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%I #10 Aug 09 2017 23:07:40
%S 1,10,9,8,7,6,5,4,3,2,20,19,18,17,16,15,14,13,12,11,110,109,108,107,
%T 106,105,104,103,102,101,100,99,98,97,96,95,94,93,92,91,90,89,88,87,
%U 86,85,84,83,82,81,80,79,78,77,76,75,74,73,72,71,70,69,68,67,66,65,64,63,62
%N a(1)=1, a(n) = 10*a(n-1) 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(n) = maximal positive integer < a(n-1) not yet in the sequence, if it exists, else a(n) = 10*a(n-1).
%C Also: a(1)=1, a(n) = a(n-1) - 1, if a(n-1) - 1 > 0 and has not been encountered so far, else a(n) = 10*a(n-1).
%C A permutation of the positive integers. The sequence is self-inverse, in that a(a(n)) = n.
%F The following formulas are given for a general parameter p > 2 considering the recurrence rule above (i.e., a(n) = p*a(n-1)...; p=10 for this sequence).
%F G.f.: g(x) = (x(1-2x)/(1-x) + px^2*f'(x^((2p-1)/(p-1))) + ((2p-1)/p^2)*(f'(x^(1/(p-1))) - px - 1)/(1-x) where f(x) = Sum_{k>=0} x^(p^k) and f'(z) = derivative of f(x) at x = z.
%F a(n) = ((3p-1)*p^(r/2) - p - 1)/(p-1) - n if both r and s are even, else a(n) = ((p^2 + 2p - 1)*p^((s-1)/2) - p - 1)/(p-1) - n, where r = ceiling(2*log_p(((p-1)n + p)/(2p-1))) and s = ceiling(2*log_p(((p-1)n + p)/p) - 1).
%F a(n) = (p^floor(1 + (k+1)/2) + (2p-1)*p^floor(k/2) - p - 1)/(p-1) - n, where k=r if r is odd, else k=s (with respect to r and s above; formally, k = ((r+s) - (r-s)*(-1)^r)/2).
%Y For parameters p=2 to p=9 see A132666 - A132673.
%Y For a similar recurrence rule concerning Fibonacci and Lucas numbers see A132664 and A132665.
%K nonn
%O 1,2
%A _Hieronymus Fischer_, Aug 24 2007, Sep 15 2007
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