|
|
A166011
|
|
Least common multiple of prime(n)-3 and prime(n)+3.
|
|
3
|
|
|
5, 0, 8, 20, 56, 80, 140, 176, 260, 416, 476, 680, 836, 920, 1100, 1400, 1736, 1856, 2240, 2516, 2660, 3116, 3440, 3956, 4700, 5096, 5300, 5720, 5936, 6380, 8060, 8576, 9380, 9656, 11096, 11396, 12320, 13280, 13940, 14960, 16016, 16376, 18236, 18620
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
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
|
|
|
MAPLE
|
|
|
MATHEMATICA
|
f[n_]:=LCM[n-3, n+3]; lst={}; Do[p=Prime[n]; AppendTo[lst, f[p]], {n, 5!}]; lst
|
|
PROG
|
(PARI) a(n) = lcm(prime(n)-3, prime(n)+3); \\ Michel Marcus, Apr 22 2016
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|