cp's OEIS Frontend

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 = 1, e = 0, and (ii) means that m is in this sequence if and only if x^m + x^5 + 1 is irreducible and 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 = 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.

A more intuitive version of A344141.

Every term other than the first is a member of A129771.

In A057496 it is stated that if x^n + x^3 + x^2 + x + 1 is irreducible, then so is x^n + x^3 + 1. It follows that no term can be equal to 15.

It is conjectured that no term can be of the form P_m(2^k), where P_m(x) = Product_{i>=0} (1 + x^(2^(d_i)))^(c_i) if the binary representation of m is m = Sum_{i>=0} c_i * 2^(d_i), k is an odd number. See my conjecture in A344177.

A more intuitive version of A344142.

In A057496 it is stated that if x^n + x^3 + x^2 + x + 1 is irreducible, then so is x^n + x^3 + 1. It follows that no term can be equal to 15.

It is conjectured that an irreducible polynomial of degree n with 5 terms exists for every n. It follows from the conjecture that for n >= 2, a(n) is of the form 2^k + 1 or an odd number with Hamming weight 4.

It is conjectured that no term can be of the form P_m(2^k), where P_m(x) = Product_{i>=0} (1 + x^(2^(d_i)))^(c_i) if the binary representation of m is m = Sum_{i>=0} c_i * 2^(d_i), k is an odd number. See my conjecture in A344177.