OFFSET
1,1
COMMENTS
Conjecture: Given e >= 0, odd numbers r, k > 0, a > 2^e*r*k, consider the following two statements:
(A) x^m + (x^k + 1)^(2^e*r) is irreducible over GF(2);
(B) x^m + x^(2^e*r*k) + 1 is irreducible over GF(2),
then:
(i) (A) implies (B);
(ii) if (B) is true and (A) is false, then:
(a) gcd(m,r) > 1;
(b) if prime p | gcd(m,r*k), then p*ord_p(2) | m;
(c) if e > 0, then m is odd.
Here ord(2,p) is the multiplicative order of 2 modulo p.
In other words, assuming that (B) is true, (A) is false if and only if (a), (b), (c) hold. (For the "if" part, note that if d = gcd(m,2^e*r) > 1 then x^m + (x^k + 1)^(2^e*r) must be reducible, since it is divisible by x^(m/d) + (x^k + 1)^(2^e*r/d).)
Here is the case r = 5, k = 3, e = 0, and (ii) means that m is in this sequence if and only if x^m + x^15 + 1 is irreducible and m is a multiple of 20. Note that when m is divisible by 20, the result (b) above for p = 3 (if m is divisible by 3 then m is a multiple of 6) is automatically true.
EXAMPLE
20 is a term because x^20 + x^15 + 1 is irreducible over GF(2) but x^20 + x^15 + x^12 + x^3 + 1 is not: x^20 + x^15 + x^12 + x^3 + 1 = (x^4 + x^3 + 1)*(x^8 + x^6 + x^3 + x^2 + 1)*(x^8 + x^7 + x^3 + x^2 + 1).
PROG
(PARI) isA344200(n) = polisirreducible(Mod(x^n+x^15+1, 2)) && !polisirreducible(Mod(x^n+x^15+x^12+x^3+1, 2))
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Jianing Song, May 11 2021
EXTENSIONS
a(7)-a(8) from Jinyuan Wang, May 15 2021
STATUS
approved