The On-Line Encyclopedia of Integer Sequences (OEIS)

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A164042

Primes p such that 2*p^2+4*p+1 is also prime.
(history; published version)

#23 by Charles R Greathouse IV at Thu Sep 08 08:45:47 EDT 2022

PROG

(MAGMAMagma) [p: p in PrimesUpTo(1500) | IsPrime(2*p^2+4*p+1)]; // Vincenzo Librandi, Apr 08 2013

Discussion

Thu Sep 08

08:45

OEIS Server: https://oeis.org/edit/global/2944

#22 by Harvey P. Dale at Tue Sep 06 16:13:27 EDT 2022

STATUS

editing

approved

#21 by Harvey P. Dale at Tue Sep 06 16:13:25 EDT 2022

MATHEMATICA

Select[Prime[Range[250]], PrimeQ[2#^2+4#+1]&] (* Harvey P. Dale, Sep 06 2022 *)

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approved

editing

#20 by Bruno Berselli at Tue Jun 10 06:15:39 EDT 2014

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approved

#19 by Bruno Berselli at Tue Jun 10 06:14:24 EDT 2014

COMMENTS

Let p in this sequence. If 2^h = 2*(p+1)/3; 2*p+1 = b (prime); and c=2*p^2+4*p+1 (prime) then p*b*2^h and c*2^h are amicable numbers. Ex. p=23, 2^h=16, c=1151, b=47 and 23*47*16 = 17296 and 1151*16= 18416; (17296,18416 are amicable numbers). - Vincenzo Librandi, Jun 09 2014

If a(k) is of the form 3·2^(h-1)-1 and 2*a(k)+1 is prime, then 2^h*a(k)*(2*a(k)+1) and 2^h*(2*a(k)^2+4*a(k)+1) are a pair of amicable numbers. - Vincenzo Librandi, Jun 09 2014

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#18 by Vincenzo Librandi at Tue Jun 10 00:31:02 EDT 2014

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proposed

#17 by Vincenzo Librandi at Tue Jun 10 00:30:54 EDT 2014

COMMENTS

Conjecture: Let p in this sequence. If 2^h = 2*(p+1)/3; 2*p+1 = b (prime); and c=2*p^2+4*p+1 (prime) then p*b*2^h and c*2^h are amicable numbers. Ex. p=23, 2^h=16, c=1151, b=47 and 23*47*16 = 17296 and 1151*16= 18416; (17296,18416 are amicable numbers). - Vincenzo Librandi, Jun 09 2014

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proposed

editing

#16 by Michel Marcus at Mon Jun 09 14:59:42 EDT 2014

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proposed

Discussion

Mon Jun 09

18:52

Bruno Berselli: Vincenzo, why conjecture?  If n>1 and  3·2^(n-1)-1,  3·2^n-1,  9·2^(2n-1)-1  are prime numbers, then   2^n·(3·2^(n-1)-1)·(3·2^n-1)  and  2^n·(9·2^(2n-1)-1)  are a pair of amicable numbers --- this is well known. In your formulation, we consider the case that p (a term of the sequence) is equal to 3·2^(n-1)-1, instead c = 2·p^2+4·p+1 = 9·2^(2h-1)-1  is prime for definition. If b = 2p+1 (=3·2^h-1) is also prime, then p*b*2^h and c*2^h are a pair of amicable numbers. Right?

Tue Jun 10

00:30

Vincenzo Librandi: Yes.

#15 by Michel Marcus at Mon Jun 09 14:59:19 EDT 2014

MATHEMATICA

lst={}; Do[p=Prime@n; a=2*p^2+4*p+1; If[PrimeQ@a, AppendTo[lst, p]], {n, 7!}]; lst [From _(* _Vladimir Joseph Stephan Orlovsky_, Aug 12 2009] *)

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editing

#14 by Vincenzo Librandi at Mon Jun 09 14:27:02 EDT 2014

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proposed