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%I #36 Apr 08 2020 07:26:06
%S 6,6,10,20,48,54,338,816,816,816,816,816,37514,37514,37514,37514,
%T 268836,268836,591360,855368,1475128,1475128,1475128,1475128,1475128,
%U 1475128,127632241,472077979,472077979,472077979,472077979,472077979,472077979,16072818386
%N a(n) is the smallest k such that in the prime power factorization of k! at least the first n positive exponents are even.
%C The sequence is nondecreasing and, by Berend's theorem, a(n) --> infinity as n goes to infinity.
%C The distinct terms 6, 10, 20, 48, 54, 338, 816, 37514, 268836, ... repeat 2, 1, 1, 1, 1, 1, 5, 4, 2, ... times.
%D P. Erdős, P. L. Graham, Old and new problems and results in combinatorial number theory, L'Enseignement Mathématique, Imprimerie Kunding, Geneva, 1980.
%H Giovanni Resta, <a href="/A240620/b240620.txt">Table of n, a(n) for n = 1..46</a> (first 36 terms 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.
%H Y.-G. Chen, <a href="http://dx.doi.org/10.1016/S0022-314X(03)00013-1">On the parity of exponents in the standard factorization of n!</a>, J. Number Theory, 100 (2003), 326-331.
%e Prime power factorizations of k! for k = 2, 3, 4, 5, 6 are 2, 2*3, 2^3*3, 2^3*3*5, 2^4*3^2*5. Thus the least k having at least 1 first even exponent is 6, and 6 is also the least k having at least 2 first even exponents. So a(1) = a(2) = 6.
%o (PARI) isokp(n,k) = {my(fk = k!, f = factor(fk)); if (#f~ < n, return (0)); if (f[n,1] != prime(n), return (0)); for (j=1, n, if (f[j,2] % 2, return(0));); return(1);}
%o a(n) = {my(k=1); while (! isokp(n,k), k++); k;} \\ _Michel Marcus_, Feb 04 2016
%Y Cf. A240537, A240606, A240619.
%K nonn
%O 1,1
%A _Vladimir Shevelev_, Apr 09 2014
%E a(17)-a(18) corrected and a(19)-a(34) added by _Hiroaki Yamanouchi_, Sep 05 2014
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