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%I #8 Jul 29 2019 13:14:04
%S 2,5,71,43103
%N When the positive integers are written as products of primes in nondecreasing order, a(n) is the least prime to occur more frequently in n-th position than in any other position.
%C In such products of primes, prime(m) occurs in n-th position A281890(m,n) times in every interval of A002110(m)^n positive integers, as explained in A281890. A002110(m) = primorial(m), product of first m primes.
%C For n >= 2, a(n) is the least prime to occur more frequently in n-th position than (n-1)-th position.
%C Primes p satisfying a(n) <= p < a(n+1) appear to occur more frequently in n-th position than in any other position.
%C The next term, a(5), is estimated to be ~ 6*10^11.
%F a(1) = prime(1) = 2.
%F For n >= 2, a(n) = min{ k : k = prime(m), A281890(m,n) > A002110(m) * A281890(m,n-1) }.
%e a(1) = prime(1) = 2, since 2 occurs in n-th position when an integer divisible by 2^n is written as a product of primes in nondecreasing order, thus more frequently in 1st position than in other positions.
%e Prime(2) = 3 occurs more often in 1st position than 2nd position, specifically once for every 6 consecutive integers (since A281890(2,1) = 1 and primorial(2) = 6) compared with 5 times for every 36 consecutive integers (since A281890(2,2) = 5 and primorial(2)^2 = 36). As 2 and 3 each occur more frequently in 1st position than 2nd position, a(2) > 3.
%e Prime(3) = 5 occurs in 1st position A281890(3,1) = 2 times in primorial(3) = 30, in 2nd position A281890(3,2) = 62 times in 30^2, in 3rd position A281890(3,3) = 1322 times in 30^3, and decreasingly frequently in subsequent positions. 2/30 < 62/30^2 and 62/30^2 > 1322/30^3. Thus 5 occurs most frequently in 2nd position and is the first prime to do so, so a(2) = 5.
%Y Cf. A002110, A027746, A126283, A281889, A281890.
%K nonn,more
%O 1,1
%A _Peter Munn_, Jul 25 2019
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