login
a(1) = 2. For n>1, a(n) = smallest m such that m == 0 (mod prime(n)), m + 1 == 0 (mod prime(n+1)) and m-1 == 0 (mod prime(n-1)).
0

%I #10 Jun 24 2014 01:08:33

%S 2,9,55,21,792,1937,170,4047,115,9454,31930,23902,2665,44978,65189,

%T 122483,134992,170983,220028,101104,85556,27887,296725,629141,154327,

%U 546208,46865,950588,1043893,1548891,70739,702946,2389965,1513988

%N a(1) = 2. For n>1, a(n) = smallest m such that m == 0 (mod prime(n)), m + 1 == 0 (mod prime(n+1)) and m-1 == 0 (mod prime(n-1)).

%C A subsequence of the mean of three consecutive numbers divisible by three consecutive primes (A-number?).

%e a(7) = 170. prime(7) = 17, prime(7-1) = 13, prime(7+1)=19 and 17 | 170, 19 | 171, 13 | 169 and 170 is the smallest such number.

%t (* First do <<NumberTheory`NumberTheoryFunctions` *) a[1]=2; a[n_] := ChineseRemainder[{1, 0, Prime[n+1]-1}, Prime/@Range[n-1, n+1]]

%K easy,nonn

%O 1,1

%A _Amarnath Murthy_, Dec 23 2002

%E More terms from _Jim Nastos_, Feb 19 2003 and _Dean Hickerson_, Feb 23 2003