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%I #27 Nov 01 2017 18:32:38
%S 2,10,22,596,606,4356,4402,26857390,26956576,39877528,39877576,
%T 7435118530,7435118622,8036090130,8036485212
%N Smallest positive k such that exactly half of the first k positive integers have n or fewer divisors.
%C For prime p, lim_{p -> inf} a(p-1)/a(p) = 1. For example, a(10)/a(11) = 39877528/39877576 = 0.999998796..., which is so close to 1 because of the relative scarcity of numbers having exactly 11 divisors (i.e., numbers that are 10th powers of primes): at k = a(10) = 39877528, there are exactly a(10)/2 = 19938764 numbers in [1..k] that have 10 or fewer divisors, and only 3 (namely, 2^10, 3^10, and 5^10) that have exactly 11 divisors, so the number of numbers in [1..k] having 11 or fewer divisors is only slightly more than k/2, and the proportion falls to exactly 1/2 relatively quickly as k is increased beyond a(10).
%C a(16) > 10^13. - _Giovanni Resta_, Apr 12 2017
%C Given that the proportion of numbers in [10^n + 1, 10^n + 10^5] having 16 or fewer divisors for n = 12..20 is, respectively, about 61.2%, 59.4%, 57.6%, 56.0%, 54.7%, 53.3%, 51.9%, 50.9%, 49.9%, it seems likely that a(16) is roughly 10^20. - _Jon E. Schoenfield_, Oct 30 2017
%e For each of the first several positive integers k, the table below shows its number of divisors d(k), the number of numbers in [1..k] with exactly n divisors, and the number of numbers in [1..k] with n or fewer divisors. (In the "d(k) = n" portion of the table, a "." indicates that the value is unchanged relative to the row above it.)
%e .
%e Number of numbers in [1..k] ...
%e -------------------------------
%e with d(k) = n with d(k) <= n
%e -------------- --------------
%e k d(k) n = 1 2 3 4 n = 1 2 3 4
%e == ==== ============== ==============
%e 1 1 1 0 0 0 1 1 1 1
%e 2 2 . 1 . . *1* 2 2 2
%e 3 2 . 2 . . 1 3 3 3
%e 4 3 . . 1 . 1 3 4 4
%e 5 2 . 3 . . 1 4 5 5
%e 6 4 . . . 1 1 4 5 6
%e 7 2 . 4 . . 1 5 6 7
%e 8 4 . . . 2 1 5 6 8
%e 9 3 . . 2 . 1 5 7 9
%e 10 4 . . . 3 1 *5* 7 10
%e .
%e Asterisks in the "d(k) <= n" portion of the table highlight the first two terms of this sequence:
%e a(1) = 2 because the number of positive integers with d(k) <= 1 first drops to exactly k/2 at k = 2.
%e a(2) = 10 because the number of positive integers with d(k) <= 2 first drops to exactly k/2 at k = 10.
%t k=1; cnt=Array[0&,7]; n=0; Reap[While[k <= 7, Do[cnt[[i]]++, {i, DivisorSigma[0, ++n], 7}]; If[ cnt[[k]] == n/2, Sow[n]; k++]]][[2, 1]] (* _Giovanni Resta_, Mar 25 2017 *)
%o (PARI) a(n) = my(k = 2); while (sum(k=1, k, numdiv(k) <= n) != k/2, k += 2); k; \\ _Michel Marcus_, Mar 26 2017
%o (PARI) a(n) = my(t=0,i=0); while(1, i++; if(numdiv(i) <=n, t++); if(2*t==i, return(i))) \\ _David A. Corneth_, Mar 26 2017
%Y Cf. A000005 (number of divisors).
%K nonn,more
%O 1,1
%A _Jon E. Schoenfield_, Mar 25 2017
%E a(12)-a(15) from _Giovanni Resta_, Mar 25 2017
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