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%I #25 Dec 08 2016 10:11:50
%S 5,0,8,20,56,80,140,176,260,416,476,680,836,920,1100,1400,1736,1856,
%T 2240,2516,2660,3116,3440,3956,4700,5096,5300,5720,5936,6380,8060,
%U 8576,9380,9656,11096,11396,12320,13280,13940,14960,16016,16376,18236,18620
%N Least common multiple of prime(n)-3 and prime(n)+3.
%C From _Altug Alkan_, Apr 22 2016: (Start)
%C 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.
%C 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.
%C 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)
%H G. C. Greubel, <a href="/A166011/b166011.txt">Table of n, a(n) for n = 1..5000</a>
%p A166011:=n->lcm(ithprime(n)+3,ithprime(n)-3): seq(A166011(n), n=1..100); # _Wesley Ivan Hurt_, Apr 22 2016
%t f[n_]:=LCM[n-3,n+3]; lst={};Do[p=Prime[n];AppendTo[lst,f[p]],{n,5!}]; lst
%t LCM[#+3,#-3]&/@Prime[Range[50]] (* _Harvey P. Dale_, Aug 09 2015 *)
%o (PARI) a(n) = lcm(prime(n)-3, prime(n)+3); \\ _Michel Marcus_, Apr 22 2016
%Y Cf. A066830, A084921, A166010.
%K nonn,easy
%O 1,1
%A _Vladimir Joseph Stephan Orlovsky_, Oct 04 2009
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