A129389 Numbers k such that the mean of 5 consecutive squares starting with k^2 is prime.

1, 7, 13, 19, 31, 37, 43, 55, 79, 97, 103, 109, 115, 121, 145, 169, 217, 223, 235, 241, 247, 253, 271, 295, 301, 307, 319, 343, 349, 361, 367, 373, 385, 415, 421, 427, 439, 445, 451, 475, 499, 511, 547, 553, 559, 571, 601, 607, 649, 673, 679, 697, 709, 751

Offset: 1

Author

Zak Seidov, Apr 12 2007

Keywords

• nonn

Comments

Sum of 5 consecutive squares starting with k^2 is equal to 5*(6 + 4*k + k^2) and mean is (6 + 4*k + k^2) = (k+2)^2 + 2. Hence a(n) = A067201(n+2).

Also, numbers k such that A000217(k) + A000217(k+3) is prime. - Bruno Berselli, Apr 17 2013

Examples

			(1^2 + ... + 5^2)/5 = 11, which is prime;
(7^2 + ... + 11^2)/5 = 83, which is prime;
(13^2 + ... + 17^2)/5 = 227, which is prime.
		

Links

• Bruno Berselli, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A056899, A067201, A129388.

Cf. A000217, A128815 (numbers n such that A000217(n)+A000217(n+2) is prime). [Bruno Berselli, Apr 17 2013]

Programs

Magma

[n: n in [1..800] | IsPrime(n^2+4*n+6)]; /* or, from the second comment: */ A000217:=func; [n: n in [1..800] | IsPrime(A000217(n)+A000217(n+3))]; // Bruno Berselli, Apr 17 2013

Mathematica

Select[Range[800], PrimeQ[#^2 + 4 # + 6] &] (* Bruno Berselli, Apr 17 2012 *)

SageMath

[n for n in (1..1000) if is_prime(n^2+4*n+6)] # G. C. Greubel, Feb 04 2024