A158509 Prime(n-1) such that prime(n-1)+5 is square and equal to prime(n+1)-1.

11, 191, 495611, 1628171, 1893371, 7806431, 14167691, 15085451, 16048031, 20124191, 23951231, 27415691, 28558331, 30096191, 41396351, 46022651, 55591931, 87871871, 88811771, 99720191, 103347551, 130051211, 139617851, 152670731, 181333151, 198922811, 222427391

Offset

1,1

Comments

Conjecture: If the condition holds, prime(n-1) and prime(n) are twin primes of the form 10k+1 and 10k+3. I.e., the last digits of the primes are 1 and 3. The 1 ending is evident in the entries here.

The conjecture is true: For p = prime(n-1) > 2 in this sequence we know that {p+1, p+3, p+5} are all even, p+6 = prime(n+1), p+5 = r^2 is a square, and p+4 = r^2-1 = (r-1)*(r+1) is composite. So p+2 = prime(n). Then p mod 10 = 1 because no prime > 5 has rightmost digit 5 and no square has rightmost digit 2. - Alois P. Heinz, Sep 07 2016

Subsequence of A001359 and of A030430. - Alois P. Heinz, Sep 07 2016

Links

Alois P. Heinz, Table of n, a(n) for n = 1..10000

S. M. Ruiz, Integer then equal.

Sebastian Martin Ruiz and others, Integers then Equals, digest of 7 messages in primenumbers Yahoo group, Mar 14 - Mar 20, 2009.

Formula

a(n) = prime(A158460(n)-1). - R. J. Mathar, Sep 07 2016

Example

For n = 6, prime(6-1)=11,11+5=16;prime(6+1)=17,17-1=16. So prime(6-1)=11 is the first entry in the sequence.

Mathematica

Transpose[Select[Partition[Prime[Range[7400000]], 3, 1], Last[#]-First[#] == 6 && IntegerQ[Sqrt[First[#]+5]]&]][[1]] (* Harvey P. Dale, Dec 22 2013 *)

Prog

(PARI) integerequal(m, n, a, b) =
{
local(x, p1, p2);
for(x=m, n,
p1=prime(x-1);
if(issquare(p1+a),
p2=prime(x+1); if((p1+a)==(p2-b),
print(x", "p1", "prime(x))
)
)
)
}

Crossrefs

Cf. A001359, A030430, A158460.

Sequence in context: A171553 A333759 A068649 * A072290 A258295 A213691

Adjacent sequences: A158506 A158507 A158508 * A158510 A158511 A158512

Keyword

nonn

Author

Cino Hilliard, Mar 20 2009

Status

approved