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 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.. 20
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS See comment in A240537. According to Berend's theorem, the sequence is unbounded. REFERENCES P. ErdÅ‘s, 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

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Last modified April 23 05:42 EDT 2021. Contains 343199 sequences. (Running on oeis4.)