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A255669 Primes p such that p divides the concatenation of the next two primes.

%I A255669 #22 May 22 2025 10:21:42
%S A255669 3,7,61,167
%N A255669 Primes p such that p divides the concatenation of the next two primes.
%C A255669 No additional terms up to the 5-millionth prime. Is the sequence finite and complete?
%C A255669 No additional terms up to the billionth prime. - _Chai Wah Wu_, Mar 10 2015
%C A255669 a(5) > 10^18. If the reasonable assumption nextprime(p) < p + (log p)^2 holds, then a(5) > 10^53. However, the 192-digits prime
%C A255669 7046979865771812080536912751677852348993288590604026845637583892...
%C A255669 6174496644295302013422818791946308724832214765100671140939597315...
%C A255669 4362416107382550335570469798657718120805369127516778523489932887 is in the sequence. - _Giovanni Resta_, May 08 2015
%e A255669 The three primes beginning with 61 are 61, 67, and 71, and 61 evenly divides 6771.
%t A255669 divQ[{a_,b_,c_}]:=Divisible[FromDigits[Flatten[IntegerDigits/@{b,c}]],a]; Transpose[Select[Partition[Prime[Range[500]],3,1],divQ]][[1]]
%o A255669 (Python)
%o A255669 from sympy import nextprime
%o A255669 A255669_list, p1, p2, l = [], 2, 3, 10
%o A255669 for n in range(10**8):
%o A255669     p3 = nextprime(p2)
%o A255669     if p3 >= l: # this test is sufficient by Bertrand-Chebyshev theorem
%o A255669         l *= 10
%o A255669     if not ((p2 % p1)*l + p3) % p1:
%o A255669         A255669_list.append(p1)
%o A255669     p1, p2 = p2, p3 # _Chai Wah Wu_, Mar 09 2015
%K A255669 nonn,base
%O A255669 1,1
%A A255669 _Harvey P. Dale_, Mar 01 2015