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%I #28 Apr 08 2020 04:19:39
%S 1,6,3,5,10,24,27,169,924,3168,720,3208,408,35421,50878,73920,18757,
%T 204513,134418,295680,427684,2746710,6867848,14476645,7278558,3668406,
%U 737564,245340483,1931850660,1514239096,3228582476,1325085081,16188866895,33517640073
%N 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.
%C Conjecture: 1) All a(n)>0; 2) a(2*n+1)>a(2*n).
%C Conjecture (2) is wrong because a(24) = 7278558 >= a(25) = 3668406.
%C a(35) > 10^11; a(36) = 8036409193. - _Hiroaki Yamanouchi_, Sep 29 2014
%D P. Erdős, P. L. Graham, Old and new problems and results in combinatorial number theory, L'Enseignement Mathematique, Imprimerie Kunding, Geneva, 1980.
%H Giovanni Resta, <a href="/A241786/b241786.txt">Table of n, a(n) for n = 0..44</a> (terms a(0)-a(34) and a(36) from Hiroaki Yamanouchi)
%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.
%e a(2)=3, since (2*3)!= 2^4*3^2*5, and here the number of the first even exponents is 2.
%o (PARI) nbev(n) = {f = factor(n); nbe = 0; i = 1; while ((i <= #f~) && ((f[i, 2] % 2) == 0), i++; nbe++); nbe;}
%o a(n) = {k = 0; while(nbev((2*k)!) != n, k++); k;} \\ _Michel Marcus_, Apr 30 2014
%Y Cf. A240537, A240606, A240620.
%K nonn
%O 0,2
%A _Vladimir Shevelev_, Apr 28 2014
%E More terms from _Peter J. C. Moses_, May 06 2014
%E a(21)-a(33) from _Hiroaki Yamanouchi_, Sep 29 2014
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