A212200 Multiplicative order of n in nim-multiplication.

1, 3, 3, 15, 15, 15, 15, 5, 15, 5, 15, 15, 5, 5, 15, 85, 85, 255, 255, 85, 85, 255, 255, 85, 85, 255, 255, 255, 255, 85, 85, 255, 255, 255, 255, 85, 255, 85, 255, 255, 255, 255, 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

Offset: 1

Author

N. J. A. Sloane, May 03 2012

Keywords

• nonn

Comments

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

Apparently, all terms belong to A001317, and A001317(k) appears 2^k times. - Rémy Sigrist, Jun 14 2020

From Jianing Song, Aug 10 2022: (Start)

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:

- 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;

- 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)}.

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.

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)

Examples

			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.
		

References

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

Links

• Alex Ryba, Table of n, a(n) for n = 1..65999

Crossrefs

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