This site is supported by donations to The OEIS Foundation. Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing. Other ways to donate

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A263182 Smallest k such that k//A002275(n)//k is prime, where // denotes concatenation and A002275(n) is the n-th 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 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) LINKS Chai Wah Wu, Table of n, a(n) for n = 0..1023 FORMULA a(A004023(n)-2) = 1. - Chai Wah Wu, Nov 04 2019 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^n-1)/9, k=1); while(!ispseudoprime(eval(Str(k, rep, k))), k++); k CROSSREFS Cf. A002275, A004023, A006519, A056251. Sequence in context: A006486 A273946 A070518 * A045525 A030774 A024685 Adjacent sequences:  A263179 A263180 A263181 * A263183 A263184 A263185 KEYWORD nonn,base AUTHOR Felix Fröhlich, Oct 11 2015 EXTENSIONS a(16)-a(46) from Chai Wah Wu, Nov 04 2019 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified December 10 20:38 EST 2019. Contains 329909 sequences. (Running on oeis4.)