OFFSET
1,1
COMMENTS
It is not known if there are infinitely many Sophie Germain pairs with this property.
The sequence is infinite under Dickson's conjecture. Aside from a(1) = 5, all terms are 29 or 179 mod 210. - Charles R Greathouse IV, Feb 05 2014
LINKS
Abhiram R Devesh and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 135 terms from Devesh)
Eric Weisstein's World of Mathematics, Sophie Germain Prime
Wikipedia, Sophie Germain prime
EXAMPLE
a(1): p = 5; (2*p)+1 = 11
Prime triples (5,7,13);(11,13,19)
a(2): p = 29; (2*p)+1=59
Prime triples (29,31,37);(59,61,67)
MATHEMATICA
sgpQ[n_]:=Module[{sg=2n+1}, AllTrue[Flatten[{sg+{0, 2, 8}, n+{2, 8}}], PrimeQ]]; Select[Prime[ Range[ 300000]], sgpQ] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Feb 02 2016 *)
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]
....## 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)
....## 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)
....if npcc==n and npcc2==n and npcc8==n:
........print "For length ", n, " the series is : ", cc, " , ", cc2, " and ", cc8
....p1=pf.nextp(p1)
(PARI) is(n)=isprime(n) && isprime(n+2) && isprime(n+8) && isprime(2*n+1) && isprime(2*n+3) && isprime(2*n+9) \\ Charles R Greathouse IV, Feb 05 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Abhiram R Devesh, Feb 04 2014
STATUS
approved