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A160679 Square root of n under Nim (or Conway) multiplication.

%I A160679 #25 Aug 21 2025 15:30:22
%S A160679 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,
%T A160679 19,18,23,22,20,21,57,56,58,59,62,63,61,60,55,54,52,53,48,49,51,50,39,
%U A160679 38,36,37,32,33,35,34,41,40,42,43,46,47,45,44,124,125,127,126,123,122,120
%N A160679 Square root of n under Nim (or Conway) multiplication.
%C A160679 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).
%H A160679 Paul Tek, <a href="/A160679/b160679.txt">Table of n, a(n) for n = 0..576</a>
%H A160679 G. P. Michon, <a href="http://www.numericana.com/answer/fields.htm#on2">Nim-multiplication in Conway's algebraically complete field On2</a>
%H A160679 <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%H A160679 <a href="/index/Ni#Nimmult">Index entries for sequences related to Nim-multiplication</a>
%H A160679 <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%F A160679 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:
%F A160679 a(0) = 0
%F A160679 a(n) = NIM(n, a(NIM(n, a(n, TIM(n,n)) )
%F A160679 Note: TIM(n,n) = A006042(n)
%F A160679 From _Jianing Song_, Aug 10 2022: (Start)
%F A160679 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).
%F A160679 Also for n > 0, a(n) = A335162(n, (A212200(n)+1)/2). (End)
%e A160679 a(2) = 3 because TIM(3,3) = 2
%e A160679 More generally, a(x)=y because A006042(y)=x.
%Y A160679 Cf. A006042 (Nim-squares). A051917 (Nim-reciprocals), A335162, A212200.
%K A160679 easy,nonn,changed
%O A160679 0,3
%A A160679 _Gerard P. Michon_, Jun 25 2009