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.

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, 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, 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

Offset: 1

Author

Vladimir Shevelev and Robert G. Wilson v, Apr 24 2015

Keywords

• nonn
• base

Comments

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.

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.

Links

• Vladimir Shevelev and 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.

Crossrefs

Cf. A232210, A257460, A257461.

Programs

Mathematica

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]

Formula

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.