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a(n) = binomial(n*Omega(n),Omega(n)) / n.
1

%I #15 Dec 16 2023 20:08:47

%S 1,1,1,7,1,11,1,253,17,19,1,595,1,27,29,39711,1,1378,1,1711,41,43,1,

%T 138415,49,51,3160,3403,1,3916,1,25637001,65,67,69,477191,1,75,77,

%U 657359,1,7750,1,8515,8911,91,1,132563501,97,11026,101,11935,1,1633355

%N a(n) = binomial(n*Omega(n),Omega(n)) / n.

%C Omega(.) = A001222(.) is the number of prime divisors of n (counted with multiplicity).

%C binomial(nk,k)= n*binomial(nk-1,k-1) ensures that all entries are integers.

%C Subcases for this sequence:

%C If n is prime, Omega(n) = 1, and a(n) = binomial (n,1) / n = 1.

%C If n and n+1 are products of two primes (A070552), then Omega(n) = Omega(n+1) = 2, and binomial(n*Omega(n), Omega(n)) / n = binomial(2*n, 2) / n = 2*n-1 and binomial(2*(n+1), 2) / (n+1) = 2*n+1, and we obtain two consecutive numbers of the form (x, x+2), for example (17,19), (27,29), (41,43),... at n =9, 14...

%C Chaining this property: If n, n+1, and n+2 are semiprimes (A056809) , we find three consecutive numbers of the form (x, x+2,x+4), for example (65, 67, 69), (169, 171, 173), at n=33, 85.

%C At places where Omega(n)=3, we find the subsequence A060544, for example a(8) = A060544(8).

%C At places where Omega(n)=4, we find the subsequence A015219.

%H Harvey P. Dale, <a href="/A178034/b178034.txt">Table of n, a(n) for n = 1..1000</a>

%e a(8) = binomial(8*Omega(8),Omega(8))/8 = binomial(8*3,3)/8 = 2024/8 = 253.

%p A178034 := proc(n)

%p local o ;

%p o := numtheory[bigomega](n) ;

%p binomial(n*o,o)/n ;

%p end proc: # _R. J. Mathar_, Jul 08 2012

%t bon[n_]:=Module[{o=PrimeOmega[n]},Binomial[n*o,o]/n]; Array[bon,60] (* _Harvey P. Dale_, Jul 22 2014 *)

%o (PARI) a(n)=my(b=bigomega(n));binomial(n*b,b)/n \\ _Charles R Greathouse IV_, Oct 25 2012

%Y Cf. A001358, A038456

%K nonn

%O 1,4

%A _Michel Lagneau_, May 17 2010