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A257459 Let b_k=1...1 consist of k>0 1'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. 3

%I #9 May 03 2015 10:11:48

%S 2,1,5,1,17,1,8,1,2,6,1,0,2,1,3,9,18,4,210,6,7,3,2,6,1,2,1,2,1,2,4,3,

%T 2,24,3,1,1,6,5,11,2,1,11,1,12,6,1,7,3,39,2,2,1,2,9,3,5,1,6,2,3,2,180,

%U 3,15,17,24,1,5,1,2,2,1,64,7,6,3,24,2,1,2,1,6,16,1,9,8,6,17,4,6,2,1,9,30,2,6,44,1,6

%N Let b_k=1...1 consist of k>0 1'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.

%C The only unknown terms less than 10000, tested to 15000, are for n: 284, 714, 1257, 1618, 2248, 2450, 2779, 3886, 3891, 4007, 4359, 4784, 4912, 5364, 6108, 6356, 6371, 7570, 7668, 8446, 9606.

%C Prime(12)=37 and b_k for k == 2 (mod 3), the concatenation is divisible by 3; for k == 1 (mod 3), the concatenation is divisible by either 7 or 13; and finally for k == 0 (mod 3), the concatenation is divisible by 37.

%H Vladimir Shevelev and Robert G. Wilson v, <a href="/A257459/a257459.txt">Table of n, a(n) for n = 1..10000 with -1 for those entries where a(n) has not yet been found. </a>

%F a(n)=k for the least k such that p(n)*10^k+(10^k-1)/9 is prime, where p(n) is the n_th prime.

%t f[n_] := Block[{k = 1, p = Prime[n]}, While[ !PrimeQ[p*10^k + (10^k - 1)/9], k++]; k]; f[12] = 0; Array[f, 100]

%Y Cf. A232210, A257460, A257461.

%K nonn,base

%O 1,1

%A _Vladimir Shevelev_ and _Robert G. Wilson v_, Apr 24 2015

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