OFFSET
1,1
COMMENTS
Each of these polynomials from n=1 up to n=12 is primitive: if you make a finite field of order 2^(2^n) as GF(2)[x]/<P(x)> then x generates the field's multiplicative group. A natural conjecture is that this is true for all n.
LINKS
Simon Tatham, Table of n, a(n) for n = 1..11
EXAMPLE
For n=3, giving 2^n=8 and 2^(2^n)=256: let x be the nimber *255. Then the powers of x (under nim-multiplication) are *1, *255, *156, *61, *205, *200, *38, *71, *179. Under nim-addition, the subset of these powers *1 + *61 + *200 + *71 + *179 sum to *0. That is, 1+x^3+x^5+x^7+x^8 = 0. No sum of the powers up to and including x^7 is zero. So the polynomial 1+x^3+x^5+x^7+x^8 over GF(2) is the minimal polynomial of *255. Therefore the sequence entry for n=3 is the integer obtained by reinterpreting this polynomial as one over the integers and evaluating it at 2, i.e. 1+2^3+2^5+2^7+2^8 = 425.
CROSSREFS
KEYWORD
nonn
AUTHOR
Simon Tatham, Mar 16 2025
STATUS
approved
