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EXAMPLE
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For k=18018, tau(k)=48: the 48 divisors of k are 1, 2, 3, 6, 7, 9, 11, 13, 14, 18, 21, 22, 26, 33, 39, 42, 63, 66, 77, 78, 91, 99, 117, 126, 143, 154, 182, 198, 231, 234, 273, 286, 429, 462, 546, 693, 819, 858, 1001, 1287, 1386, 1638, 2002, 2574, 3003, 6006, 9009, 18018.
The least common multiple of the first 8 divisors, (1, 2, 3, 6, 7, 9, 11, 13), is again 18018, but the least common multiple of the first 7 divisors, (1, 2, 3, 6, 7, 9, 11), is less than 18018.
Since tau#(k)=8 (see A222084 for the definition of tau#(n)), tau(k)/tau#(k) = 48/8 = 6, and since 18018 is the minimum number k to have this ratio, a(6)=18018.
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MAPLE
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with(numtheory);
local a, b, c, d, j, n, t, v;
v:=array(1..100); for j from 1 to 100 do v[j]:=0; od; t:=0;
for n from 1 to q do
a:=ifactors(n)[2]; b:=nops(a); c:=0;
for j from 1 to b do if a[j][1]^a[j][2]>c then c:=a[j][1]^a[j][2]; fi; od;
a:=op(sort([op(divisors(n))])); b:=nops(divisors(n));
for j from 1 to b do if a[j]=c then break; fi; od;
if type(tau(n)/j, integer) then if tau(n)/j=t+1
then t:=t+1; lprint(t, n); while v[t+1]>0 do t:=t+1; lprint(t, v[t]); od;
else if tau(n)/j>t+1 then if v[tau(n)/j]=0 then v[tau(n)/j]:=n; fi; fi;
fi; fi; od; end:
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