A328455 Least prime p such that all digits of p*prime(n) are the same (or -1 if p does not exist).

2, 2, 11, 11, 2, -1, -1, -1, -1, -1, -1, 3, 271, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 11, -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, 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

Offset: 1

Author

Ivan N. Ianakiev, Oct 16 2019

Keywords

• sign
• base

Comments

From Bernard Schott, Oct 22 2019: (Start)

Proposition:

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

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

* 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.

Proof:

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

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

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

* 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.

Some examples:

Case 1: prime(n) is a repunit.

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

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).

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

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

For m = 3, 111 = 3 * 37.

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

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

For m = 4, 1111 = 11 * 101.

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

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

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

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

(End)

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

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

Examples

			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).
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
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Programs

Maple

f:= proc(n) local o,p,q;
  p:= ithprime(n);
  o:= numtheory:-order(10,p);
  q:= (10^o-1)/(9*p);
  if isprime(q) then q elif q = 1 then 2 else -1 fi
end proc:
2,2,11,11,seq(f(n),n=5..100); # Robert Israel, Nov 19 2019

Mathematica

a[1]=a[2]=2;a[3]=a[4]=11; a[n_]:= Which[Union[IntegerDigits[Prime[n]]]=={1},2,
Module[{i=1},While[!Divisible[(10^i-1),9*Prime[n]],i++]; k=(10^i-1)/(9*Prime[n]);
PrimeQ[k]],k,True,-1]; a/@Range[85] (* Ivan N. Ianakiev, Oct 24 2019 *)

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

Extensions

More terms from Bernard Schott and David A. Corneth, Oct 22 2019