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A391374

Number of positive integers k such that k's arithmetic mean of divisors is at most n.

1, 3, 6, 8, 11, 14, 18, 20, 23, 26, 30, 35, 37, 41, 45, 47, 49, 55, 60, 62, 68, 69, 72, 79, 82, 84, 90, 94, 96, 102, 106, 111, 113, 115, 116, 124, 127, 134, 138, 141, 144, 153, 153, 156, 169, 171, 173, 178, 183, 184, 186, 190, 191, 199, 201, 208, 216, 216, 218, 227

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OFFSET

1,2

COMMENTS

a(n) is the number of integers k such that sigma(k)/tau(k) = A000203(k)/A000005(k) <= n.

a(n) is finite for all n since sigma(k) >= k+1 and tau(k) <= 2*sqrt(k) implies no qualifying k can exceed 4*n^2.

LINKS

Table of n, a(n) for n=1..60.

FORMULA

a(n) = |{k in Z+ : sigma(k) <= n*tau(k)}|.

EXAMPLE

For n=2, we need sigma(k)/tau(k) <= 2. The qualifying integers are 1, 2, 3 so a(2) = 3.