) = a(n-1) + n if the minimal positive integer not yet in the sequencer is greater than a(n-1), else a(n) = a(n-1)-1.

%C Also: a(1)=1, a(2)=3, a(n) = maximal positive number < a(n-1) not yet in the sequence, if it exists, else a(n) = a(n-1) + n.

%C Also: a(1)=1, a(2)=3, a(n) = a(n-1) - 1, if a(n-1) - 1 > 0 and has not been encountered so far, else a(n) = a(n-1) + n.

%C A permutation of the positive integers. The sequence is self-inverse, in that a(a(n)) = n.

%F G.f.: g(x) = (F'(x) - x^2 - 1/(1-x))/(1-x) where F(x) = Sum_{k>=0} x^Fibonacci(k). F(x) is the g.f. of the Fibonacci indicator sequence (see A104162) and F'(x) = derivative of F(x).

%F a(n) = Fibonacci(Fibonacci_inverse(n+1) + 2) - n - 3 = A000045(A130233(n+1) + 2) - n - 3.

%F a(n) = A000032(floor(log_phi(sqrt(5)*(n+1) + 1) + 2)) - n - 3, where phi = (1 + sqrt(5))/2 is the golden ratio.

%F a(n) = A000032(floor(log_phi(sqrt(5)*n + 2*phi) + 2)) - n - 3.

%Y Cf. A000045, A104152, A130233.

%Y For an analog concerning Lucas numbers see A132664.

%Y See A132666-A132674 for sequences with a similar recurrence rule.

%K nonn

%O 1,2

%A _Hieronymus Fischer_, Sep 15 2007