

A007530


Prime quadruples: numbers k such that k, k+2, k+6, k+8 are all prime.
(Formerly M3816)


111



5, 11, 101, 191, 821, 1481, 1871, 2081, 3251, 3461, 5651, 9431, 13001, 15641, 15731, 16061, 18041, 18911, 19421, 21011, 22271, 25301, 31721, 34841, 43781, 51341, 55331, 62981, 67211, 69491, 72221, 77261, 79691, 81041, 82721, 88811, 97841, 99131
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OFFSET

1,1


COMMENTS

Except for the first term, 5, all terms == 11 (mod 30).  Zak Seidov, Dec 04 2008
Some further values: For k = 1, ..., 10, a(k*10^3) = 11721791, 31210841, 54112601, 78984791, 106583831, 136466501, 165939791, 196512551, 230794301, 265201421.  M. F. Hasler, May 04 2009
k is the first prime of 2 consecutive twin prime pairs.  Daniel Forgues, Aug 01 2009
The prime quadruples of form p + (0, 2, 6, 8) have the quadruple congruence class (1, +1, 1, +1) (mod 6).  Daniel Forgues, Aug 12 2009
s = (p+8)(p) = 8 is the smallest s giving an admissible prime quadruple form, for which the only admissible form is p + (0, 2, 6, 8), since (0, 2, 6, 8) is the only form not covering all the congruence classes for any prime <= 4. Since s is smallest, these prime quadruples are prime constellations (or prime quadruplets), i.e., they contain consecutive primes.  Daniel Forgues, Aug 12 2009
Except for the first term, 5, all prime quadruples are of the form (15k4, 15k2, 15k+2, 15k+4), with k >= 1, and so are centered on 15k.  Daniel Forgues, Aug 12 2009
Starting at a(2) and examining the first 50 terms, (a(n)+4)/15 is a prime in 8 cases and a semiprime in 21; the last 18 terms have 2 primes and 11 semiprimes. Do the number of semiprimes continue to occur greater than mere chance?  J. M. Bergot, Apr 27 2015


REFERENCES

H. Rademacher, Lectures on Elementary Number Theory. Blaisdell, NY, 1964, p. 4.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS



FORMULA



EXAMPLE

a(1)=5 is the start of the first prime quadruplet, {5,7,11,13}.
a(2)=11 is the start of the second prime quadruplet, {11,13,17,19}, and all other prime quadruplets differ from this one by a multiple of 30.
a(100)=470081 is the start of the 100th prime quadruplet;
a(500)=4370081 is the start of the 500th prime quadruplet.
a(167)=1002341 is the least quadruplet prime beyond 10^6. (End)


MATHEMATICA

A007530 = Select[Range[1, 10^5  1, 2], Union[PrimeQ[# + {0, 2, 6, 8}]] == {True} &] (* Alonso del Arte, Sep 24 2011 *)
Select[Prime[Range[10000]], AllTrue[#+{2, 6, 8}, PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Mar 11 2019 *)


PROG

(PARI) A007530( n, print_all=0, s=2 )={ my(p, q, r); until(!n, until( p+8==s=nextprime(s+2), p=q; q=r; r=s); print_all && print1(p", ")); p} \\ The optional 3rd argument can be used to obtain large values by starting from some precomputed point instead of zero, using a(n+k) = A007530(k+1, , a(n)) (or A007530(k, , a(n)1) for k>0); e.g., you get a(10^4+k) using A007530(k+1, , 265201421) (value of a(10^4) from the comments section).  M. F. Hasler, May 04 2009
(PARI) forprime(p=2, 10^5, if(isprime(p+2) && isprime(p+6) && isprime(p+8), print1(p, ", "))) \\ Felix Fröhlich, Jun 22 2014
(Magma) [ p: p in PrimesUpTo(11000) IsPrime(p+2) and IsPrime(p+6) and IsPrime(p+8)] // Vincenzo Librandi, Nov 18 2010
(Python)
from sympy import primerange
def aupto(limit):
p, q, r, alst = 2, 3, 5, []
for s in primerange(7, limit+9):
if p+2 == q and p+6 == r and p+8 == s: alst.append(p)
p, q, r = q, r, s
return alst


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS

Values up to a(1000) checked with the given PARI code by M. F. Hasler, May 04 2009


STATUS

approved



