

A263182


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


2



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
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OFFSET

0,2


COMMENTS

Theorem: a(2^r*s) >= 10^(2^r1) 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^(r1) 1))*(10^(2^(r1) + 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 2adic 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^r1). 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)


LINKS



FORMULA



EXAMPLE

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


MATHEMATICA

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


PROG

(PARI) a(n) = my(rep=(10^n1)/9, k=1); while(!ispseudoprime(eval(Str(k, rep, k))), k++); k


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



