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Number of primes p(i) for i=1 to 20*10^n such that p(i+1)=p(i)+6 and p(i+2)=p(i)+12 and p(i+3)=p(i)+18, 4 consecutive primes in arithmetic progression 0 6 12 18.
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%I #7 Nov 03 2024 13:48:31

%S 1,12,64,416,2748,19348

%N Number of primes p(i) for i=1 to 20*10^n such that p(i+1)=p(i)+6 and p(i+2)=p(i)+12 and p(i+3)=p(i)+18, 4 consecutive primes in arithmetic progression 0 6 12 18.

%e for i=1 to 20*10^1 (200) only one prime p(54)=251

%e 251,257,263,269 consecutive primes in arithmetic progression 0,6,12,18

%e for i=1 to 20*10^2 (2000) 12 primes in arithmetic progression 0,6,12,18

%e for i=1 to 20*10^3 (20000) 64 primes in arithmetic progression 0,6,12,18

%t Table[Length[Select[Partition[Prime[Range[20*10^n]],4,1],Differences[#]=={6,6,6}&]],{n,6}] (* _Harvey P. Dale_, Nov 03 2024 *)

%K more,nonn

%O 1,2

%A _Pierre CAMI_, Mar 02 2005

%E Definition clarified by _Harvey P. Dale_, Jul 01 2015