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A242775
Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the concatenation b_k and prime(n) is prime, or a(n)=0 if there is no such prime.
5
0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 1, 1, 4, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 3, 2, 1, 2, 7, 3, 1, 3, 2, 2, 8, 1, 1, 7, 2, 1, 1, 5, 3, 2, 2, 2, 3, 1, 3, 8, 5, 1, 1, 4, 3, 1, 4, 5, 3, 6, 1, 2, 1, 2, 1, 3, 1, 2, 2, 1, 3, 1, 6, 3, 1, 3, 4, 2, 3, 8, 4, 1, 3, 34, 1
OFFSET
1,8
COMMENTS
Conjecture: for n>=4, a(n)>0.
Records >=1: 1,2,4,7,8,34,... correspond to primes 7,19,41,127,157,443,...
LINKS
EXAMPLE
For n<=3, a(n) = 0, because 3..32, 3..33 and 3..35 can never be prime, whatever the number of 3's that are concatenated.
For n=4, prime(n)=7, 37 is prime. So a(4)=1.
PROG
(PARI) a(n) = {if (n<=3, return (0)); p = prime(n); k = 1; while (! isprime(p = eval(concat("3", Str(p)))), k++); k; } \\ Michel Marcus, Sep 17 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Sep 13 2014
EXTENSIONS
More terms from Peter J. C. Moses, Sep 14 2014
STATUS
approved