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%I #24 Oct 13 2021 10:32:29
%S 0,1,1,4,1,2,1,2,179,0,1,2,1,4,5,28,1,3590,1,4,0,0,1,0,25,122,0,46,1,
%T 0,1,0,71,4,569,2,1,20,5,0,1,2,1,8,0,0,1,0,193,2,0,0,1,0,0,2,5,4,1,0,
%U 1,2,0,4,5,938,1,2,119,58,1,116,1,0,125,346,5,2,1,2,0,0,1,0,0,32
%N Least number k > 0 such that concatenating k consecutive natural numbers beginning with n is prime, or 0 if no such number exists.
%C The current zero values are only conjectural: a(n) > 5000 - n for all a(n) = 0 shown. [Edited by _M. F. Hasler_, Apr 27 2017]
%C A positive value for a(1) will satisfy A075019(a(1)) = A007908(a(1)). - _Michel Marcus_, Jul 09 2014
%C Probably a(n) > 0 for all n. Appending k integers gives a number of size ~10^(k log_10 k) and so the expected number of primes with k < x is about the integral of 1/(k log k) up to x which is log log x. This diverges, so by the Borel-Cantelli lemma we expect that there will be a prime eventually. (Corrections for the particular base at hand affect the expected number but not its order of growth.) On the other hand, log log x grows slowly so finding the values of a(1), a(10), a(21), etc. may be hard. - _Charles R Greathouse IV_, Jul 10 2014 [Corrected by _Pontus von Brömssen_, Oct 12 2021]
%e 14 is not prime. 1415 is not prime. 141516 is not prime. 14151617 is prime. Thus a(14) = 4 since 4 consecutive numbers were concatenated.
%o (PARI) a(n) = {p=""; tot=0; for(i=n,5000, p=concat(p,Str(i)); tot++; if(ispseudoprime(eval(p)), return(tot)))}
%o n=1;while(n<100,print1(a(n),", ");n++)
%Y Cf. A007908, A075019, A281571 (base-2 variant).
%K nonn,base,more,hard
%O 1,4
%A _Derek Orr_, Jun 27 2014
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