A232210 Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the concatenation prime(n)b_k is prime, or a(n)=0 if there is no such prime.

1, 0, 1, 1, 1, 14, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 3, 1, 2, 6, 2, 2, 1, 1, 2, 1, 4, 4, 23, 1, 2, 1, 6, 2, 2, 5, 1, 10, 2, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 2, 4, 2, 1, 1, 1, 2, 4, 1, 2, 5, 4, 2, 3, 1, 1, 5, 4, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 6, 4, 2, 14, 2, 4, 1, 3

Offset

1,6

Comments

Conjecture: for n>=3, a(n)>0.

Records are 1,14,23,50,252,4752,...

The corresponding primes are 2,13,131,653,883,1279,...

These primes beginning with the second one we call "stubborn primes".

Counter-conjecture: a(2889)=0. - Hans Havermann, Oct 15 2014

If a(n)=1, then the resulting primes are in A092993 and form A055782; if a(n)=2, then they form sequence 4133,4733,5333,7933,..., etc. - Vladimir Shevelev, Oct 16 2014

If a prime p divides Pb_k, then it also divides Pb_{k+m(p-1)} for all m>=0. This follows from Fermat's little theorem applied to b_x=(10^x-1)/3 with x=p-1. - M. F. Hasler, Oct 20 2014

Links

Hans Havermann, Table of n, a(n) for n = 1..2888 (first 200 terms from Michel Marcus)

Hans Havermann, "Google+ discussion relating to this sequence"

Vladimir Shevelev, "Stubborn primes"

Robert G. Wilson v, Table of n, a(n) for n = 1..10000 with -1 for those entries where a(n) has not yet been found

Example

For n=1, start with prime(1)=2 and get already at the first step the prime 23. So a(1)=1.
For n=2, starting with prime(2)=3, one never gets a prime by appending further digits "3", therefore a(2)=0.
For n=3, n=4, n=5, one gets after the first step the primes 53, 73, 113, and therefore a(n)=1.
For n=6, start with prime(6)=13; one has to append 14 "3"s in order to get a new prime, so a(6)=14.
For n=2889, start with prime(2889) = 26293. (Do not mix up with prime(2899) = 26393...!) Appending 2k-1 or 6k-4 or 6k-2 or 18k-6 or 36k-18 or 180k-144 digits "3" yields a number divisible by 11 resp. 7 resp. 13 resp. 19 resp. 101 resp. 31. For 18k-12 and 36k (with k <> 1 (mod 5)) digits "3" there is no simple pattern and both yield sometimes large primes in the factorization, but (so far) always composite numbers 26293...3 (up to several thousand digits). - M. F. Hasler, Oct 16 2014

Mathematica

f[n_] := Block[{k = 1, p = Prime@ n}, While[ !PrimeQ[p*10^k + (10^k - 1)/3], k++]; k]; f[2] = 0; Array[f, 100] (* Robert G. Wilson v, Apr 24 2015 *)
m3[n_]:=Module[{k=10n+3}, While[!PrimeQ[k], k=10k+3]; IntegerLength[k]-IntegerLength[ n]]; Join[{1, 0}, m3/@Prime[Range[3, 90]]] (* Harvey P. Dale, Feb 11 2018 *)

Prog

(PARI) a(n) = {if (n==2, return (0)); p = prime(n); k = 1; while (! isprime(p = p*10+3), k++); k; } \\ Michel Marcus, Sep 13 2014

Crossrefs

Cf. A055782, A092993, A242775, A247341, A247342, A248919.

Sequence in context: A002393 A185284 A262705 * A040199 A173747 A040200

Adjacent sequences: A232207 A232208 A232209 * A232211 A232212 A232213

Keyword

nonn,base

Author

Vladimir Shevelev, Sep 13 2014

Extensions

More terms from Peter J. C. Moses, Sep 13 2014

Status

approved