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A132669 a(1)=1, a(n) = 5*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. 5

%I #11 Aug 09 2017 23:07:39

%S 1,5,4,3,2,10,9,8,7,6,30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,

%T 14,13,12,11,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37,

%U 36,35,34,33,32,31,155,154,153,152,151,150,149,148,147,146,145,144,143

%N a(1)=1, a(n) = 5*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) = 5*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) = 5*a(n-1).

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

%F G.f.: g(x) = (x(1-2x)/(1-x) + 5x^2*f'(x^(9/4)) + (9/25)*(f'(x^(1/4)) - 5x - 1))/(1-x) where f(x) = Sum_{k>=0} x^(5^k) and f'(z) = derivative of f(x) at x = z.

%F a(n) = (14*5^(r/2) - 6)/4 - n, if both r and s are even, else a(n) = (34*5^((s-1)/2) - 6)/4 - n, where r = ceiling(2*log_5((4n+5)/9)) and s = ceiling(2*log_5((4n+5)/5)) - 1.

%F a(n) = (5^floor(1 + (k+1)/2) + 9*5^floor(k/2) - 6)/4 - 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).

%F a(n) = A133629(m) + A133629(m+1) + 1 - n, where m:=max{ k | A133629(k) < n }.

%F a(A133629(n) + 1) = A133629(n+1).

%F a(A133629(n)) = A133629(n-1) + 1 for n > 0.

%Y For formulas concerning a general parameter p (with respect to the recurrence rule ... a(n) = p*a(n-1) ...) see A132374.

%Y For p=2 to p=10 see A132666 through A132674.

%Y Cf. A087503.

%K nonn

%O 1,2

%A _Hieronymus Fischer_, Sep 15 2007, Sep 23 2007

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