A240606 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.

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, 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, 0, 0, 4, 5, 0, 0, 2

Offset: 1

Author

Vladimir Shevelev, Apr 09 2014

Keywords

• nonn

Comments

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

Examples

			(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.
		

References

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

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• D. Berend, Parity of exponents in the factorization of n!, J. Number Theory, 64 (1997), 13-19.
• Y.-G. Chen, On the parity of exponents in the standard factorization of n!, J. Number Theory, 100 (2003), 326-331.

Crossrefs

Cf. A240537.

Programs

Mathematica

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

PARI

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

Sage

def a(n):
    f = list(factor(factorial(2*n)))
    c = -1
    for pf in f:
        c = c + 1
        if pf[1] % 2:
            return c   # Ralf Stephan, Apr 09 2014

Extensions

More terms and example from Ralf Stephan, Apr 09 2014