A241786 Smallest k such that the number of the first even exponents in prime power factorization of (2*k)! is n, or a(n)=0 if there is no such k.

1, 6, 3, 5, 10, 24, 27, 169, 924, 3168, 720, 3208, 408, 35421, 50878, 73920, 18757, 204513, 134418, 295680, 427684, 2746710, 6867848, 14476645, 7278558, 3668406, 737564, 245340483, 1931850660, 1514239096, 3228582476, 1325085081, 16188866895, 33517640073

Offset: 0

Author

Vladimir Shevelev, Apr 28 2014

Keywords

• nonn

Comments

Conjecture: 1) All a(n)>0; 2) a(2*n+1)>a(2*n).

Conjecture (2) is wrong because a(24) = 7278558 >= a(25) = 3668406.

a(35) > 10^11; a(36) = 8036409193. - Hiroaki Yamanouchi, Sep 29 2014

Examples

			a(2)=3, since (2*3)!= 2^4*3^2*5, and here the number of the first even exponents is 2.
		

References

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

Links

• Giovanni Resta, Table of n, a(n) for n = 0..44 (terms a(0)-a(34) and a(36) from Hiroaki Yamanouchi)
• D. Berend, Parity of exponents in the factorization of n!, J. Number Theory, 64 (1997), 13-19.

Crossrefs

Cf. A240537, A240606, A240620.

Programs

PARI

nbev(n) = {f = factor(n); nbe = 0; i = 1; while ((i <= #f~) && ((f[i, 2] % 2) == 0), i++; nbe++); nbe;}
a(n) = {k = 0; while(nbev((2*k)!) != n, k++); k;} \\ Michel Marcus, Apr 30 2014

Extensions

More terms from Peter J. C. Moses, May 06 2014

a(21)-a(33) from Hiroaki Yamanouchi, Sep 29 2014