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 A160679 Square root of n under Nim (or Conway) multiplication 1
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS 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). LINKS Paul Tek, Table of n, a(n) for n = 0..576 FORMULA 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)) ) Note: TIM(n,n) = A006042(n) EXAMPLE a(2) = 3 because TIM(3,3) = 2 More generally, a(x)=y because A006042(y)=x. CROSSREFS A006042 (Nim-squares). A051917 (Nim-reciprocals). Sequence in context: A265345 A154448 A099896 * A233276 A304083 A276441 Adjacent sequences:  A160676 A160677 A160678 * A160680 A160681 A160682 KEYWORD easy,nonn AUTHOR Gerard P. Michon, Jun 25 2009 STATUS approved

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Last modified February 18 02:57 EST 2020. Contains 332006 sequences. (Running on oeis4.)