%I #36 May 18 2021 08:44:51

%S 20,60,100,140,260,900,7980,13500

%N Numbers m > 15 such that x^m + x^15 + 1 is irreducible over GF(2) while x^m + x^15 + x^12 + x^3 + 1 = x^m + (x^3 + 1)^5 is not.

%C Conjecture: Given e >= 0, odd numbers r, k > 0, a > 2^e*r*k, consider the following two statements:

%C (A) x^m + (x^k + 1)^(2^e*r) is irreducible over GF(2);

%C (B) x^m + x^(2^e*r*k) + 1 is irreducible over GF(2),

%C then:

%C (i) (A) implies (B);

%C (ii) if (B) is true and (A) is false, then:

%C (a) gcd(m,r) > 1;

%C (b) if prime p | gcd(m,r*k), then p*ord_p(2) | m;

%C (c) if e > 0, then m is odd.

%C Here ord(2,p) is the multiplicative order of 2 modulo p.

%C 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).)

%C 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.

%e 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).

%o (PARI) isA344200(n) = polisirreducible(Mod(x^n+x^15+1, 2)) && !polisirreducible(Mod(x^n+x^15+x^12+x^3+1, 2))

%Y Similar sequences: A344177 (r=3, k=1), A344198 (r=3, k=3), A344199 (r=3, k=5), A344197 (r=5, k=1), this sequence (r=5, k=3).

%K nonn,hard,more

%O 1,1

%A _Jianing Song_, May 11 2021

%E a(7)-a(8) from _Jinyuan Wang_, May 15 2021