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The author has shown that a(n) == 2 (mod p_n) if p_n == 3 (mod 4). He has also established the following general theorem:
Let p be any odd prime and let d be any quadratic non-residue modulo p. Then we have the congruence
Product_{x=1..(p-1)/2} (x^2 - d) == 2*(-1)^((p+1)/2) (mod p).
This can be proved as follows: By Wilson's theorem we have (((p-1)/2)!)^2 == (-1)^((p+1)/2) (mod p), and thus we reduce the desired congruence to
Product_{0<k<p,(k/p)=-1} (1 - k) == 2 (mod p). (*)
Clearly
Product_{1<k<p, (k/p)=1} (1 - k)
== Product_{j=2..(p-1)/2} (1 - j^2)
= (-1)^((p+1)/2)*((p-1)/2)!)^2*(p+1)/(2p-2)
== -1/2 (mod p),
and Product_{k=2..p-1} (1 - k) = (-1)^(p-2)*(p-2)! == -1 (mod p) by Wilson's theorem. Therefore (*) follows.
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