A129389 -id:A129389 - cp's OEIS frontend

A098062 Primes of the form n^2 + 4n + 8.

13, 29, 53, 173, 229, 293, 733, 1093, 1229, 1373, 2029, 2213, 3253, 4229, 4493, 5333, 7229, 7573, 9029, 9413, 10613, 13229, 13693, 15629, 18229, 18773, 21613, 24029, 26573, 27893, 31333, 33493, 37253, 41213, 42853, 46229, 47093, 54293, 55229

Offset: 1

Giovanni Teofilatto, Sep 12 2004

Or, primes that are equal to the mean of 7 consecutive squares. - Zak Seidov, Apr 14 2007
Sum of 7 consecutive squares starting with m^2 is equal to 7*(13 + 6*m + m^2) and mean is (13 + 6*m + m^2)=(m+3)^2+4. Hence a(n)=A005473(n+1). Note that only nonnegative m's are considered. - Zak Seidov, Apr 14 2007
a(n)==1 (mod 4).
a(n)= A005473(n+1). - Zak Seidov, Apr 12 2007

			13 = (0^2 + ... + 6^2)/7, 29 = (2^2 + ... + 8^2)/7 = 29, 53 = (4^2 + ... + 10^2)/7 = 53.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A005473, A056899, A067201, A007591, A129389, A129412.

[a: n in [0..250] | IsPrime(a) where a is n^2 + 4*n + 8]; // Vincenzo Librandi, Jul 17 2012

Select[ Table[ n^2 + 4n + 8, {n, 240}], PrimeQ[ # ] &] (* Robert G. Wilson v, Sep 14 2004 *)

for(n=0,240,if(isprime(p=n^2+4*n+8),print1(p,","))) \\ Klaus Brockhaus

Edited, corrected and extended by Robert G. Wilson v and Klaus Brockhaus, Sep 14 2004

Edited by N. J. A. Sloane, Jul 02 2008 at the suggestion of R. J. Mathar

A129388 Primes that are equal to the mean of 5 consecutive squares.

Original entry on oeis.org

11, 83, 227, 443, 1091, 1523, 2027, 3251, 6563, 9803, 11027, 12323, 13691, 15131, 21611, 29243, 47963, 50627, 56171, 59051, 62003, 65027, 74531, 88211, 91811, 95483, 103043, 119027, 123203, 131771, 136163, 140627, 149771, 173891, 178931

Offset: 1

Zak Seidov, Apr 12 2007

The sum of 5 consecutive squares starting with k^2 is equal to 5*(6 + 4*k + k^2) and the mean is (6 + 4*k + k^2) = (k+2)^2 + 2. Hence a(n)= A056899(n+2).

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

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A056899, A067201, A102305, A129389.

[a: n in [1..600] | IsPrime(a) where a is  n^2 + 2*n + 3 ]; // Vincenzo Librandi, Mar 22 2013

Select[Table[n^2 + 2 n + 3, {n, 1, 600}], PrimeQ] (* Vincenzo Librandi, Mar 22 2013 *)

A102305=[n^2+2*n+3 for n in range(1,1001)]
[n^2+2*n+3 for n in (1..600) if is_prime(A102305[n-1])] # G. C. Greubel, Feb 03 2024

A129412 Numbers k such that mean of 7 consecutive squares starting with k^2 is prime.

Original entry on oeis.org

0, 2, 4, 10, 12, 14, 24, 30, 32, 34, 42, 44, 54, 62, 64, 70, 82, 84, 92, 94, 100, 112, 114, 122, 132, 134, 144, 152, 160, 164, 174, 180, 190, 200, 204, 212, 214, 230, 232, 240, 242, 250, 252, 262, 264, 272, 274, 284, 290, 300, 304, 310, 314, 344, 354, 370, 372

Offset: 1

Zak Seidov, Apr 14 2007

Sum of 7 consecutive squares starting with k^2 is equal to 7*(13 + 6*k + k^2) and mean is (13 + 6*k + k^2) = (k+3)^2+4. Hence a(n) = A007591(n+1)-3.

			(0^2+...+6^2)/7=13 prime, (2^2+...+8^2)/7=29 prime, (4^2+...+10^2)/7=53 prime.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A005473, A056899, A067201, A007591, A129389, A098062.

Select[Range[0,400],PrimeQ[Mean[Range[#,#+6]^2]]&]  (* Harvey P. Dale, Apr 23 2011 *)

is(n)=isprime(n^2+6*n+13) \\ Charles R Greathouse IV, Jun 13 2017

cp's OEIS Frontend

A098062 Primes of the form n^2 + 4n + 8.

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A129388 Primes that are equal to the mean of 5 consecutive squares.

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A129412 Numbers k such that mean of 7 consecutive squares starting with k^2 is prime.

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