

A355919


Let n = p_1*p_2*...*p_k be the prime factorization of n, with the primes sorted in descending order; let b(n) = 7^(p_1  1)*13^(p_2  1)*19^(p_3  1)*...*A002476(k)^(p_k  1). Sequence lists m such that b(m) > A343771(m).


1



128, 256, 512, 1024, 2048, 3072, 4096, 6144, 8192, 12288, 16384, 24576, 32768, 49152, 65536, 73728, 98304, 131072, 147456, 196608, 262144, 294912
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OFFSET

1,1


COMMENTS

{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_11)1) * (p_{r+1})^(q1), 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^21) * (p_2)^(q1) * ... * (p_{e1})^(q1) < b(q^e).
It seems that q^(N+1) is the smallest qrough term in this sequence.


LINKS



EXAMPLE

128 is a term since b(128) = 7 * 13 * 19 * 31 * 37 * 43 * 61 > A343771(128) = 7^3 * 13 * 19 * 31 * 37 * 43.


PROG

(PARI) b(n) = my(f=factor(n), w=omega(n), p=1, product=1); forstep(i=w, 1, 1, for(j=1, f[i, 2], p=nextprime(p+1); while(!(p%6==1), p=nextprime(p+1)); product *= p^(f[i, 1]1))); product


CROSSREFS



KEYWORD

nonn,more


AUTHOR



EXTENSIONS



STATUS

approved



