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A307124 a(n) is twice the square of the product of the first n primes each decreased by one.

%I A307124 #45 Sep 08 2022 08:46:21
%S A307124 2,8,128,4608,460800,66355200,16986931200,5503765708800,
%T A307124 2663822603059200,2088436920798412800,1879593228718571520000,
%U A307124 2435952824419268689920000,3897524519070829903872000000,6875233251640943950430208000000
%N A307124 a(n) is twice the square of the product of the first n primes each decreased by one.
%C A307124 a(n) is also the number of pairs (c, d) for which d !== c (mod p) and d !== -c (mod p) for every prime p up to and including the n-th prime, where 1 <= c, d <= the product of the first n primes.
%C A307124 In some cases, c - d and c + d are prime with (c - d) + (c + d) = 2c, as in Goldbach's conjecture.
%H A307124 Sally M. Moite, <a href="http://vixra.org/abs/1906.0069">Partition of the Primorial Square By Remainder Agreement Counts</a>, viXra:1906.0069 (2019). Additional comments. Unedited.
%F A307124 From _Jon E. Schoenfield_, Mar 26 2019: (Start)
%F A307124 a(n) = 2*Product_{j=1..n} (prime(j)-1)^2.
%F A307124 a(n) = 2*A005867(n)^2. (End)
%e A307124 a(4) = 2 * 1^2 * 2^2 * 4^2 * 6^2 = 4608.
%e A307124 For n = 2, 2 * 3 = 6, the pairs (c, d) are (1, 6), (2, 3), (3, 2), (3, 4), (4, 3), (5, 6), (6, 1) and (6, 5), so a(2) = 8.
%t A307124 a[n_]:=2 Product[(Prime[i] - 1)^2, {i, 1, n}]; Array[a,10]
%t A307124 2*FoldList[Times,(Prime[Range[15]]-1)^2] (* _Harvey P. Dale_, Jul 20 2022 *)
%o A307124 (PARI) a(n) = 2*prod(k=1, n, prime(k)-1)^2; \\ _Michel Marcus_, Mar 27 2019
%o A307124 (Magma)  [2*(&*[(NthPrime(i)-1)^2: i in [1..n]]): n in [1..14]]; // _Marius A. Burtea_, Jun 19 2019
%Y A307124 Cf. A002110, A005867, A061742.
%K A307124 nonn
%O A307124 1,1
%A A307124 _Sally Myers Moite_, Mar 26 2019
%E A307124 First Mathematica program edited by _Harvey P. Dale_, Jul 20 2022