A092307 Primes p such that there are no primes q, 3 < q < p, such that (q-1) divides (p-1).

5, 7, 11, 23, 47, 59, 83, 107, 167, 179, 227, 239, 263, 347, 359, 383, 443, 467, 479, 503, 563, 587, 647, 659, 719, 827, 839, 863, 887, 983, 1019, 1187, 1223, 1259, 1283, 1307, 1319, 1367, 1439, 1487, 1499, 1523, 1619, 1787, 1823, 1847, 1907, 2027, 2039, 2063

Offset: 1

Author

T. D. Noe, Feb 12 2004

Keywords

• nonn

Comments

Using a sieve, these primes can be generated quickly. In the set of primes < 10^9, the density of these primes is about 1/10. It is easy to show that this sequence contains all "safe" primes (A005385).

Primes p such that 6p is the denominator of some Bernoulli number. - T. D. Noe, Sep 26 2006

Except for 5 and 7, primes p such that 12p is the denominator of B(p - 1)/(p - 1) where B(n) is the Bernoulli number. [Peter Luschny, Dec 24 2008]

Primes p such that A027642(p-1) = 6p. Composites m such that A027642(m-1) = 6m are Carmichael numbers 310049210890163447, 18220439770979212619, ... - Amiram Eldar and Thomas Ordowski, May 26 2021

Examples

			11 is in the sequence because 10 is not a multiple of either 4 or 6.
13 is not in the sequence because, although 12 is not a multiple of 6 or 10, it is a multiple of 4.
		

Links

• T. D. Noe, Table of n, a(n) for n=1..1000
• P. Luschny, Von Staudt prime number, definition and computation. [From _Peter Luschny_, Dec 24 2008]

Crossrefs

Cf. A090801 (distinct numbers appearing as denominators of Bernoulli numbers)

Cf. A092308 (for p=prime(n), the number of primes q such that q-1 divides p-1).

Cf. A005385 (primes p such that (p-1)/2 is also prime).

Cf. A152951. [From Peter Luschny, Dec 24 2008]

Programs

Maple

For p>7: seq(`if`(denom(bernoulli(n-1)/(n-1))=12*n,n,NULL),n=2..500); # Peter Luschny, Dec 24 2008

Mathematica

t = Table[p = Prime[n]; Length[Select[Divisors[p - 1] + 1, PrimeQ]], {n, 311}]; Prime[Flatten[Position[t, 3]]]
npqQ[n_]:=NoneTrue[Prime[Range[3,PrimePi[n]-1]],Mod[n-1,#-1]==0&]; Select[ Prime[ Range[3,400]],npqQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 26 2019 *)

Perl

use ntheory ":all"; forprimes { say if (bernfrac($-1))[1] == 6*$ } 1000; # Dana Jacobsen, Dec 29 2015

Perl

use ntheory ":all"; forprimes { my $p=$; say if vecnone { $ > 3 && $ < $p-1 && is_prime($+1) } divisors($p-1); } 5,1000; # Dana Jacobsen, Dec 29 2015

Formula

Let h(x) = 12x(x + log(exp(-x) -1) - log(x)) and [x^n]S(h) denote the coefficient of x^n in the series expansion of h. Consider for n > 1 the relation n = denominator((n - 1)![x^n]S(h)). [Peter Luschny, Dec 24 2008]