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Define a(0)=3; then a(n)=k*a(n-1)^2-1=least prime first of twin primes.
1

%I #3 Mar 31 2012 13:22:04

%S 3,107,206081,20894934252011

%N Define a(0)=3; then a(n)=k*a(n-1)^2-1=least prime first of twin primes.

%C k sequence in A112880

%e 12*3^2-1=107, 107 and 109 twin primes so a(1)=107

%e 18*107^2-1=206081, 206081 and 206083 twin primes so a(2)=206081

%e a(3)=492*(18*(12*3^2-1)^2-1)^2-1

%e a(4)=702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1

%e a(5)=2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1

%e a(6)=2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1

%e a(7)=130650*(2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1

%e a(8)=980910*(130650*(2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1

%e a(9)=5528418*(980910*(130650*(2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1

%e a(10)=21357012*(5528418*(980910*(130650*(2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1

%Y Cf. A112880.

%K hard,nonn

%O 0,1

%A _Pierre CAMI_, Oct 01 2005