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A232210
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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.
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10
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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
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OFFSET
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1,6
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COMMENTS
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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".
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
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LINKS
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EXAMPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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