A356136 - OEIS

%I #22 Oct 18 2022 13:26:57

%S 2,27,-1,665,-1,57675,-1,57230,-1

%N a(n) is the smallest number k > 1 such that, in the interval 1..k, there are as many integers that have exactly 2n divisors as there are primes (or -1 if no such number exists).

%C Every number with an odd number of divisors is a square, so there is no odd m > 1 for which there exists an integer k > 1 such that, in the interval 1..k, there are as many integers with exactly m divisors as there are primes.

%C For integers k > 1, the ratio (number of integers in 1..k with exactly 6 divisors) / (number of primes <= k) reaches a maximum of 109/162 = 0.672839... at k = 964. That ratio never reaches 1, so a(3) = -1.

%C Do there exist any odd values of n > 1 such that a(n) > -1?

%C a(12) = 3266925 = 3 * 5^2 * 43 * 1013; a(16) = 42268198 = 2 * 7 * 19 * 131 * 1213.

%C a(24) = 7230602050. - _David A. Corneth_, Oct 16 2022

%C a(32) = 942227458280. - _Giovanni Resta_, Oct 18 2022

%e a(2) = 27 = 3^3 because 27 is the smallest number k > 1 such that, in the interval 1..k, there are as many integers with exactly 2*2=4 divisors as there are primes: 9 numbers with 4 divisors (6, 8, 10, 14, 15, 21, 22, 26, 27) and 9 primes (2, 3, 5, 7, 11, 13, 17, 19, 23).

%e a(4) = 665 = 5 * 7 * 19: in 1..665, there are 121 integers with exactly 8 divisors and there are 121 primes, and 665 is the smallest k > 1 for which those two counts are equal.

%Y Cf. A000010.

%K sign,hard,more

%O 1,1

%A _Jon E. Schoenfield_, Oct 14 2022

%E a(1)=2 prepended by _Ivan N. Ianakiev_, Oct 16 2022