A377071 - OEIS

%I #5 Oct 29 2024 11:21:38

%S 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,

%T 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,

%U 4,3,10,1,6,1,3,4,4,3,10,1,6,1,3,1,15,3

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

%C 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).

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

%H Michael De Vlieger, <a href="/A377071/b377071.txt">Table of n, a(n) for n = 1..10000</a>

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

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

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

%e 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.

%e 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.

%e 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.

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

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

%K nonn,easy

%O 1,6

%A _Michael De Vlieger_, Oct 25 2024