OFFSET
1,1
COMMENTS
From Altug Alkan, Apr 22 2016: (Start)
For n > 1, a(n) is (p-3)*(p+3)/2 where p is the n-th prime. The reason is that the greatest common divisor of p-3 and p+3 is always 2 where p is the n-th prime and n > 2.
Proof: Let us assume that q is the greatest common divisor of p-3 and p+3. Because of the fact that any divisor of a and b must divide a-b, we know that q must divide 6. Note that q cannot be a multiple of 3 because p is prime, that is, q must be 1 or 2. Since we know that p-3 and p+3 are always even numbers for odd prime p, q must be 2 because we define it as the greatest common divisor.
If the greatest common divisor of p-3 and p+3 is always 2 where p is the n-th prime and n > 2, then the least common multiple of p-3 and p+3 must be (p-3)*(p+3)/2 where p is the n-th prime and n > 2 because of the general identity lcm(a, b) * gcd(a, b) = a*b. Note that for p = 3, (p-3)*(p+3)/t always is equal to 0 for any nonzero integer t, so it can be said that a(n) is (p-3)*(p+3)/2 where p is the n-th prime and n > 1. (End)
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..5000
MAPLE
A166011:=n->lcm(ithprime(n)+3, ithprime(n)-3): seq(A166011(n), n=1..100); # Wesley Ivan Hurt, Apr 22 2016
MATHEMATICA
f[n_]:=LCM[n-3, n+3]; lst={}; Do[p=Prime[n]; AppendTo[lst, f[p]], {n, 5!}]; lst
LCM[#+3, #-3]&/@Prime[Range[50]] (* Harvey P. Dale, Aug 09 2015 *)
PROG
(PARI) a(n) = lcm(prime(n)-3, prime(n)+3); \\ Michel Marcus, Apr 22 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Joseph Stephan Orlovsky, Oct 04 2009
STATUS
approved