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%I #9 Mar 08 2013 23:24:43
%S 5,8,12,18,19,21,25,33,36,44,53,37,53,64,14,31,32,69,71,76,77,108,120,
%T 39,93,105,123,125,157,170,52,91,93,99,190,192,89,225,238,121,72,158,
%U 251,238,251,270,205,50,209,282,284,286,287,288,289,361,385,370,281,282,340,342,344,346,309,310,312,367,460,275
%N When A114183 decreases in value for the n-th time, dropping to k (say), a(n) is the number of steps earlier that floor(k/2) appeared in A114183.
%C The fact that, when a number k occurs in A114183, floor(k/2) has already appeared, is a key step in the proof that A114183 is a permutation of the natural numbers. This fact is obvious if k is the result of a doubling step. The present sequence is an attempt to gain insight into why it is true when k occurs at a square root step.
%H N. J. A. Sloane, <a href="/A222802/b222802.txt">Table of n, a(n) for n = 1..7322</a>
%e The first 50 terms of A114183 are:
%e 1, 2, 4, 8, 16, 32, 5, 10, 3, 6, 12, 24, 48, 96, 9, 18, 36, 72, 144, 288, 576, 1152, 33, 66, 132, 11, 22, 44, 88, 176, 13, 26, 52, 7, 14, 28, 56, 112, 224, 448, 21, 42, 84, 168, 336, 672, 25, 50, 100, 200.
%e The sequence decreases from 32 to 5, from 10 to 3, from 96 to 9, and so on.
%e The values of k are therefore 5, 3, 9, 33, 11, 13, 7, 21, 25, ...
%e and the corresponding values of floor(k/2) are 2, 1, 4, 16, 5, 6, 3, 10, 12, ...
%e Since 2 appeared in A114183 5 steps before 5, a(1) = 5,
%e since 1 appeared 8 steps before 3, a(2) = 8,
%e since 4 appeared 12 steps before 9, a(3) = 12, and so on.
%Y Cf. A114183, A221715, A221716, A213220.
%K nonn
%O 1,1
%A _N. J. A. Sloane_, Mar 08 2013