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Number of ways of factoring n with all factors greater than 1
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What is the relation between the number of ways of factoring n with all factors greater than 1 and the number of proper divisors of n?
Sequences
Number of ways of factoring n with all factors greater than 1 (a(1)=1 either by convention or by considering 1 as the empty product of factors greater than 1). (Cf. A001055)
- {1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 2, 5, 1, 4, 1, 4, 2, 2, 1, 7, 2, 2, 3, 4, 1, 5, 1, 7, 2, 2, 2, 9, 1, 2, 2, 7, 1, 5, 1, 4, 4, 2, 1, 12, 2, 4, 2, 4, 1, 7, 2, 7, 2, 2, 1, 11, 1, 2, 4, 11, 2, 5, 1, 4, 2, 5, 1, 16, 1, ...}
For n > 0, a(n) = number of proper divisors (actually number of aliquot divisors, also number of divisors greater than 1) of n. (Cf. A032741(n), n > 0)
- {0, 1, 1, 2, 1, 3, 1, 3, 2, 3, 1, 5, 1, 3, 3, 4, 1, 5, 1, 5, 3, 3, 1, 7, 2, 3, 3, 5, 1, 7, 1, 5, 3, 3, 3, 8, 1, 3, 3, 7, 1, 7, 1, 5, 5, 3, 1, 9, 2, 5, 3, 5, 1, 7, 3, 7, 3, 3, 1, 11, 1, 3, 5, 6, 3, 7, 1, 5, 3, 7, 1, 11, 1, ...}
d(n) (also called tau(n) or sigma_0(n)), the number of divisors of n, n >= 1. (Cf. A000005)
- {1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 6, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 10, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 7, 4, 8, 2, 6, 4, 8, 2, 12, 2, ...}
See also
- Number of proper divisors (proper divisors of n being divisors d, 1 < d < n)
- Number of aliquot divisors (aliquot divisors of n being divisors d, 1 ≤ d < n)
- Number of divisors greater than 1 (divisors greater than 1 of n being divisors d, 1 < d ≤ n)
- Number of divisors (divisors d of n, 1 ≤ d ≤ n)