A250032 a(n) is the numerator of the density of natural numbers m such that gcd(m,floor(m/n))>1.

1, 1, 1, 11, 7, 19, 16, 117, 269, 877, 1003, 11243, 4261, 56163, 61883, 199663, 107339, 919889, 2009948, 38444267, 41354174, 43432679, 46078049, 266161243, 379669754, 387106183, 407127338, 1258564159, 1322304979, 19229195413, 40830611677, 634491904301, 2638247862269, 2717256540199, 2823435623209, 2886468920107, 1006725304509

Offset: 1

Author

Stanislav Sykora, Nov 16 2014

Keywords

• nonn
• frac

Comments

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).

Examples

			When n=1, S includes all natural numbers except 1, so d(1)=1. Hence a(1)=1 and A250033(1)=1.
When n=2, S includes all even numbers greater than 2, so d(2)=1/2. Hence a(2)=1 and A250033(2)=2.
When n=10, the subset S is A248500 and d(10)=877/2100. Hence a(10)=877 and A250033(10)=2100.
When n=16, S is A248502 and d(16)=199663/480480. Hence a(16)=199663 and A250033(16)=480480.
		

Links

• Stanislav Sykora, Table of n, a(n) for n = 1..1000
• S. Sykora, On some number densities related to coprimes, Stan's Library, Vol. V, Nov 2014, DOI: 10.3247/SL5Math14.005

Crossrefs

Cf. A229099, A248500, A248502, A250031, A250033, A250034.

Programs

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)};
s(n)=1+s_aux(n,1,1);
a=vector(1000,n,numerator(s(n-1)/n))

Formula

For n>1, a(n)/A250033(n) = s(n-1)/n, where s(n) = A250034(n)/A072155(n).

lim(n->infinity)a(n)/A250033(n) = 1-1/zeta(2) = A229099.