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A340669
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Permutation of the nonnegative integers formed by negation in complex base i-1.
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3
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0, 29, 58, 7, 116, 25, 14, 3, 232, 21, 50, 239, 28, 17, 6, 235, 464, 13, 42, 471, 100, 9, 478, 467, 56, 5, 34, 63, 12, 1, 470, 59, 928, 957, 26, 935, 84, 953, 942, 931, 200, 949, 18, 207, 956, 945, 934, 203, 112, 941, 10, 119, 68, 937, 126, 115, 24, 933, 2, 31
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OFFSET
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0,2
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COMMENTS
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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.
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.
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
In terms of bit flips, in the following "^^" is each 01 or 11 and F marks the bits flipped above them.
n = 1506 = binary 00001 0 1 11 100 01 0
FFF^^ F ^^ FFF ^^
a(n) = 29914 = binary 11101 0 0 11 011 01 0
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PROG
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(PARI) { a(n) = for(i=0, if(n, logint(n, 2)),
if(bittest(n, i),
if(bittest(n, i+1), n=bitxor(n, 4<<i); i+=2,
n=bitxor(n, 28<<i); i+=4))); n; }
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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