A303971 Primes p such that 2*p + 1 and (5*p^2 + 4*p + 1)/2 are prime.

3, 5, 29, 53, 83, 173, 233, 239, 359, 653, 719, 743, 1013, 1583, 1889, 2129, 2399, 2939, 2969, 3299, 3359, 3413, 3449, 3539, 3863, 4073, 5399, 5639, 6323, 6983, 7433, 7643, 7649, 8243, 10613, 11369, 11519, 11699, 12119, 12653, 12923, 13463, 13619, 13649, 14303, 14489, 15629, 16253, 17333

Offset

1,1

Comments

(5*p^2 + 4*p + 1)/2 is equivalent to (A005384(k)^2 + A005385(k)^2)/2 for Sophie Germain primes and their safe primes whenever a particular k produces a prime.

a(n) == 5 (mod 6) for n > 1. a(n) == 23 or 29 (mod 30) for n > 2. - Robert Israel, May 08 2018

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

For p = A005384(3) = 5, (5*5^2 + 4*5 + 1)/2 = 73, which is prime, so 5 is in the sequence.

Maple

select(p -> isprime(p) and isprime(2*p+1) and isprime((5*p^2+4*p+1)/2),
[3, 5, seq(seq(30*i+j, j=[23, 29]), i=0..1000)]); # Robert Israel, May 08 2018

Mathematica

Select[Prime[Range[2000]], AllTrue[{2#+1, (5#^2+4#+1)/2}, PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, May 19 2019 *)

Prog

(PARI) select(p->p<>2 && isprime(2*p+1) && isprime((5*p^2+4*p+1)/2), primes(3000)) \\ Andrew Howroyd, May 03 2018

Crossrefs

Subsequence of A005384 and so of A000040.

Cf. A005385.

Sequence in context: A346659 A067200 A106089 * A136085 A153413 A257718

Adjacent sequences: A303968 A303969 A303970 * A303972 A303973 A303974

Keyword

nonn,easy

Author

J. M. Bergot, May 03 2018

Status

approved