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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)

Define "largely abundant numbers" to be numbers k such that sigma(k) >= sigma(j) for all j < k. This sequence gives all the largely abundant numbers that are not highly abundant numbers.

Analogous to A244353 as A002093 is analogous to A002182.

No more terms below 10^10.

a(22) > 10^13. - Giovanni Resta, Jul 02 2019