cp's OEIS Frontend

Numbers m >= 4 that are in A057461 but not in A057496.

In A057496 it is stated that if x^m + x^3 + x^2 + x + 1 is irreducible, then so is x^m + x^3 + 1. It seems that if x^m + x^3 + 1 is irreducible and m is not a multiple of 6, then so is x^m + x^3 + x^2 + x + 1. In other words, it seems that this sequence consists of the terms in A057461 that are multiples of 6.

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 = 3, k = 1, e = 0, and (ii) means that m is in this sequence if and only if x^m + x^3 + 1 is irreducible and m is a multiple of 6.

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 = 3, k = 3, e = 0, and (ii) means that m is in this sequence if and only if x^m + x^9 + 1 is irreducible and m is a multiple of 6.

If the conjecture is true, this is also numbers m >= 10 such that x^m + x^9 + 1 is irreducible over GF(2) while x^m + x^9 + x^8 + x + 1 is not (the case r = 9, k = 1, e = 0).

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 = 3, k = 5, e = 0, and (ii) means that m is in this sequence if and only if x^m + x^15 + 1 is irreducible, m is a multiple of 6, if m is divisible by 5 then m is a multiple of 20.

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.