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%I #15 May 13 2013 01:50:01
%S 3,23,223,31223,231223,31231223,2931231223,372931231223,
%T 17372931231223,1317372931231223,1971317372931231223,
%U 1571971317372931231223,891571971317372931231223,79891571971317372931231223,25179891571971317372931231223,4325179891571971317372931231223
%N a(1) = 3 ; a(n+1) is the prime number obtained by the concatenation {p, a(n)} where p is the smallest prime prefix.
%C By Xylouris' version of Linnik's theorem, a(n) << 3^(6.2^n + cn) for some constant c. [_Charles R Greathouse IV_, Dec 27 2011]
%e a(1) = 3;
%e a(2) = 23 because 2 is the smallest prime prefix and 23 is prime;
%e a(3) = 223 because 2 is the smallest prime prefix and 223 is prime;
%e a(4) = 31223 because 31 is the smallest prime prefix and 31223 is prime.
%p a0:=3: printf(`%d, `,a0):for it from 1 to 20 do: i:=0:for n from 1 to 1000 while(i=0) do:p0:=ithprime(n):n0:=length(a0):x:=p0*10^n0+a0: if type(x,prime)=true then printf(`%d, `,x):i:=1:a0:=x:else fi:od:od:
%Y Cf. A173291.
%K nonn,base
%O 1,1
%A _Michel Lagneau_, Dec 27 2011
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