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%I #24 Dec 05 2015 16:37:57
%S 1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,2,1,2,1,1,1,3,1,1,1,2,1,1,1,1,1,1,
%T 1,3,1,1,1,3,1,1,1,2,2,1,1,4,1,2,1,2,1,3,1,3,1,1,1,3,1,1,2,1,1,1,1,2,
%U 1,1,1,4,1,1,2,2,1,1,1,4,1,1,1,3,1,1,1,3,1,3,1,2,1,1,1,5,1,2,2,3
%N Number of values of k, n included, such that the distinct prime divisors of k are the same as those of n and Omega(k) = Omega(n), where Omega(n) is the number of prime divisors of n with multiplicity (A001222); a(1) = 1.
%C a(n) > 1 if 1 < omega(n) < Omega(n), where omega(n) is the number of distinct prime divisors of n (A001221).
%C a(n) = 1 if n is a squarefree number (A005117) or a prime power p^e where p is a prime and e >= 2 (A246547).
%H Robert Israel, <a href="/A264857/b264857.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = C(A001222(n)-1, A001221(n)-1). - _Robert Israel_, Nov 27 2015
%e The numbers k with the same prime divisors 2 and 3 are (6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...) and those for which A001222(k) = 3 are (12, 18), so a(12) = a(18) = 2.
%e The numbers k with the same prime divisor 5 are (5, 25, 125, 625, 3125, ...) and those for which A001222(k) = 4 are (625), so a(625) = 1.
%p seq(binomial(numtheory:-bigomega(n)-1, nops(numtheory:-factorset(n))-1) ,n=1..200); # _Robert Israel_, Nov 27 2015
%t Table[Binomial[PrimeOmega@ n - 1, PrimeNu@ n - 1], {n, 100}] (* _Michael De Vlieger_, Nov 28 2015 *)
%o (PARI) a(n) = if(n==1, 1, binomial(bigomega(n)-1, omega(n)-1)) \\ _Altug Alkan_, Dec 04 2015
%Y Cf. A001221, A001222.
%K nonn
%O 1,12
%A _Gionata Neri_, Nov 26 2015
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