A263182 Smallest k such that k//A002275(n)//k is prime, where // denotes concatenation and A002275(n) is the n-th repunit (R_n).

1, 3, 13, 17, 1073, 19, 17, 29, 10000117, 73, 17, 3, 1007, 3, 43, 11, 1000000000000029, 1, 31, 11, 1191, 1, 1143, 31, 10000079, 21, 91, 59, 1019, 3, 67, 117, 10000000000000000000000000000077, 109, 89, 49, 1097, 41, 1053, 43, 10000047, 87, 23, 53, 1149, 83, 57

Offset: 0

Author

Felix Fröhlich, Oct 11 2015

Keywords

• nonn
• base

Comments

a(n) = 3 if n is in A056251.

From Chai Wah Wu, Nov 05 2019 : (Start)

Theorem: a(2^r*s) >= 10^(2^r-1) for all r >= 0, s > 0.

Proof: Note that if k has m digits, then k//A002275(n)//k = k*(10^(n+m)+1) + A002275(n)*10^m which is a multiple of gcd(A002275(n),10^(n+m)+1).

Next, since 10^(2^r) - 1 = (10^(2^(r-1) -1))*(10^(2^(r-1) + 1)) and 9 does not divide 10^n+1, by induction it is easy to see that 10^(2^w) + 1 is a divisor of A002275(2^r) for 1 <= w < r. Since A002275(2^r) is a divisor of A002275(2^r*s), 10^(2^w) + 1 is also a divisor of A002275(2^r*s).

For 1 <= m < 2^r, let t be the 2-adic valuation of m, i.e. 0 <= t = A007814(m) < r.

Then 10^(2^r*s+m)+1 = 10^(2^t*q)+1 = (10^(2^t))^q + 1 for some odd number q.

Since the sum of two odd powers a^q+b^q is divisible by a+b, this implies that 10^(2^r*s+m)+1 is divisible by 10^(2^t)+1.

This means that for n = 2^r*s and 1 <= m < 2^r, gcd(A002275(n),10^(n+m)+1) >= 10^(2^t)+1 > 1, i.e. k//A002275(n)//k is not prime.

Thus a(2^r*s) must have at least 2^r digits, i.e. a(2^r*s) >= 10^(2^r-1). QED

As a consequence, a(n) >= 10^(A006519(n)-1). This result is still true if some of the digits of k are leading zeros.

(End)

Examples

			R_0 = 0 and the smallest k such that k//0//k is prime is 1, so a(0) = 1.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..1023

Crossrefs

Cf. A002275, A004023, A006519, A056251.

Programs

Mathematica

Table[k = 1; While[! PrimeQ[f[n, k]], k++]; k, {n, 0, 7}] (* Michael De Vlieger, Oct 13 2015 *)

PARI

a(n) = my(rep=(10^n-1)/9, k=1); while(!ispseudoprime(eval(Str(k, rep, k))), k++); k

Formula

a(A004023(n)-2) = 1. - Chai Wah Wu, Nov 04 2019

Extensions

a(16)-a(46) from Chai Wah Wu, Nov 04 2019