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A061299
Least number whose number of divisors is A007304(n) (the n-th number that is the product of 3 distinct primes).
8
720, 2880, 46080, 25920, 184320, 2949120, 129600, 414720, 11796480, 1658880, 188743680, 3732480, 2073600, 26542080, 12079595520, 14929920, 48318382080, 106168320, 8294400, 3092376453120, 1698693120, 18662400, 238878720
OFFSET
1,1
COMMENTS
All terms are divisible by a(1) = 720, the first entry.
All terms [=a(j)], not only arguments [=j] have 3 distinct prime factors at exponents determined by the p,q,r factors of their arguments: a(pqr) = RPQ.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..3706 (terms 1..835 from David A. Corneth)
FORMULA
a(n) = A005179(A007304(n)).
Min{x; A000005(x) = pqr} p, q, r are distinct primes. If k = pqr and p > q > r then A005179(k) = 2^(p-1)*3^(q-1)*5^(r-1).
From Reinhard Zumkeller, Jul 15 2004: (Start)
A000005(a(n)) = A007304(n).
A000005(m) != A007304(n) for m < a(n).
a(n) = A005179(A007304(n)).
a(p*m*q) = 2^(q-1) * 3^(m-1) * 5^(p-1) for primes p < m < q.
a(A000040(i)*A000040(j)*A000040(k)) = 2^(A084127(k)-1) * 3^(A084127(j)-1) * 5^(A084127(i)-1) for i < j < k. (End)
EXAMPLE
For n = 5: A007304(5) = 78 = 2*3*13, A005179(78) = 184320 = (2^12)*(3^2)*(5^1) = a(5).
KEYWORD
nonn,changed
AUTHOR
Labos Elemer, Jun 05 2001
EXTENSIONS
Edited by N. J. A. Sloane, Apr 20 2007
STATUS
approved