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
Examples
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
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
Programs
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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 *) -
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
Extensions
More terms from Peter J. C. Moses, Sep 13 2014
Comments