login
a(1)=1, a(2)=2, a(n) = a(n-1) + n if the minimal positive integer not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.
7

%I #18 Sep 23 2024 04:23:37

%S 1,2,5,4,3,9,8,7,6,16,15,14,13,12,11,10,27,26,25,24,23,22,21,20,19,18,

%T 17,45,44,43,42,41,40,39,38,37,36,35,34,33,32,31,30,29,28,74,73,72,71,

%U 70,69,68,67,66,65,64,63,62,61,60,59,58,57,56,55,54,53,52,51,50,49,48

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

%C Also: a(1)=1, a(2)=2, 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)=2, 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) = (L'(x) - x^2 - 1/(1-x))/(1-x) where L(x) = Sum_{k>=0} x^Lucas(k) and Lucas(k) = A000032(k). L(x) is the g.f. of the Lucas indicator sequence (see A102460) and L'(x) = derivative of L(x).

%F a(n) = Lucas(Lucas_inverse(n+1)+2) - n - 3 = A000032(A130241(n+1) + 2) - n - 3 for n > 1.

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

%Y Cf. A000032, A102460, A130241.

%Y For an analog concerning Fibonacci numbers see A132665.

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

%K nonn

%O 1,2

%A _Hieronymus Fischer_, Sep 15 2007