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A112540 Numbers m such that (15m-4, 15m-2, 15m+2, 15m+4) is a prime quadruple. 4

%I #66 Sep 08 2022 08:45:22

%S 1,7,13,55,99,125,139,217,231,377,629,867,1043,1049,1071,1203,1261,

%T 1295,1401,1485,1687,2115,2323,2919,3423,3689,4199,4481,4633,4815,

%U 5151,5313,5403,5515,5921,6523,6609,6741,7323,7769,7953,8147,9031,9611,10485,11047

%N Numbers m such that (15m-4, 15m-2, 15m+2, 15m+4) is a prime quadruple.

%C Also (4p + 16)/60 such that (p, p+2, p+6 and p+8) is a prime quadruple for p >= 11. - _Michel Lagneau_, Jul 02 2012

%C The density of these four-prime groups is approximately equal to (log x)^-3.45 (but not (log x)^-4). - _Xueshi Gao_, Jun 01 2014

%C All of the terms of this sequence are either 1, 7 or 13 modulo 14. - _Rodolfo Ruiz-Huidobro_, Dec 27 2019

%C From _Eric Snyder_, Jun 23 2021: (Start)

%C Building on the comment of R. Ruiz-Huidobro above, all terms of the sequence are congruent to one of {-1, 0 ,1} (mod 7). The appearance of mod 14 stems from the fact that all entries in this list must be odd. Equivalently, a(n) cannot be +- 2 or +- 3 (mod 7). This can be generalized for all larger primes:

%C All primes p >= 7 can be expressed as 15k +- q in a least absolute residue system, with q in {2, 4} if k is odd, and q in {1,7} if k is even.

%C For all primes 15k +- q >= 7, four residues +-r, +-s (mod p) exist such that, if for any p >= 7, [(m == +- r (mod p) or m == +- s (mod p)), and (m != k)], then m is not in this sequence. For the different values of p = 15k +- q, the values of +-r and +-s are as follows:

%C For p = 15k +- 1 (k even), r = +- 2k, s = +- 4k

%C For p = 15k +- 2 (k odd), r = +- k, s = +- 2k

%C For p = 15k + 4 (k odd), r = +- k, s = +- (7k + 2)

%C For p = 15k - 4 (k odd), r = +- k, s = +- (7k - 2)

%C For p = 15k + 7 (k even), r = +- (4k + 2), s = +- (8k + 4)

%C For p = 15k - 7 (k even), r = +- (4k - 2), s = +- (8k - 4)

%C These can be used to create an Eratosthenes-like sieve for the prime decades. (End)

%H Dana Jacobsen, <a href="/A112540/b112540.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Snyder, <a href="http://invirtuo.cc/files/numbertheory/Decades.pdf">An Eratosthenean Sieve for the Prime Decades</a>

%e m = 7 yields the quadruple (15*7 - 4 = 101, 15*7 - 2 = 103, 15*7 + 2 = 107, 15*7 + 4 = 109), so 7 is a term of the sequence.

%p A112540:=n->`if`(isprime(15*n-4) and isprime(15*n-2) and isprime(15*n+2) and isprime(15*n+4),n,NULL); seq(A112540(n), n=1..20000); # _Wesley Ivan Hurt_, Jul 26 2014

%t Select[Range[6610], PrimeQ[15# - 4] && PrimeQ[15# - 2] && PrimeQ[15# + 2] && PrimeQ[15# + 4]&] (* _T. D. Noe_, Nov 16 2006 *)

%o (PARI) for(n=1, 1e4, if(isprime(15*n-4) && isprime(15*n-2) && isprime(15*n+2) && isprime(15*n+4), print1(n, ", "))) \\ _Felix Fröhlich_, Jul 26 2014

%o (Perl) use ntheory ":all"; say for map { (4*$_+16)/60 } sieve_prime_cluster(11,15*10000, 2,6,8); # _Dana Jacobsen_, Dec 15 2015

%o (Magma) [n: n in [0..2*10^4] | IsPrime(15*n-4) and IsPrime(15*n-2) and IsPrime(15*n+2) and IsPrime(15*n+4)]; // _Vincenzo Librandi_, Dec 28 2015

%o (Python)

%o from sympy import isprime

%o def ok(m): return all(isprime(15*m+k) for k in [-4, -2, 2, 4])

%o print(list(filter(ok, range(11111)))) # _Michael S. Branicky_, Jun 24 2021

%Y Cf. A007530, A014561.

%K nonn,easy

%O 1,2

%A _Karsten Meyer_, Dec 16 2005

%E Corrected by _T. D. Noe_, Nov 16 2006

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