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a(n) = smallest number k such that three consecutive prime numbers prime(n), prime(n+1) and prime(n+2) are divisors of k, k+1 and k+2 respectively.
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%I #15 Oct 12 2018 19:59:14

%S 8,54,20,791,1936,169,4046,114,9453,31929,23901,2664,44977,65188,

%T 122482,134991,170982,220027,101103,85555,27886,296724,629140,154326,

%U 546207,46864,950587,1043892,1548890,70738,702945,2389964

%N a(n) = smallest number k such that three consecutive prime numbers prime(n), prime(n+1) and prime(n+2) are divisors of k, k+1 and k+2 respectively.

%H Marius A. Burtea, <a href="/A180095/b180095.txt">Table of n, a(n) for n = 1..1227</a>

%e a(20) = 85555 is a term because prime(20) = 71 => 85555 = 71*1205 ; 85556 =

%e 73*1172 and 85557 = 79*1083 where 71, 73 and 79 are three consecutive primes.

%p with(numtheory):for p from 1 to 50 do: p1:=ithprime(p):p2:=ithprime(p+1):p3:=ithprime(p+2):it:=0:for n from 1 to 5000000 while(it=0) do:if irem(n,p1)=0 and irem(n+1,p2)=0 and irem(n+2,p3)=0 then it:=1:printf(`%d, `,n):else fi:od:od:

%t snk[n_]:=Module[{k=1},While[!AllTrue[{k,k+1,k+2}/n,IntegerQ],k++];k]; snk/@Partition[Prime[Range[35]],3,1] (* _Harvey P. Dale_, Feb 26 2015 *)

%o (Sage) def A180095(n): return crt([-2..0][::-1], [nth_prime(i) for i in [n..n+2]]) # _D. S. McNeil_, Jan 16 2011

%Y Cf. A077338.

%K nonn

%O 1,1

%A _Michel Lagneau_, Jan 16 2011