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
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)
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)
