A228120 - OEIS

%I #23 Feb 24 2022 02:55:09

%S 2,10,100,44200,1635400,5315050000,435834100000,5370347780200000,

%T 26078677338040210000000,5893781078397087460000000,

%U 142760638726203851727985000000000,20723419838773203524537758570000000000,9159751568737755957845689287940000000000,2354514140744040168964234431464977000000000000

%N a(n) = (1^2 + 1)*(2^2 + 1)*(3^2 + 1)*...*(((prime(n) - 1)/2)^2 + 1).

%C 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:

%C Let p be any odd prime and let d be any quadratic non-residue modulo p. Then we have the congruence

%C Product_{x=1..(p-1)/2} (x^2 - d) == 2*(-1)^((p+1)/2) (mod p).

%C 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

%C        Product_{0<k<p,(k/p)=-1} (1 - k) == 2 (mod p).   (*)

%C   Clearly

%C         Product_{1<k<p, (k/p)=1} (1 - k)

%C      == Product_{j=2..(p-1)/2} (1 - j^2)

%C       = (-1)^((p+1)/2)*((p-1)/2)!)^2*(p+1)/(2p-2)

%C      == -1/2 (mod p),

%C and Product_{k=2..p-1} (1 - k) = (-1)^(p-2)*(p-2)! == -1 (mod p) by Wilson's theorem. Therefore (*) follows.

%H Zhi-Wei Sun, <a href="/A228120/b228120.txt">Table of n, a(n) for n = 2..30</a>

%e a(3) = (1^1+1)*(2^2+1)*(3^2+1) = 100 and a(3) == 2 (mod 7).

%t a[n_]:=Product[(x^2+1),{x,1,(Prime[n]-1)/2}]

%t Table[a[n],{n,2,15}]

%Y Cf. A101686, A000040.

%K nonn

%O 2,1

%A _Zhi-Wei Sun_, Aug 11 2013