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%I #5 Mar 30 2012 18:39:12
%S 1,18,43920,4397760,122377651200,30993058252800,3921055557027840000,
%T 711860093387348628602880000,551343268152723132127641600000,
%U 567168796017902020698219516788736000000
%N 1/(2p^2)*(p-1)!*sum(j=1,p-1,1/j) where p runs through the primes >=5.
%C Always an integer.
%F a(n)=A000254(A000040(n)-1)/2/A000040(n)^2
%K nonn
%O 3,2
%A _Benoit Cloitre_, Feb 09 2003
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