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a(n) = 0 iff A014664(n) = A243905(n), i.e., iff prime(n) is a Wieferich prime (A001220). So far this is known to be the case only for prime(183) = 1093 and prime(490) = 3511, i.e., a(183) = 0 and a(490) = 0.

Keller proved that the occurrence of 2 consecutive Woodall numbers that are divisible by the same prime is restricted to primes p with even h(p), the order of 2 mod p, and that there are an infinity of such pairs.

a(n) = 0 only for n = 1, p = 2. For any odd prime, a prime q meeting the requirement does exist.

a(n) is the smallest prime q <> p such that q == 1 (mod ord_{p}(2)), where ord_{p}(2) = A002326((p-1)/2) = A014664(n). Strong conjecture: a(n) < A014664(n)^2. - Thomas Ordowski, Mar 15 2019

The maximum element in the k-th column is prime(k) - 1. By Dirichlet's theorem on arithmetic progressions, all divisors of prime(k) - 1 occur infinitely many times in the n-th column.

Also the union of 2 and squarefree semiprimes which never occur in A332952. See A245486 for more information.

Also squarefree semiprimes which never occur in A332951.

This sequence is infinite. It appears that all terms can be divisible by 2 or 3.

If A014664(i) = A014664(j) for some 1 < i < j, then 2*prime(i) is a term. See A245486 for more information.

All terms are odd.

Although the most significant digits of powers of 2 in base n are generally not periodic (the exception being when n is a power of 2), the least significant digits are. For example, 2 to an even power is congruent to 1 (mod 3) and 2 to an odd power is congruent to -1 (mod 3). This means that one can determine one of the prime factors of a Mersenne number, A000225, using the exponent. If n == 0 (mod 2), then A000225(n) == 0 (mod 3) (is a multiple of 3); if n == 0 (mod 4), then A000225(n) == 0 (mod 5); if n == 0 (mod 3), then A000225(n) == 0 (mod 7), and so on.

This general fact gives a reason for why certain Mersenne numbers are not prime (even with prime exponents). If p is congruent to 0 mod A014664(n) (the length of an n-th row) and prime(n) is less than the A000225(p), then prime(n) is a nontrivial factor of A000225(p).

This sequence uses the centered version of mod. The residue system modulo prime(n) is {-1*floor(prime(n)/2)..floor(prime(n)/2)}. This is so that this sequence will encode information about the numbers around 2^A007013(4). If a(n) = k and prime(n) < 2^A007013(4) - k, then 2^A007013(4) - k is not prime (prime(n) is a factor of 2^A007013(4) - k). For example, a(22) = -3, so prime(22) = 79 is a factor of 2^A007013(4) + 3.

The length of this sequence is the lowest value of n such that A014664(n) = A007013(4). This is because for any power of 2, 2^p, if p == 0 (mod A014664(n)), then 2^p == 1 (mod prime(n)) (prime(n) is a factor of A000225(p)). Since A007013(4) is prime, we can apply this to get: If A014664(n) = A007013(4) and prime(n) < A007013(5), then A007013(5) is not prime (prime(n) is a nontrivial factor).

For any n such that prime(n) < 5*(10^51 + 5*10^9), a(n) != 1.

A Sylow p subgroup is a subgroup of order p^r that necessarily exists when r is a maximal power of p. It is not necessarily unique, but when it is unique it is normal in G.

The condition that G of order p*2^n contains a unique Sylow p subgroup places an upper bound on the number of isomorphism classes of G; it is equivalent to stating that the minimal 2^r such that 2^r == 1 (mod p) be such that 2^r > 2^n. The condition that the maximal 2^m dividing p-1, i.e. for p == 1 (mod 2^m), is such that 2^m >= 2^n ensures a lower bound which is equal to the upper bound. See the Miles Englezou link for a proof.

If we relax the two conditions and just consider an arbitrary odd prime p and the number of isomorphism classes for |G| = p*2^n, it is likely that the set of such numbers is unique to p. Since every odd prime has a minimal 2^r such that 2^r == 1 (mod p) (a consequence of Fermat's little theorem), when 2^r = 2^n for |G| = p*2^n, the number of isomorphism classes will differ from a(n) due to the existence of groups where the Sylow p subgroup is not unique.