cp's OEIS Frontend

Analog of A037019: this is an easy way to produce a number k such that A002654(k) = n, or equivalently, a number k whose prime factors are all congruent to 1 modulo 4 and with exactly n divisors.

{b(n)} is an analog of A037019 and of A340388: all prime factors of b(n) are all congruent to 1 modulo 6 and b(n) has exactly n divisors, so A002324(b(n)) = n. By definition we have A343771(n) <= b(n), and it seems that the equality holds for most n. This sequence lists the exceptions.

Since {b(n)} agrees with A343771(n) for most n, it cannot have its own entry.

Let q be a prime, then q^e is here if and only if e >= N+1, where N is the number of primes congruent to 1 modulo 6 below 7^q (N = 6, 32, 958, ... for q = 2, 3, 5, ...).

Proof: p_1 < p_2 < ... be the primes congruent to 1 modulo 6. Suppose that A343771(q^e) = (p_1)^(q^(m_1)-1) * (p_2)^(q^(m_2)-1) * ... * (p_r)^(q^(m_r)-1) with r <= e, m_1 >= m_2 >= ... >= m_r. If m_1 >= 2, then r < e, so we can substitute (p_1)^(q^(m_1)-1) with (p_1)^(q^(m_1-1)-1) * (p_{r+1})^(q-1), which a smaller number with exactly q^e divisors, a contradiction. So we have m_1 = 1, namely A343771(q^e) = b(q^e). On the other hand, if e >= N+1, then A343771(q^e) <= (p_1)^(q^2-1) * (p_2)^(q-1) * ... * (p_{e-1})^(q-1) < b(q^e).

It seems that q^(N+1) is the smallest q-rough term in this sequence.