Number of +ve integers that have n as their greatest prime power

[hv at crypt.org]

While trying to refine a program to deliver guaranteed minimal values
for the previously discussed "sum of unit fractions" email I stumbled
on the following sequence, which seems reasonably interesting:

%I A000001
%S A000001 1,1,2,2,6,0,12,8,16,0,48,0,96,0,0,48,240,0,480,0,0,0,960,0,960,0
%T A000001 960,0,3840,0,7680,3072,0,0,0,0,18432,0,0,0,36864,0,73728,0,0,0
%U A000001 147456,0,147456,0,0,0,442368,0,0,0,0,0,884736,0,1769472,0,0,589824
%N A000001 For each prime power n, a(n) is the number of positive integers that have n as their greatest prime power.
%e A000001 a(4) = 2 since only 4 and 12 have 4 as their greatest prime power - all other multiples of 4 are divisible by 8, 9, or some prime >= 5.
%F A000001 a(1) = 1; a(p^k) = prod_{q <= p^k, q prime} { floor(k ln p / ln q) } / k when p prime, k >= 1, a(n) = 0 otherwise
%C A000001 a(n) is the number of occurrences of n in A034699
%K A000001 nonn,easy,new
%O A000001 1,3
%Y A000001 Cf A034699
%A A000001 hv at crypt.org (Hugo van der Sanden)

Can someone confirm that the formula is correct, and perhaps check that
my numbers are correct?

Hugo