

A181930


Table T(d,n), where T(d,n)/LCM(1..d) gives the probability that d is the nth divisor of an integer.


1



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, 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, 720, 392, 384, 188, 154, 70, 26, 9, 1, 0, 0, 0, 0, 0, 480, 848, 560
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OFFSET

1,9


COMMENTS

By probability is meant limit density on [1,n] as n grows without bound.
Equivalently: T(n,d) is LCM(1..d) times the asymptotic density of the set of natural numbers whose nth divisor is equal to d.
Rows are (1), (0,1), (0,1,1), (0,0,2,1),....


LINKS

David W. Wilson, Table of n, a(n) for n = 1..820


FORMULA

T(d,d) = 1.
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.
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 ntau(d) nondivisors 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 nth divisor of LCM(S). Hence every number of the form LCM(S) + k LCM(1..d) has nth divisor d, so numbers with nth divisor d have asymptotic density >= 1/LCM(1..d), and T(d,n) > 0.
SUM(d=1..inf; T(d,n)/LCM(1..d)) = 1.
SUM(n=1..d; T(d,n)/LCM(1..d)) = 1/d.
T(d,tau(d)) = LCM(1..d)/d * product( (q1)/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( (q1)/q, q prime and q<d).  Benoit Jubin, Apr 02 2012


EXAMPLE

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.


CROSSREFS

Sequence in context: A049243 A077908 A052922 * A256797 A109167 A066426
Adjacent sequences: A181927 A181928 A181929 * A181931 A181932 A181933


KEYWORD

nonn,tabl


AUTHOR

David W. Wilson, Apr 02 2012


STATUS

approved



