OFFSET
1,1
COMMENTS
If n == 1 (mod 3), either a(n) = 0 or a(n) is the sum of the squares of the primes 3, 5, ..., prime(2*n+2), because p^2 == 1 (mod 3) for all primes p > 3 so the sum of 2*n+1 such primes is divisible by 3. - Pontus von Brömssen, Jan 20 2025
If n == 1 (mod 3), a(n) > 0 iff n is in A370633. - Robert Israel, Nov 27 2025
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
MAPLE
P:= map(`^`, select(isprime, [seq(i, i=3..10^7, 2)]), 2):
PS:= ListTools:-PartialSums([0, op(P)]):
N:= nops(PS):
f:= proc(n) local k;
if n mod 3 = 1 then
if isprime(PS[2*n+2]) then return PS[2*n+2] else return 0 fi
fi;
for k from 1 to N-(2*n+1) do
if isprime(PS[k+2*n+1]-PS[k]) then return PS[k+2*n+1]-PS[k] fi
od;
FAIL
end proc:
map(f, [$1..100]); # Robert Israel, Nov 27 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Jan 20 2025
EXTENSIONS
Escape clause and more terms added by Pontus von Brömssen, Jan 20 2025
STATUS
approved
