A281889 a(n) is the least integer k such that more than half of all integers are divisible by a product of n integers chosen from 2..k.

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Author

Peter Munn, Feb 01 2017

Keywords

• nonn
• hard
• more

Comments

The n chosen integers need not be distinct.

By "more than half of all integers" we mean more precisely "more than half of the integers in -m..m, for all sufficiently large m (depending on n)", and similarly with 1..m for "more than half of all positive integers".

Equivalently, a(n) is the least prime p such that more than half of all positive integers can be written as a product of primes of which n or more are not greater than p. (In this sense, a(n) might be called the median n-th least prime factor of the integers.)

The number of integers that satisfy the "product of primes" criterion for p = prime(m) is the same in every interval of primorial(m)^n integers and is A281891(m,n). Primorial(m) = A002110(m), product of the first m primes.

a(n) is the least k = prime(m) such that 2 * A281891(m,n) > A002110(m)^n.

a(n) is the least k such that more than half of all positive integers equate to the volume of an orthotope with integral sides at least n of which are orthogonal with length between 2 and k inclusive.

The next term is estimated to be a(5) ~ 3*10^18.

Examples

			For n=1, we have a(1) = 3 since for all m > 1, more than half of the integers in -m..m are divisible by an integer chosen from 2..3, i.e., either 2 or 3. We must have a(1) > 2, because the only integer in 2..2 is 2, but in each interval -2m-1..2m+1, only 2m+1 integers are even, so 2 is not a divisor of more than half of all integers in the precise sense given above.
		

Crossrefs

Other sequences about medians of prime factors: A126282, A126283, A284411, A290154.

Cf. A002110, A014673, A016088, A027746, A115561, A281891, A309366.