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%I #19 Dec 02 2014 16:45:27
%S 1,1,1,11,7,19,16,117,269,877,1003,11243,4261,56163,61883,199663,
%T 107339,919889,2009948,38444267,41354174,43432679,46078049,266161243,
%U 379669754,387106183,407127338,1258564159,1322304979,19229195413,40830611677,634491904301,2638247862269,2717256540199,2823435623209,2886468920107,1006725304509
%N a(n) is the numerator of the density of natural numbers m such that gcd(m,floor(m/n))>1.
%C Let m be any natural number, and P(m) a relational expression on m (i.e., a property of m) evaluating to either 0 (false) or 1 (true). This defines a subset S of natural numbers N for which P(m)=1. When there exists a limit d=limit(M->infinity, Sum(m=1..M, P(m))/M), d is said to be the limit mean density (or just density) of the subset S in N. Now, choose an integer parameter n and set P(m)=gcd(m,floor(m/n))>1. This makes the property P, the corresponding subset S, and the density d all dependent upon n. The reference proves that for any n>0, the density d(n) exists and is a rational number. The value of a(n) is the numerator of d(n), while A250033(n) is the denominator of d(n).
%H Stanislav Sykora, <a href="/A250032/b250032.txt">Table of n, a(n) for n = 1..1000</a>
%H S. Sykora, On some number densities related to coprimes, Stan's Library, Vol. V, Nov 2014, DOI: <a href="http://dx.doi.org/10.3247/SL5Math14.005">10.3247/SL5Math14.005</a>
%F For n>1, a(n)/A250033(n) = s(n-1)/n, where s(n) = A250034(n)/A072155(n).
%F lim(n->infinity)a(n)/A250033(n) = 1-1/zeta(2) = A229099.
%e When n=1, S includes all natural numbers except 1, so d(1)=1. Hence a(1)=1 and A250033(1)=1.
%e When n=2, S includes all even numbers greater than 2, so d(2)=1/2. Hence a(2)=1 and A250033(2)=2.
%e When n=10, the subset S is A248500 and d(10)=877/2100. Hence a(10)=877 and A250033(10)=2100.
%e When n=16, S is A248502 and d(16)=199663/480480. Hence a(16)=199663 and A250033(16)=480480.
%o (PARI) s_aux(n,p0,inp)={my(t=0/1,tt=0/1,in=inp,pp);while(1,pp=p0*prime(in);tt=n\pp;if(tt==0,break,t+=tt/pp-s_aux(n,pp,in++)));return(t)};
%o s(n)=1+s_aux(n,1,1);
%o a=vector(1000,n,numerator(s(n-1)/n))
%Y Cf. A229099, A248500, A248502, A250031, A250033, A250034.
%K nonn,frac
%O 1,4
%A _Stanislav Sykora_, Nov 16 2014
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