A228120 a(n) = (1^2 + 1)*(2^2 + 1)*(3^2 + 1)*...*(((prime(n) - 1)/2)^2 + 1).

2, 10, 100, 44200, 1635400, 5315050000, 435834100000, 5370347780200000, 26078677338040210000000, 5893781078397087460000000, 142760638726203851727985000000000, 20723419838773203524537758570000000000, 9159751568737755957845689287940000000000, 2354514140744040168964234431464977000000000000

Offset

2,1

Comments

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.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 2..30

Example

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

Mathematica

a[n_]:=Product[(x^2+1), {x, 1, (Prime[n]-1)/2}]
Table[a[n], {n, 2, 15}]

Crossrefs

Cf. A101686, A000040.

Sequence in context: A101686 A324241 A188193 * A377919 A074109 A291101

Adjacent sequences: A228117 A228118 A228119 * A228121 A228122 A228123

Keyword

nonn

Author

Zhi-Wei Sun, Aug 11 2013

Status

approved