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6, 14, 38, 42, 57, 65, 70, 93, 106, 114, 118, 138, 154, 158, 182, 186, 190, 205, 210, 217, 218, 222, 266, 277, 281, 285, 309, 326
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internal format)
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OFFSET
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1,1
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COMMENTS
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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).
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LINKS
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EXAMPLE
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14 belongs to the sequence, because it is squarefree, and 1+2^(2k)+3^(2k)+4^(2k)+5^(2k)+6^(2k)+7^(2k)+8^(2k)+9^(2k)+10^(2k)+11^(2k)+12^(2k)+13^(2k)+14^(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.
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MATHEMATICA
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lim = 330; listq = {};
Do[M = (Transpose[FactorInteger[m (m + 1) (2 m + 1)]][[1]] - 1)/2;
L = 1; Do[L = LCM[L, j], {j, M}];
list = Select[Range[m-1], PrimeQ[#] && Mod[L, (# - 1)/2] == 0 &];
Do[If[ Mod[Quotient[m, p] - Mod[m, p], p] == 0,
AppendTo[listq, m]], {p, list}]
, {m, Select[Range[lim],
SquareFreeQ[# (# + 1) (2 # + 1)] &]}]; DeleteDuplicates[listq]
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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