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%I #19 Jan 02 2023 12:30:54
%S 0,29,58,7,116,25,14,3,232,21,50,239,28,17,6,235,464,13,42,471,100,9,
%T 478,467,56,5,34,63,12,1,470,59,928,957,26,935,84,953,942,931,200,949,
%U 18,207,956,945,934,203,112,941,10,119,68,937,126,115,24,933,2,31
%N Permutation of the nonnegative integers formed by negation in complex base i-1.
%C Complex base i-1 of Khmelnik and Penney uses an integer n>=0 to represent a complex integer z(n) = A318438(n) + A318439(n)*i. a(n) is the negation of z in this representation, so that z(a(n)) = -z(n). Every z is uniquely represented, so this is a self-inverse permutation.
%C Khmelnik's table 4 is carries applied to z which become states and transitions by bits of n and certain 0<->1 bit flips in n. The result is the transformation in the formulas below. Bit flips may extend into 0-bits above the most significant bit of n causing the bit length of a(n) to be greater than the bit length of n.
%H Kevin Ryde, <a href="/A340669/b340669.txt">Table of n, a(n) for n = 0..8192</a>
%H Joerg Arndt, <a href="https://jjj.de/fxt/demo/bits/index.html#radix-m1pi">The fxt demos: bit wizardry. radix(-1+i)</a>
%H Solomon I. Khmelnik, <a href="http://lib.izdatelstwo.com/Papers2/s4.djvu">Specialized Digital Computer for Operations with Complex Numbers</a> (in Russian), Questions of Radio Electronics, volume 12, number 2, 1964.
%H Kevin Ryde, <a href="http://user42.tuxfamily.org/dragon/index.html">Iterations of the Dragon Curve</a>, see index MinusNeg.
%H Andrey Zabolotskiy, <a href="http://list.seqfan.eu/oldermail/seqfan/2016-September/016716.html">English translation of theorems from Khmelnik</a>, September 2016
%H Andrey Zabolotskiy, <a href="https://pastebin.com/raw/uHJW13zf">Python Code by bit flips or conversion</a> and <a href="https://pastebin.com/raw/aCnECyj8">Python Code by Khmelnik's Pi and Beta</a>, September 2016
%F a(n) is formed by transforming n as follows. Write n in binary with four high 0-bits and consider bits from least to most significant. At a 01 pair (high 0, low 1), apply an 0<->1 flip to three bits immediately above this pair. At a 11 pair, flip one bit immediately above this pair. Repeat, each time seeking the next higher 01 or 11 pair above the bits just flipped.
%e For n=1506, location z(1506) = 11-35*i. Its negation is -(11-35*i) = z(29914) so a(1506) = 29914. And being self-inverse conversely a(29914) = 1506.
%e In terms of bit flips, in the following "^^" is each 01 or 11 and F marks the bits flipped above them.
%e n = 1506 = binary 00001 0 1 11 100 01 0
%e FFF^^ F ^^ FFF ^^
%e a(n) = 29914 = binary 11101 0 0 11 011 01 0
%o (PARI) { a(n) = for(i=0,if(n,logint(n,2)),
%o if(bittest(n,i),
%o if(bittest(n,i+1), n=bitxor(n,4<<i); i+=2,
%o n=bitxor(n,28<<i); i+=4))); n; }
%Y Cf. A318438, A318439, A340670.
%K nonn,base
%O 0,2
%A _Kevin Ryde_, Jan 15 2021
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