%I #32 Aug 10 2022 07:52:53

%S 1,3,3,15,15,15,15,5,15,5,15,15,5,5,15,85,85,255,255,85,85,255,255,85,

%T 85,255,255,255,255,85,85,255,255,255,255,85,255,85,255,255,255,255,

%U 255,255,85,255,85,255,85,85,255,255,255,255,255,255,255,255,255,255,85,85,255,255,51,255,255,255,51,255,255,17,255,85,255,17,255,85

%N Multiplicative order of n in nim-multiplication.

%C For n <= 255, computed using _R. J. Mathar_'s Maple programs from A051775. a(256) = 21845 from _J. H. Conway_ and Alex Ryba, May 04 2012

%C Apparently, all terms belong to A001317, and A001317(k) appears 2^k times. - _Rémy Sigrist_, Jun 14 2020

%C From _Jianing Song_, Aug 10 2022: (Start)

%C The observation above is incorrect. Note that {0,1,...,2^2^k-1} together with the nim operations makes a field isomorphic to GF(2^2^k). This means that:

%C - Every number is a divisor of a number of the form 2^2^k-1, and every divisor of 2^2^k-1 for some k appears;

%C - If d is a divisor of 2^2^k-1 for some k, then d appears phi(d) times among {a(1),a(2),...,a(2^2^m-1)} for all m >= k, phi = A000010. This means that if d > 1, and k is the smallest number such that d | 2^2^k-1, then d can only appear among {a(2^2^(k-1)),...a(2^2^k-1)}.

%C So the correct result should be: all terms are divisors of numbers of the form 2^2^k-1, and each divisor d appears phi(d) times.

%C For example, 641 would appear 640 times in this sequence, among {a(2^32),...,a(2^64-1)}, although to determine their positions is hard. (End)

%D J. H. Conway, On Numbers and Games, Academic Press, Chapter 6.

%H Alex Ryba, <a href="/A212200/b212200.txt">Table of n, a(n) for n = 1..65999</a>

%e The nim-products 4*4*...*4 are (cf. A051775): 4, 4^2=6, 4^3=4*6=14, 4^4=4*14=5, 4^5=2, 4^6=8, ..., 4^14=15, 4^15=1, so 4 has order a(4) = 15.

%Y Cf. A001317, A051775, A051776, A212201, A212202, A212203, A212204.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, May 03 2012