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

1,1

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 b<n. - Jens Kruse Andersen, Feb 03 2003

Links

Table of n, a(n) for n=1..43.

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

Example

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.

Prog

(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

Crossrefs

Cf. A030996, A052077, A052078, A052080.

Sequence in context: A160125 A151868 A344913 * A291483 A181295 A166299

Adjacent sequences: A052076 A052077 A052078 * A052080 A052081 A052082

Keyword

nonn,base,hard

Author

Patrick De Geest, Jan 15 2000

Extensions

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

Status

approved