A360661 Number of factorizations of n into a prime number of factors > 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

Table of n, a(n) for n=1..100.

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.

Crossrefs

Cf. A001055, A038499, A339890.

Cf. A001248, A001358, A006881, A007304, A008578, A030078, A030514, A054753, A080258.

Sequence in context: A337345 A294904 A055212 * A028422 A064577 A371729

Adjacent sequences: A360658 A360659 A360660 * A360662 A360663 A360664

Keyword

nonn

Author

Ilya Gutkovskiy, Feb 15 2023

Status

approved