%I #23 Jul 19 2015 11:30:02

%S 0,0,2,0,3,1,0,0,2,4,0,3,0,0,2,0,1,1,0,2,0,0,1,5,0,0,6,0,1,3,0,0,1,1,

%T 0,3,0,0,4,2,0,0,1,0,2,2,0,5,0,0,1,0,2,1,0,0,1,1,0,3,0,0,1,0,6,1,0,2,

%U 0,0,4,5,0,0,2

%N Let the prime factorization of (2*n)! be 2^e_1*3^e_2*5^e_3*...; then a(n) = maximal k such that e_1, ..., e_k are all even..

%C See comment in A240537. According to Berend's theorem, the sequence is unbounded.

%D P. Erdős, P. L. Graham, Old and new problems and results in combinatorial number theory, L'Enseignement Mathematique, Imprimerie Kunding, Geneva, 1980.

%H Charles R Greathouse IV, <a href="/A240606/b240606.txt">Table of n, a(n) for n = 1..10000</a>

%H D. Berend, <a href="http://dx.doi.org/10.1006/jnth.1997.2106">Parity of exponents in the factorization of n!</a>, J. Number Theory, 64 (1997), 13-19.

%H Y.-G. Chen, <a href="http://dx.doi.org/10.1016/S0022-314X(03)00013-1">On the parity of exponents in the standard factorization of n!</a>, J. Number Theory, 100 (2003), 326-331.

%e (2*10)! = 2432902008176640000 = 2^18 * 3^8 * 5^4 * 7^2 * 11 * 13 * 17 * 19, and the first 4 exponents are even, so a(10) = 4.

%t Map[Count[First[Split[Mod[Last[Transpose[FactorInteger[(2*#)!]]],2]]],0]&,Range[75]] (* _Peter J. C. Moses_, Apr 09 2014 *)

%o (Sage) def a(n):

%o f = list(factor(factorial(2*n)))

%o c = -1

%o for pf in f:

%o c = c + 1

%o if pf[1] % 2:

%o return c # _Ralf Stephan_, Apr 09 2014

%o (PARI) fv(n,p)=my(s); while(n\=p, s+=n); s

%o a(n)=n*=2; my(s); forprime(p=2,,if(fv(n,p)%2, return(s), s++)) \\ _Charles R Greathouse IV_, Apr 09 2014

%Y Cf. A240537.

%K nonn

%O 1,3

%A _Vladimir Shevelev_, Apr 09 2014

%E More terms and example from _Ralf Stephan_, Apr 09 2014