A344380 -id:A344380 - cp's OEIS frontend

A344378 Positive integers m for which there exists a positive even integer 2k such that Sum_{j=1..m} j^(2k) has no prime divisor smaller than 2*m + 3.

1, 2, 5, 10, 21, 29, 30, 33, 34, 41, 46, 61, 66, 69, 77, 78, 82, 86, 101, 102, 105, 109, 110, 113, 129, 133, 141, 142, 145, 165, 173, 177, 178, 185, 194, 201, 209, 213, 214, 221, 226, 230, 246, 254, 257, 258, 273, 282, 286, 290, 298, 313, 317, 321, 322, 329, 330

Offset: 1

René Gy, May 16 2021

nonn

a(n)*(a(n)+1)*(2a(n)+1) must be squarefree, so this is a subsequence of A172186. It is the complement of A344380 in A172186.

Let L= LCM_j[(p_j-1)/2], where p_j run through the set of the prime divisors of a(n)*(a(n)+1)*(2a(n)+1). For a given member a(n) any admissible k must be a multiple of L, and for any prime p smaller than a(n) such that (p-1)/2 divides L, it holds that p does not divide a(n)-Floor[a(n)/p]. But the converse is not true: 397 is squarefree and satisfies the former condition, but Sum_{j=1..397} j^(2k) is always divisible either by 17 or by 73. 397 is the smallest "false positive" with the above test. Other "false positives" are rather scarce: 397,469,478,561,885,1002,1554,1658,1702,1977,... - René Gy, Apr 15 2025

			2 belongs to the sequence since 1 + 2^(2*2) = 17 is a prime number which is larger than 2*2 + 1 = 5.
5 belongs to the sequence because 1 + 2^20 + 3^20 + 4^20 + 5^20 = 96470431101379 = 137*704163730667 has no prime divisor smaller than 2*5 + 3 = 13.

René Gy, When the sum of the first n consecutive even (2k>0) powers is a prime number?, Math StackExchange.

Cf. A172186, A344380.

Incorrect Mathematica program removed by Jinyuan Wang, Mar 18 2025

cp's OEIS Frontend

A344378 Positive integers m for which there exists a positive even integer 2k such that Sum_{j=1..m} j^(2k) has no prime divisor smaller than 2*m + 3.

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