cp's OEIS Frontend

A380472 a(n) = gcd_{primes P >= prime(n+1)} Product_{i=1..n} (P^2-i^2).

%I A380472 #70 Jul 03 2025 09:17:37
%S A380472 1,24,360,40320,1814400,479001600,43589145600,20922789888000,
%T A380472 3201186852864000,2432902008176640000,562000363888803840000,
%U A380472 620448401733239439360000,201645730563302817792000000,304888344611713860501504000000,132626429906095529318154240000000,263130836933693530167218012160000000
%N A380472 a(n) = gcd_{primes P >= prime(n+1)} Product_{i=1..n} (P^2-i^2).
%C A380472 a(n) is the GCD of all numbers of the form Product_{i=1..n} (P^2-i^2) where P is a prime larger than or equal to the (n+1)-th prime.
%F A380472 E.g.f.: Sum_{n >= 0} a(n)/(2*n)!*z^(2*n) = (1 + 12*z^2 + 12*z^4 + 20*z^6 + 3*z^8)/(1 - z^4)^3.
%F A380472 a(n) = (2*n+2)!*(3/4-(-1)^n/4).
%F A380472 b(n) = (2*n+1)*(4*n+1) = a(2n)/a(2n-1) for n>=1 gives the odd ratios of a(n) (A014634).
%F A380472 c(n) = 4*2*n*(4*n-1) = a(2n-1)/a(2n-2) for n>=1 gives the even ratios of a(n) (4 times A014635).
%F A380472 Sum_{n>=0} 1/a(n) = 3*cosh(1)/2 - cos(1)/2 - 1. - _Amiram Eldar_, Jul 03 2025
%e A380472 a(1) = 24 because 24 = GCD{P^2-1^2} GCD is taken on all numbers of the form P^2-1^2 with P a prime and P>3. This implies that for all primes P>3, P^2-1 is divisible by 24.
%e A380472 a(2) = 360 because 360 = GCD{(P^2-1^2)(P^2-2^2)} GCD is taken on all numbers of the form (P^2-1^2)(P^2-2^2) with P a prime and P>5. This implies that for all primes P>5, (P^2-1^2)(P^2-2^2) is divisible by 360.
%e A380472 a(3) = 40320 because 40320 = GCD{(P^2-1^2)(P^2-2^2)(P^2-3^2)}.
%e A380472 b(5) = 231 = a(7)/a(6).
%e A380472 c(2) = 112 = a(3)/a(2).
%p A380472 seq((2*n + 2)!*(3/4 - (-1)^n/4), n = 0..20)
%t A380472 Table[(2*n + 2)!*(3/4 - (-1)^n/4), {n, 0, 20}]
%Y A380472 Cf. A014634 (odd ratio), A014635 (even ratio, multiplied by 4), A084920.
%K A380472 nonn,easy
%O A380472 0,2
%A A380472 _Lily Bétaz_, _Martin Chevalier_, and _Basile Fusil_, Jun 23 2025