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A360661
Number of factorizations of n into a prime number of factors > 1.
1
0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 3, 0, 1, 1, 3, 0, 3, 0, 3, 1, 1, 0, 5, 1, 1, 2, 3, 0, 4, 0, 5, 1, 1, 1, 7, 0, 1, 1, 5, 0, 4, 0, 3, 3, 1, 0, 9, 1, 3, 1, 3, 0, 5, 1, 5, 1, 1, 0, 9, 0, 1, 3, 7, 1, 4, 0, 3, 1, 4, 0, 12, 0, 1, 3, 3, 1, 4, 0, 9, 3, 1, 0, 9, 1, 1, 1, 5, 0, 9, 1, 3, 1, 1, 1, 13, 0, 3, 3, 7
OFFSET
1,8
COMMENTS
From Bernard Schott, Mar 25 2023: (Start)
a(n) depends only on the prime signature of n.
a(n) = 0 iff n is in A008578 (1 with primes).
a(n) = 1 iff n is in A001358 (semiprimes).
a(n) = 2 iff n is in A030078 (p^3).
a(n) = 3 iff n is in A080258 (p^4 or p*q^2).
a(n) = 4 iff n is in A007304 (p*q*r). (End)
LINKS
Sean A. Irvine, Java program (github)
Eric Weisstein's World of Mathematics, Unordered Factorization
FORMULA
From Bernard Schott, Mar 25 2023: (Start)
a(A000040(n)) = 0.
a(A001248(n)) = a(A006881(n)) = 1.
a(A030514(n)) = a(A054753(n)) = 3. (End)
EXAMPLE
a(2) = 0 since 2 = 2 is the unique factorization of 2.
a(4) = 1 since 4 = 2^2 = 2 * 2.
a(6) = 1 since 6 = 2 * 3.
a(8) = 2 since 8 = 2^3 = 2 * 4 = 2 * 2 * 2.
a(12) = 3 since 12 = 3 * 2^2 = 2 * 6 = 3 * 4 = 2 * 2 * 3.
a(16) = 3 since 16 = 2^4 = 2 * 8 = 4 * 4 = 2 * 2 * 4.
a(30) = 4 since 30 = 2 * 3 * 5 = 2 * 15 = 3 * 10 = 5 * 6.
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Feb 15 2023
STATUS
approved