A377071 a(n) = binomial(bigomega(n) + omega(n) - 1, omega(n) - 1), where bigomega = A001222 and omega = A001221.

1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 4, 1, 3, 3, 1, 1, 4, 1, 4, 3, 3, 1, 5, 1, 3, 1, 4, 1, 10, 1, 1, 3, 3, 3, 5, 1, 3, 3, 5, 1, 10, 1, 4, 4, 3, 1, 6, 1, 4, 3, 4, 1, 5, 3, 5, 3, 3, 1, 15, 1, 3, 4, 1, 3, 10, 1, 4, 3, 10, 1, 6, 1, 3, 4, 4, 3, 10, 1, 6, 1, 3, 1, 15, 3

Offset: 1

Author

Michael De Vlieger, Oct 25 2024

Keywords

• nonn
• easy

Comments

Number of permutations of the integer partitions of omega(n) supplemented with zeros such that there are bigomega(n) parts, whose sum equals bigomega(n).

a(n) = cardinality of { m : rad(m) | n, bigomega(m) = bigomega(n) }, where rad = A007947.

Examples

			For n = 6, omega(6) = 2, bigomega(6) = 2, we have 3 exponent combinations [2,0], [1,1], [0,2]. Raising prime factors {2, 3} to these exponents yields {4, 6, 9}, i.e., row 6 of A377070.
For n = 10, omega(10) = 2, bigomega(10) = 2, we have 3 exponent combinations [2,0], [1,1], [0,2]. Raising prime factors {2, 5} to these exponents yields {4, 10, 25}, i.e., row 10 of A377070.
For n = 12, omega(12) = 2, bigomega(12) = 3, we have 4 exponent combinations [3,0], [2,1], [1,2], [0,3]. Raising prime factors {2, 3} to these exponents yields {8, 12, 18, 27}, i.e., row 6 of A377070.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A000961, A001221, A001222, A007947, A024619, A376248, A377070.

Programs

Mathematica

Array[Binomial[#2 + #1 - 1, #1 - 1] & @@ {PrimeNu[#], PrimeOmega[#]} &, 120]

Formula

a(n) is the length of row n of A377070.

For prime power p^k, k >= 0, a(p^k) = 1.

For n in A024619, a(n) > 1.