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%I #25 Jan 06 2024 04:01:13
%S 1,2,3,4,6,8,12
%N Positive integers k such that tau(k) >= k/2, where tau(k) = A000005(k).
%C There are only 7 positive integers which meet this constraint. These happen to be proper divisors of 24. Inductively, there can only be a finite number of integers which meet this constraint. 1 has a perfect tau(k) / k ratio at 1. Every time a j-th power of a prime is multiplied by it, its ratio is multiplied by (j + 1)/p^j. Although 2 also achieves a perfect score, the scores must degrade after 2 because the above ratio is less than 1 otherwise.
%e tau(1) = 1 >= 0.5
%e tau(2) = 2 >= 1
%e tau(3) = 2 >= 1.5
%e tau(4) = 3 >= 2
%e so 1, 2, 3, 4 are in the sequence.
%e tau(5) = 2 < 2.5
%e so 5 is not in the sequence.
%t Select[Range[10^3], 2 DivisorSigma[0, #] >= # &] (* _Michael De Vlieger_, Jan 20 2019 *)
%o (PARI) for (n = 1, 100, if (sigma(n, 0) >= n / 2, print1(n, ", ")));
%Y Cf. A000005, A368523.
%Y Equals A018253 without 24.
%K nonn,full,fini
%O 1,2
%A _Keith J. Bauer_, Jan 12 2019
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