OFFSET
1,1
COMMENTS
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
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
EXAMPLE
13 = (0^2 + ... + 6^2)/7, 29 = (2^2 + ... + 8^2)/7 = 29, 53 = (4^2 + ... + 10^2)/7 = 53.
MATHEMATICA
Select[ Table[ n^2 + 4n + 8, {n, 240}], PrimeQ[ # ] &] (* Robert G. Wilson v, Sep 14 2004 *)
PROG
(PARI) for(n=0, 240, if(isprime(p=n^2+4*n+8), print1(p, ", "))) \\ Klaus Brockhaus
(Magma) [a: n in [0..250] | IsPrime(a) where a is n^2 + 4*n + 8]; // Vincenzo Librandi, Jul 17 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Giovanni Teofilatto, Sep 12 2004
EXTENSIONS
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
STATUS
approved