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A181930
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Table T(d,n), where T(d,n)/lcm(1..d) gives the probability that d is the n-th divisor of an integer.
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1
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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
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1,9
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COMMENTS
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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 n-th divisor is equal to d.
Rows are (1), (0,1), (0,1,1), (0,0,2,1), ....
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LINKS
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FORMULA
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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 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.
Sum_{d>=1} 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( (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
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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