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A182194 a(1)=2, a(n)=a(n-1)^2 if the minimal natural number > 1 not yet in the sequence is greater than a(n-1), else a(n)=a(n-1)-1.

%I A182194 #10 Aug 09 2017 23:07:40
%S A182194 2,4,3,9,8,7,6,5,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,100,
%T A182194 99,98,97,96,95,94,93,92,91,90,89,88,87,86,85,84,83,82,81,80,79,78,77,
%U A182194 76,75,74,73,72,71,70,69,68,67,66,65,64,63,62,61,60,59,58,57,56,55,54
%N A182194 a(1)=2, a(n)=a(n-1)^2 if the minimal natural number > 1 not yet in the sequence is greater than a(n-1), else a(n)=a(n-1)-1.
%C A182194 A reordering of the natural numbers > 1.
%C A182194 The sequence is quasi self-inverse in that a(a(n-1)-1)=n.
%H A182194 Hieronymus Fischer, <a href="/A182194/b182194.txt">Table of n, a(n) for n = 1..10200</a>
%F A182194 a(n)=a(n-1)-1, if a(n-1)-1 > 1 is not in the set {a(k)| 1<=k<=n-1}, else a(n)=a(n-1)^2.
%F A182194 a(a(n)-1)=n+1.
%F A182194 If we define b(1)=2, b(2)=3, b(k)=b(k-2)^2+1, we get the sequence 2, 3, 5, 10, 26, 101, 677, 10202, 458330, 104080805, …. The b(k) are those terms a(n) of the original sequence for which a(n+1)=a(n)^2.
%F A182194 With these b(k) we obtain for k>1:
%F A182194 a(b(k)-2)=b(k-1),
%F A182194 a(b(k)-1)=b(k-1)^2.
%F A182194 a(b(k))=b(k-1)^2 - 1.
%F A182194 a(n)=b(m)+b(m-1)-n-2, where m is the least index such that b(m)>n+1 (valid for n>=1).
%e A182194 a(2)=4=a(1)^2, since 3>2=a(1) is the minimal number not yet in the sequence (because of a(1)=2);
%e A182194 a(15)=19=a(14)-1, since the minimal number not yet in the sequence (=10) is <=a(14)=20.
%e A182194 a(10^4)=b(8)+b(7)-10^4-2=877.
%e A182194 a(10^6)=b(10)+b(9)-10^6-2= 103539133.
%Y A182194 Cf. A020703, A038722, A132666, A132664, A132665, A132674, A210882.
%K A182194 nonn
%O A182194 1,1
%A A182194 _Hieronymus Fischer_, Apr 30 2012