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a(0)=11, then a(n) is the smallest prime p not already used such that (first two digits of p) = (last two digits of a(n-1)).
2

%I #9 Jan 07 2022 19:32:32

%S 11,113,13,131,31,311,1117,17,173,73,733,331,313,137,37,373,739,397,

%T 97,971,71,719,19,191,911,1123,23,233,337,379,79,797,977,773,7321,211,

%U 1129,29,293,937,3719,193,9311,1151,5113,139,3911,1153,53,5323,239,3917,179

%N a(0)=11, then a(n) is the smallest prime p not already used such that (first two digits of p) = (last two digits of a(n-1)).

%C The next-to-last digit of p is not allowed to be zero.

%C E.g., after a(5)=311, we don't take 1103 and 1109 but 1117.

%H Zak Seidov, <a href="/A175684/b175684.txt">Table of n, a(n) for n = 0..1000</a>

%t s={11};id2={1,1};Do[Do[p=Prime[k];If[FreeQ[s,p],idp=IntegerDigits[p];If[id2==Take[idp,2]&&idp[[ -2]]>0,AppendTo[s,p];id2=Take[idp,-2];Break[]]],{k,6,10^7}],{1000}];s

%K base,nonn

%O 0,1

%A _Zak Seidov_, Aug 09 2010