%I #23 Mar 13 2018 04:14:51
%S 1,0,1,0,1,1,0,0,2,1,0,4,4,3,1,0,0,0,4,5,1,0,16,20,12,6,5,1,0,0,0,48,
%T 20,26,10,1,0,0,96,40,52,44,36,11,1,0,0,0,72,48,66,34,22,9,1,0,576,
%U 720,392,384,188,154,70,26,9,1,0,0,0,0,0,480,848,560
%N Table T(d,n), where T(d,n)/lcm(1..d) gives the probability that d is the n-th divisor of an integer.
%C By probability is meant limit density on [1,n] as n grows without bound.
%C Equivalently: T(n,d) is lcm(1..d) times the asymptotic density of the set of natural numbers whose n-th divisor is equal to d.
%C Rows are (1), (0,1), (0,1,1), (0,0,2,1), ....
%H David W. Wilson, <a href="/A181930/b181930.txt">Table of n, a(n) for n = 1..820</a>
%F T(d,d) = 1.
%F T(d,n) = 0 if n < tau(d) = A000005(d). If d is a divisor of n, then every divisor of d is also a divisor of n, and d is therefore at least the tau(n)-th divisor of n.
%F T(d,n) > 0 for all tau(d) <= n <= d. To show this, let S be the set of the divisors of d along with the smallest n-tau(d) non-divisors of d. Then S has n elements, the largest being d. The elements of S are the smallest divisors of lcm(S), so d is the n-th divisor of lcm(S). Hence every number of the form lcm(S) + k*lcm(1..d) has n-th divisor d, so numbers with n-th divisor d have asymptotic density >= 1/lcm(1..d), and T(d,n) > 0.
%F Sum_{d>=1} T(d,n)/lcm(1..d) = 1.
%F Sum_{n=1..d} T(d,n)/lcm(1..d) = 1/d.
%F T(d,tau(d)) = (lcm(1..d)/d) * Product( (q-1)/q, q prime and there is an a with q^a < d and q^a does not divide d). In particular, if p is prime, then T(p,2) = (lcm(1..p)/p) * product( (q-1)/q, q prime and q < d). - _Benoit Jubin_, Apr 02 2012
%e T(5,4) = 4th element of the fifth row (0,4,4,3,1) = 3. T(5,4)/lcm(1..5) = T(5,4)/A003418(5) = T(5,4)/60 = 1/20 is the probability that 5 is the 4th divisor of an integer.
%K nonn,tabl
%O 1,9
%A _David W. Wilson_, Apr 02 2012
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