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%I #16 Oct 02 2023 20:53:38
%S 1,4,5,6,7,11,20,21,25,26,27,28,30,31,32,33,36,37,38,39,40,42,43,44,
%T 45,63,64,66,67,68,69,73,125,126,127,128,130,131,132,133,135,136,137,
%U 154,155,156,159,160,161,163,164,165,167,168,170,172,173,174,177,178,179
%N Numbers j such that the concatenation of the last digit of p(j) and the first digit of prime(j+1) is a prime.
%C There are roughly 5/(18m log 10) * 10^m terms of this sequence up to 10^m: all primes between 10^k and 2*10^k, half the primes between 3*10^k and 4*10^k, 3/5 of the primes between 7*10^k and 8*10^k, and 1/4 of the primes between 9*10^k and 10*10^k for all 1 < k < m, using the prime number theorem in arithmetic progressions. Thus the "probability" that a random number up to 10^m is in this sequence is 0.12/m.
%H Harvey P. Dale, <a href="/A086101/b086101.txt">Table of n, a(n) for n = 1..1000</a>
%e 20 is a term because prime(20)=71, prime(21)=73, and 17 is a prime.
%t cldfdQ[{a_,b_}]:=PrimeQ[FromDigits[Join[{Mod[a,10]},{First[IntegerDigits[b]]}]]]; Position[ If[cldfdQ[#],1,0]&/@Partition[Prime[Range[200]],2,1],1]//Flatten (* _Harvey P. Dale_, Apr 26 2022 *)
%K easy,nonn,base
%O 1,2
%A _Zak Seidov_, Jul 09 2003
%E Comment from _Charles R Greathouse IV_, Apr 27 2010
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