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%I #28 May 26 2021 03:22:06
%S 1,2,5,10,21,29,30,33,34,41,46,61,66,69,77,78,82,86,101,102,105,109,
%T 110,113,129,133,141,142,145,165,173,177,178,185,194,201,209,213,214,
%U 221,226,230,246,254,257,258,273,282,286,290,298,313,317,321,322,329,330
%N 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.
%C a(n)*(a(n)+1)*(2a(n)+1) must be squarefree, so A344378 is a subsequence of A172186. A344378 is the complement of A344380 in A172186.
%H René Gy, <a href="https://math.stackexchange.com/q/4122583/130022">When the sum of the first n consecutive even (2k>0) powers is a prime number?</a>, Math StackExchange.
%e 2 belongs to the sequence since 1 + 2^(2*2) = 17 is a prime number which is larger than 2*2 + 1 = 5.
%e 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.
%t lim = 330; listu = {};
%t listn = Select[Range[1, lim],
%t SquareFreeQ[# (# + 1) (2 # + 1)] &]; listL = {};
%t Do[M = (Transpose[FactorInteger[m (m + 1) (2 m + 1)]][[1]] - 1)/2;
%t L = 1; Do[L = LCM[L, j], {j, M}];
%t AppendTo[listL, L], {m, listn}]; list = Transpose[{listn, listL}];
%t Do[n = l[[1]]; L = l[[2]];
%t listp = Select[Range[n-1],
%t PrimeQ[#] && Mod[L, (# - 1)/2] == 0 &]; lp = Length[listp];
%t j = 1; While[j \[LessSlantEqual] lp, p = listp[[j]];
%t If[Mod[n - Floor[n/p], p] == 0, j = lp + 2, j = j + 1]];
%t If[j != lp + 2, AppendTo[listu, n]];
%t , {l, list}]; listu
%Y Cf. A172186, A344378.
%K nonn
%O 1,2
%A _René Gy_, May 16 2021
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