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%I #24 Mar 04 2021 09:20:40
%S 2,2,0,4,15,0,7,2,0,4,129,0,5,50,0,128,3,0,23,38,0,9998,17,0,25,2,0,
%T 16,341,0,569,42,0,14,1203,0,2465,102,0,212,1161,0,197
%N Concatenation of n consecutive ascending numbers starting from a(n) produces the smallest possible prime of this form, 0 if no such prime exists.
%C Next term a(44)=10^348-32 (only probable prime with 15324 digits). a(110)=9999968. If n is divisible by 22 then either a(n)=0 or a(n)=10^x-b for some b<n. - _Jens Kruse Andersen_, Feb 03 2003
%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_078.htm">Puzzle 78. The least prime by concatenating K consecutive integers</a>, The Prime Puzzles and Problems Connection.
%e For n = 7 we have a(7) = 7 so the seven consecutive ascending numbers 7,8,9,10,11,12 and 13 concatenated together gives the smallest possible prime of this form, 78910111213.
%o (PARI) isok(vc) = {my(x=""); for (i=1, #vc, x = concat(x, Str(vc[i]))); ispseudoprime(eval(x));}
%o a(n) = if (n % 3, for(i=1, oo, my(vc = vector(n, k, k+i-1)); if (isok(vc), return(i))), 0); \\ _Michel Marcus_, Mar 04 2021
%Y Cf. A030996, A052077, A052078, A052080.
%K nonn,base,hard
%O 1,1
%A _Patrick De Geest_, Jan 15 2000
%E Terms a(7)-a(43) calculated by _Carlos Rivera_ and _Felice Russo_
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