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

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%I A377071 #5 Oct 29 2024 11:21:38
%S A377071 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 A377071 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 A377071 4,3,10,1,6,1,3,4,4,3,10,1,6,1,3,1,15,3
%N A377071 a(n) = binomial(bigomega(n) + omega(n) - 1, omega(n) - 1), where bigomega = A001222 and omega = A001221.
%C A377071 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 A377071 a(n) = cardinality of { m : rad(m) | n, bigomega(m) = bigomega(n) }, where rad = A007947.
%H A377071 Michael De Vlieger, <a href="/A377071/b377071.txt">Table of n, a(n) for n = 1..10000</a>
%F A377071 a(n) is the length of row n of A377070.
%F A377071 For prime power p^k, k >= 0, a(p^k) = 1.
%F A377071 For n in A024619, a(n) > 1.
%e A377071 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 A377071 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 A377071 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 A377071 Array[Binomial[#2 + #1 - 1, #1 - 1] & @@ {PrimeNu[#], PrimeOmega[#]} &, 120]
%Y A377071 Cf. A000005, A000961, A001221, A001222, A007947, A024619, A376248, A377070.
%K A377071 nonn,easy
%O A377071 1,6
%A A377071 _Michael De Vlieger_, Oct 25 2024