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A222802 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.

%I A222802 #9 Mar 08 2013 23:24:43
%S A222802 5,8,12,18,19,21,25,33,36,44,53,37,53,64,14,31,32,69,71,76,77,108,120,
%T A222802 39,93,105,123,125,157,170,52,91,93,99,190,192,89,225,238,121,72,158,
%U A222802 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 A222802 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 A222802 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 A222802 N. J. A. Sloane, <a href="/A222802/b222802.txt">Table of n, a(n) for n = 1..7322</a>
%e A222802 The first 50 terms of A114183 are:
%e A222802 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 A222802 The sequence decreases from 32 to 5, from 10 to 3, from 96 to 9, and so on.
%e A222802 The values of k are therefore 5, 3, 9, 33, 11, 13, 7, 21, 25, ...
%e A222802 and the corresponding values of floor(k/2) are 2, 1, 4, 16, 5, 6, 3, 10, 12, ...
%e A222802 Since 2 appeared in A114183 5 steps before 5, a(1) = 5,
%e A222802 since 1 appeared 8 steps before 3, a(2) = 8,
%e A222802 since 4 appeared 12 steps before 9, a(3) = 12, and so on.
%Y A222802 Cf. A114183, A221715, A221716, A213220.
%K A222802 nonn
%O A222802 1,1
%A A222802 _N. J. A. Sloane_, Mar 08 2013