A380472 a(n) = gcd_{primes P >= prime(n+1)} Product_{i=1..n} (P^2-i^2).
1, 24, 360, 40320, 1814400, 479001600, 43589145600, 20922789888000, 3201186852864000, 2432902008176640000, 562000363888803840000, 620448401733239439360000, 201645730563302817792000000, 304888344611713860501504000000, 132626429906095529318154240000000, 263130836933693530167218012160000000
Offset: 0
Examples
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.
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.
a(3) = 40320 because 40320 = GCD{(P^2-1^2)(P^2-2^2)(P^2-3^2)}.
b(5) = 231 = a(7)/a(6).
c(2) = 112 = a(3)/a(2).
Programs
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Maple
seq((2*n + 2)!*(3/4 - (-1)^n/4), n = 0..20)
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Mathematica
Table[(2*n + 2)!*(3/4 - (-1)^n/4), {n, 0, 20}]
Formula
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.
a(n) = (2*n+2)!*(3/4-(-1)^n/4).
b(n) = (2*n+1)*(4*n+1) = a(2n)/a(2n-1) for n>=1 gives the odd ratios of a(n) (A014634).
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).
Sum_{n>=0} 1/a(n) = 3*cosh(1)/2 - cos(1)/2 - 1. - Amiram Eldar, Jul 03 2025
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