OFFSET
1,1
COMMENTS
Terms belong to A172186 but not to A344378. Even though a(n)*(a(n)+1)*(2*a(n)+1) is squarefree, Sum_{j=1..a(n)} j^(2k) always has a prime divisor which is smaller than 2*a(n)+3, whatever k. For the integers m such that m*(m+1)*(2*m+1) is nonsquarefree, Sum_{j=1..m} j^(2k) always has a prime divisor which is smaller than 2*m+3, whatever k, because it is divisible by any prime p such that p^2 divides m*(m+1)*(2*m+1).
LINKS
René Gy, When the sum of the first n consecutive even (2k>0) powers is a prime number?, Math StackExchange.
EXAMPLE
14 belongs to the sequence, because it is squarefree, and Sum_{j=1..14} j^(2k) is always divisible by 29 when 14 does not divide k, and when 14 divides k, it is divisible by 13 or by 7.
CROSSREFS
KEYWORD
nonn
AUTHOR
René Gy, May 16 2021
EXTENSIONS
More terms added and incorrect Mathematica program removed by Jinyuan Wang, Mar 07 2025
STATUS
approved
