A328455 - OEIS

%I #71 Jun 06 2021 07:59:43

%S 2,2,11,11,2,-1,-1,-1,-1,-1,-1,3,271,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,

%T -1,11,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,

%U -1,-1,-1,-1,4649,-1,-1,-1,-1,-1,41,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1

%N Least prime p such that all digits of p*prime(n) are the same (or -1 if p does not exist).

%C From _Bernard Schott_, Oct 22 2019: (Start)

%C Proposition:

%C For n >= 5, there exist terms a(n) <> -1 iff

%C * prime(n) is a prime repunit with m 1's and m is in A004023, then a(n) = 2, or

%C * prime(n) is a divisor of a semiprime repunit with m 1's and m is in A046413, then a(n) is the other prime factor of this semiprime.

%C Proof:

%C We must solve a(n) * prime(n) = repdigit = k * repunit, 1<= k <= 9, with a(n) least prime p that is solution.

%C According to the unicity of prime factorization, only two possibilities:

%C * prime(n) is repunit, then a(n) = k = 2 (smallest prime).

%C * prime(n) is not a repunit, then a(n) * prime(n) must be a repunit that is semiprime, then k = 1 and a(n) is the other factor of this semiprime.

%C Some examples:

%C Case 1: prime(n) is a repunit.

%C The first few values of m are 2, 19, 23, ...

%C If this repunit is the k-th prime, then a(k) = 2; it is the case for prime(5) = 11 with a(5) = 2 (see example).

%C Case 2: prime(n) is a divisor of repunit.

%C The first few values of m are 3, 4, 5, 7, 11, 17, 47, 59, 71, ...

%C For m = 3, 111 = 3 * 37.

%C    As 37 = prime(12), a(12) = 3 with 3 * 37 = 111.

%C    As 3 = prime(2) and a(2) = 2 < 37, 37 is not right here.

%C For m = 4, 1111 = 11 * 101.

%C    As 101 = prime(26), a(26) = 11 with 11 * 101 = 11111.

%C    As 11 = prime(5) and a(5) = 2 < 101, 101 is not right here.

%C For m = 5, 11111 = 41 * 271, so as prime(13) = 41 and prime(58) = 271, then a(13) = 271, and (58) = 41.

%C For m = 7, 1111111 = 239 * 4649, so a(52) = 4649 and a(628) = 239.

%C (End)

%C a(n) is positive for n in {1,2,3,4,5,12,13,26,52,58,628,2431,2968,42536,...}. - _Ivan N. Ianakiev_, Oct 26 2019

%C If n > 4, a(n) = 2 if A077573(n) = prime(n), A077573(n)/prime(n) if that is prime, otherwise -1. - _Robert Israel_, Nov 19 2019

%H Robert Israel, <a href="/A328455/b328455.txt">Table of n, a(n) for n = 1..10000</a>

%e Prime(5) is 11 and the least prime p such that all the digits of p*prime(5) are the same is 2 (as 2*11 = 22).

%e a(6) = -1 as repdigits are of the form k*(10^m - 1)/9, 1 <= k <= 9. We need the repdigit to be a semiprime of the form 13*p for some prime p. We need m = 6*t for some t >= 1. So (7*13) || (10^m - 1)/9, i.e., (10^m - 1)/9 can't be a semiprime and a(6) = -1. - _David A. Corneth_, Oct 16 2019

%p f:= proc(n) local o,p,q;

%p   p:= ithprime(n);

%p   o:= numtheory:-order(10,p);

%p   q:= (10^o-1)/(9*p);

%p   if isprime(q) then q elif q = 1 then 2 else -1 fi

%p end proc:

%p 2,2,11,11,seq(f(n),n=5..100); # _Robert Israel_, Nov 19 2019

%t a[1]=a[2]=2;a[3]=a[4]=11; a[n_]:= Which[Union[IntegerDigits[Prime[n]]]=={1},2,

%t Module[{i=1},While[!Divisible[(10^i-1),9*Prime[n]],i++]; k=(10^i-1)/(9*Prime[n]);

%t PrimeQ[k]],k,True,-1]; a/@Range[85] (* _Ivan N. Ianakiev_, Oct 24 2019 *)

%o (PARI) a(n) = if(n<=5, return([2, 2, 11, 11, 2][n])); my(p=prime(n)); for(i=1, oo, if((10^i-1)/9%p==0, c=(10^i-1)/(9*p); if(isprime(c), return(c), return(-1)))) \\ _David A. Corneth_, Oct 22 2019

%K sign,base

%O 1,1

%A _Ivan N. Ianakiev_, Oct 16 2019

%E More terms from _Bernard Schott_ and _David A. Corneth_, Oct 22 2019