OFFSET
1,1
COMMENTS
It is not known if there are infinitely many Sophie Germain pairs with this property.
LINKS
Abhiram R Devesh and Dana Jacobsen, Table of n, a(n) for n = 1..1155 [first 155 terms from Abhiram R Devesh]
Eric Weisstein's World of Mathematics, Sophie Germain Prime
Wikipedia, Sophie Germain Prime
EXAMPLE
a(1): p = 5; (2*p)+1 = 11; prime quadruplets (5,7,13,17); (11,13,19,23).
a(2): p = 29; (2*p)+1 = 59; prime quadruplets (29,31,37,41); (59,61,67,71).
PROG
(Python)
p1=2
n=2
count=0
while p1>2:
....## Generate the a chain of numbers with length 4
....cc=[]
....cc.append(p1)
....for i in range(1, n):
........cc.append((2**(i)*p1+((2**i)-1)))
....## chain entries + 2
....cc2=[c+2 for c in cc]
....## chain entries + 8
....cc8=[c+8 for c in cc]
....## chain entries + 12
....cc12=[c+12 for c in cc]
....## check if cc is a Sophie Germain Pair or not
....## pf.isp_list returns True or false for a given list of numbers
....## if they are prime or not
....##
....pcc=pf.isp_list(cc)
....pcc2=pf.isp_list(cc2)
....pcc8=pf.isp_list(cc8)
....pcc12=pf.isp_list(cc12) ....## Number of primes for cc
....npcc=pcc.count(True)
....## Number of primes for cc2
....npcc2=pcc2.count(True)
....## Number of primes for cc8
....npcc8=pcc8.count(True)
....## Number of primes for cc12
....npcc12=pcc12.count(True)
....if npcc==n and npcc2==n and npcc8==n and npcc12==n:
........print "For length ", n, " the series is : ", cc, " , ", cc2, " , ", cc8, " and " cc12
....p1=pf.nextp(p1)
(PARI) forprime(p=1, 1e9, my(t=2*p+1); if(isprime(t) && isprime(p+2) && isprime(p+8) && isprime(p+12) && isprime(t+2) && isprime(t+8) && isprime(t+12), print1(p, ", "))) \\ Felix Fröhlich, Jul 26 2014
(Perl) use ntheory ":all"; my @p = sieve_prime_cluster(1, 2e9, 2, 8, 12); my %h; undef @h{@p}; for (@p) { say if exists $h{2*$_+1} } # Dana Jacobsen, Oct 03 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Abhiram R Devesh, Feb 05 2014
STATUS
approved