OFFSET
1,1
COMMENTS
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).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
EXAMPLE
11 = (1^2 + ... + 5^2)/5;
83 = (7^2 + ... + 11^2)/5;
227 = (13^2 + ... + 17^2)/5.
MATHEMATICA
Select[Table[n^2 + 2 n + 3, {n, 1, 600}], PrimeQ] (* Vincenzo Librandi, Mar 22 2013 *)
PROG
(Magma) [a: n in [1..600] | IsPrime(a) where a is n^2 + 2*n + 3 ]; // Vincenzo Librandi, Mar 22 2013
(SageMath)
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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Zak Seidov, Apr 12 2007
STATUS
approved