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Includes all members of A002182.

Conjecture (false!): includes all members of A094348.

Contribution from Matthew Vandermast, Oct 10 2008: (Start)

Counterexample to conjecture: 5354228880, the smallest positive multiple of the first 23 positive integers, does not belong to A067128. It is the smallest member of A003418 (a subsequence of A094348) not to be largely composite.

Intersection of A067128 and A025487.

Includes all members of A097212. (End)

The first zero terms calculated by Peter J. C. Moses appear for n=163,186,248,...

For n=151, the required k, if it exists, should be >9800. So one can conjecture that a(151) is the first zero term.

No term is a highly composite number (A002182), since as P. Erdős noted, every highly composite number is divisible by every prime less than p; on the other hand, the sequence is a strict subsequence of A244353.

Let p=prime(k), k=k(n), be the greatest prime divisor of a(n). David A. Corneth noted that a(11)=4680 is the first term which is not divisible by both prime(k-2) and prime(k-1).

What is the next such term?

Is there a case when prime(k-1) divides a(n), but prime(k-2) does not? - Vladimir Shevelev, May 21 2016

From David A. Corneth, May 22 2016: (Start)

It seems that most terms have the property mentioned in the second comment. 105753195046699200 is the first term whose three largest distinct prime divisors are consecutive primes. No other terms below that number have that property.

There are many terms where prime(k-1) divides a(n), but prime(k-2) does not. They are 107442720, 537213600, 1074427200, 1889727840, 9448639200, 18897278400, 37794556800, ... (End)

Intersection of A067128 and A080259. - Michel Marcus, May 22 2016; edited by Michael De Vlieger, Jan 23 2024

From Vladimir Shevelev, May 23 2016: (Start)

To explain what happens, note that, if p = p_k = prime(k), k = k(n), is the greatest prime divisor of a(n), then b(n)=a(n)*prime(k+r)/prime(k) is also in the sequence, if between a(n) and b(n) there is no highly composite number(HCN). Indeed, d(a(n))=d(b(n)) and in the interval [a(n),b(n)] there is no number x with a greater d(x).

For example, between the term 3960=2^3*3^2*5*11 and 4680=3960*13/11 there is no HCN, so 4680 is also a term; however, between 3960 and 3960*17/11=6120 there is an HCN, so 6120 is not a term.

If n is a large HCN, then its greatest prime divisor is O(log n)[Erdős]. So we can also say that prime(k)=O(log n), since, by Erdős's theorem, if n_1 is the next HCN, then n < n_1 < n + n*(log n)^-c, where c > 0 is constant. So if

p_(k+r)/p_k < 1 + (log n)^-c (1)

then there is a real possibility that there are numbers a(m)*p_(k+i)/p_k, i=1,...,r, which are all terms, where a(m) is between n and the next HCN. Note that (1) means that (1+o(1))*(k+r)*log(k+r)/(k*log k) < 1 + (log n)^-c, where k=O(log n/loglog n). Since, for r < k, log(k+r) = log k + log(1+r/k) ~ log k +r/k + O((r/k)^2), then log(k+r)/log k = 1 + O(r/(k*log k). Thus (1) means r < k/(log n)^c = O(log n)^(1-c)/loglog n. Thus if c < 1, then r could be arbitrary large for sufficiently large n.

Moreover, the numerical results of David A. Corneth (cf. A273415) allow us to conjecture that indeed 0 < c < 1. (End)

By the theorem of Vladimir Shevelev mentioned in sequence A273015, such an element exists for each prime.

Is a(n + 1) / a(n) ~ 1 for large n?

Every term in this sequence also appears in A002183, where every element of this sequence occurs exactly once.

In A067128 it is asked if A034287 = A067128. If that is the case then this sequence is also the number of divisors of A034287.

A proof of the existence of a(n) for all n was given by Vladimir Shevelev, May 14 2016, as follows:

(Start)

I give a proof of the existence of k in new David's sequence A273038: "Least k such that for all m >= k, A067128(m) is divisible by n."

Let us change the notation. Suppose N in A067128 has prime power factorization (PPF) N=2^k_1*...*p_n^k_n, k_n>=1, (1)

where p_i=prime(i).

From my theorem in A273015 it follows that, when N runs through A067128, p_n in (1) is unbounded and, moreover, tends to infinity, when N tends to infinity.

Let us show that, when N runs through A067128, k_1 is also unbounded.

Indeed, suppose k_1 is bounded. Consider a number N_1 with PPF N_1=2^(k_1+x)*...*p_(n-1)^k_(n-1) such that all powers p^i , i=2,...,n-1, are the same as in (1) and satisfy 2^x

Then N_1d(N).

We want (k_1+x+1)*...*(k_(n-1)+1)>(k_1+1)*...*(k_(n-1)+1)*(k_n+1), or k_1+x+1>(k_1+1)*(k_n+1)=k_1*k_n+k_n+k_1+1, or x>(k_1+1)*k_n.

So, by (2), (k_1+1)*k_n

Since by hypothesis k_1 is bounded, for large n we can choose the required x, which gives a contradiction. So k_1 is unbounded.

Moreover, we see that k_1 tends to infinity as log_2(p_n), n=n(N), when N tends to infinity, otherwise (3) again leads to contradiction.

Suppose m=2^m_1*3^m_2*...*p_r^m_r.

We can choose k_1 > m_1. In the same way we prove that k_2 tends to infinity and choose k_2 > m_2,..., and so on. k_r tends to infinity and we choose k_r > m_r. All k_i , i=1,...,r tend to infinity at least as log_p_r(p_n), n=n(N).

So there exists a large M_m such that for all N from A067128 > M_m, m|N.

(End)