

A059055


Primes which can be written as (b^k+1)/(b+1) for positive integers b and k.


9



3, 7, 11, 13, 31, 43, 61, 73, 157, 211, 241, 307, 421, 463, 521, 547, 601, 683, 757, 1123, 1483, 1723, 2551, 2731, 2971, 3307, 3541, 3907, 4423, 4831, 5113, 5701, 6007, 6163, 6481, 8011, 8191, 9091, 9901, 10303, 11131, 12211, 12433, 13421, 13807, 14281
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OFFSET

1,1


COMMENTS

For (b^k+1)/(b+1) to be a prime, k must be an odd prime. 2=(0^0+1)/(0+1) has been excluded since neither b nor k would be positive.
From Bernard Schott, May 14 2019: (Start)
The Brazilian primes m that belong to this sequence must satisfy the Diophantine equation m = (b^q1)/(b1) = (c^k+1)/(c+1) and b,c > 1, k,q (primes) > 2. This equation seems not to be studied. There are solutions with (c = b+1, q = k = 3) as 43 = (6^31)/(61) = (7^3+1)/(7+1). So, terms of A002383 are solutions of this equation and belong to the intersection of this sequence and A085104. The Brazilian primes in the bfile are of the form (111)_b.
The primes that are not Brazilian form the second subsequence of this sequence which begins with {11, 61, 521, 547, 683, ...}. These last are of the form (c^k+1)/(c+1) with k prime >= 5, as 547 = (3^7+1)/(3+1). (End)


LINKS

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 3880 terms from T. D. Noe)
H. Dubner and T. Granlund, Primes of the Form (b^n+1)/(b+1), J. Integer Sequences, 3 (2000), #P00.2.7.


EXAMPLE

43 is in the sequence since (2^7+1)/(2+1) = 129/3 = 43; indeed also (7^3+1)/(7+1) = 344/8 = 43.


MATHEMATICA

max = 89; maxdata = (1 + max^3)/(1 + max); a = {}; Do[i = 1; While[i = i + 2; cc = (1 + m^i)/(1 + m); cc <= maxdata, If[PrimeQ[cc], a = Append[a, cc]]], {m, 2, max}]; Union[a] (* Lei Zhou, Feb 08 2012 *)


CROSSREFS

Cf. A002383, A059054.
Cf. A003424, A085104.
Sequence in context: A342183 A206945 A206946 * A243367 A145670 A004061
Adjacent sequences: A059052 A059053 A059054 * A059056 A059057 A059058


KEYWORD

nonn,easy


AUTHOR

Henry Bottomley, Dec 21 2000


STATUS

approved



