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A160679
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Square root of n under Nim (or Conway) multiplication.
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2
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0, 1, 3, 2, 7, 6, 4, 5, 14, 15, 13, 12, 9, 8, 10, 11, 30, 31, 29, 28, 25, 24, 26, 27, 16, 17, 19, 18, 23, 22, 20, 21, 57, 56, 58, 59, 62, 63, 61, 60, 55, 54, 52, 53, 48, 49, 51, 50, 39, 38, 36, 37, 32, 33, 35, 34, 41, 40, 42, 43, 46, 47, 45, 44, 124, 125, 127, 126, 123, 122, 120
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OFFSET
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0,3
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COMMENTS
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Because Conway's field On2 (endowed with Nim-multiplication and [bitwise] Nim-addition) has characteristic 2, the Nim-square function (A006042) is an injective field homomorphism (i.e., the square of a sum is the sum of the squares). Thus the square function is a bijection within any finite additive subgroup of On2 (which is a fancy way to say that an integer and its Nim-square have the same bit length). Therefore the Nim square-root function is also a field homomorphism (the square-root of a Nim-sum is the Nim-sum of the square roots) which can be defined as the inverse permutation of A006042 (as such, it preserves bit-length too).
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LINKS
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FORMULA
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Letting NIM (= XOR) TIM and RIM denote respectively the sum, product and square root in Conway's Nim-field On2, we see that the bit-length of NIM(x,TIM(x,x)) is less than that of the positive integer x. This remark turns the following relations into an effective recursive definition of a(n) = RIM(n) which uses the fact that RIM is a field homomorphism in On2:
a(0) = 0
a(n) = NIM(n, a(NIM(n, a(n, TIM(n,n)) )
For 0 <= n <= 2^2^k-1, a(n) = A335162(n, 2^(2^k-1)). This is because {0,1,...,2^2^k-1} together with the nim operations makes a field isomorphic to GF(2^2^k).
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EXAMPLE
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a(2) = 3 because TIM(3,3) = 2
More generally, a(x)=y because A006042(y)=x.
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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