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A007530
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Prime quadruples: numbers k such that k, k+2, k+6, k+8 are all prime.
(Formerly M3816)
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111
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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
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1,1
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COMMENTS
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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 (15k-4, 15k-2, 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
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REFERENCES
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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).
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LINKS
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FORMULA
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EXAMPLE
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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)
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Values up to a(1000) checked with the given PARI code by M. F. Hasler, May 04 2009
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STATUS
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approved
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