A052079 Concatenation of n consecutive ascending numbers starting from a(n) produces the smallest possible prime of this form, 0 if no such prime exists.

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, 16, 341, 0, 569, 42, 0, 14, 1203, 0, 2465, 102, 0, 212, 1161, 0, 197

Offset: 1

Author

Patrick De Geest, Jan 15 2000

Keywords

• nonn
• base
• hard

Comments

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 bJens Kruse Andersen, Feb 03 2003

Examples

			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.
		

Links

• Carlos Rivera, Puzzle 78. The least prime by concatenating K consecutive integers, The Prime Puzzles and Problems Connection.

Crossrefs

Cf. A030996, A052077, A052078, A052080.

Programs

PARI

isok(vc) = {my(x=""); for (i=1, #vc, x = concat(x, Str(vc[i]))); ispseudoprime(eval(x));}
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

Extensions

Terms a(7)-a(43) calculated by Carlos Rivera and Felice Russo