A077313 Primes of the form 2^r*5^s - 1.

3, 7, 19, 31, 79, 127, 199, 499, 1249, 1279, 1999, 4999, 5119, 8191, 12799, 20479, 31249, 49999, 51199, 79999, 81919, 131071, 199999, 524287, 799999, 1249999, 1310719, 3124999, 3276799, 4999999, 7812499, 12499999, 19999999, 20479999

Offset: 1

Author

Amarnath Murthy, Nov 04 2002

Keywords

• nonn

Comments

Primes p such that 10^p is divisible by p+1. Primes p whose fractions p/(p+1) are terminating decimals, i.e., primes p such that A158911(p)=0. Primes p such that the prime divisors of p+1 are also prime divisors of the numbers m obtained by the concatenation of p and p+1. For example, for p=19, m = 1920, the prime divisors of 20 are {2, 5} and the prime divisors of 1920 are {2, 3, 5}. - Jaroslav Krizek, Feb 25 2013

For n > 1, all terms are congruent to 1 (mod 6). - Muniru A Asiru, Sep 29 2017

Examples

			1250000 = 2*2*2*2*5*5*5*5*5*5*5 and 1250000 - 1 = A000040(96469), therefore 1249999 is a term.
List of (r, s): (2, 0), (3, 0), (2, 1), (5, 0), (4, 1), (7, 0), (3, 2), (2, 3), (1, 4), (8, 1), (4, 3), (3, 4), (10, 1), ...  - _Muniru A Asiru_, Sep 29 2017
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..4222

Crossrefs

Cf. A003592, A023509, A077497.

Programs

GAP

A:=Filtered([1..10^7],IsPrime);;    I:=[5];;
B:=List(A,i->Elements(Factors(i+1)));;
C:=List([0..Length(I)],j->List(Combinations(I,j),i->Concatenation([2],i)));;
A077313:=List(Set(Flat(List([1..Length(C)],i->List([1..Length(C[i])],j->Positions(B,C[i][j]))))),i->A[i]); # Muniru A Asiru, Sep 29 2017

Mathematica

With[{n = 10^8}, Union@ Select[Flatten@ Table[2^p*5^q - 1, {p, 0, Log[2, n/(1)]}, {q, 0, Log[5, n/(2^p)]}], PrimeQ]] (* Michael De Vlieger, Sep 30 2017 *)

Extensions

More terms from Reinhard Zumkeller, Nov 15 2002

More terms from Vladeta Jovovic, May 08 2003