OFFSET
1,1
COMMENTS
With the exception of 2^2+3^2+5^2=38 and 3^2+5^2+7^2=83 all sums of squares of 3 consecutive primes are divisible by 3 because mod(p^2,3)=1 for all primes p>3.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
EXAMPLE
a(1)=113 because (7^2+11^2+13^2)/3=(49+121+169)/3=339/3=113 is prime.
MATHEMATICA
b = {}; a = 2; Do[k = (Prime[n]^a + Prime[n + 1]^a + Prime[n + 2]^a)/3; If[PrimeQ[k], AppendTo[b, n]], {n, 1, 200}]; b (* Artur Jasinski, Sep 30 2007 *)
PROG
(PARI) v=vector(10000); i=0; p=5; q=7; forprime(r=8, 1e8, if(isprime(t=(p^2+q^2+r^2)/3), v[i++]=t; if(i==#v, return)); p=q; q=r) \\ Charles R Greathouse IV, Feb 14 2011
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Hugo Pfoertner, Jun 14 2003
EXTENSIONS
Edited by N. J. A. Sloane, Jun 30 2008 at the suggestion of R. J. Mathar.
STATUS
approved