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

1,3

Comments

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

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.

Example

(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.

Mathematica

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

Prog

(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
(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

Crossrefs

Cf. A240537.

Sequence in context: A298610 A186492 A137448 * A335889 A324379 A035165

Adjacent sequences: A240603 A240604 A240605 * A240607 A240608 A240609

Keyword

nonn

Author

Vladimir Shevelev, Apr 09 2014

Extensions

More terms and example from Ralf Stephan, Apr 09 2014

Status

approved