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1/(2p^2)*(p-1)!*sum(j=1,p-1,1/j) where p runs through the primes >=5.
0

%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