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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).

m and m+1 squarefree implies that m*(m+1) is a squarefree oblong number and that m*(m+1)/2 is a squarefree triangular number. - Daniel Forgues, Aug 18 2012

Numbers m such that A002378(m) is squarefree. - Thomas Ordowski, Sep 01 2015

Original title was: "Numbers c such that (c^2-1)c is square free and gcd(c-1,c,c+1)=1", but (c^2-1)c is never squarefree for odd c, and gcd(n,n+1) is always = 1. - M. F. Hasler, Nov 03 2013

These numbers c with distribution a+b=c such that a=(c-1)/2 (see A172186) and b=(c+1)/2 (see A179019) have minimal possible values with function L(a,b,c) = log(c)/log(N(a,b,c)) = log(c)/log((c^2-1)c/4).

This function is minimal orbital in hypothesis (a,b,c).

There are no numbers or distributions which have value L less than log(c)/log((c^2-1)*c/4).

Equivalently, odd squarefree numbers c such that (c^2 - 1)/4 is also squarefree. - Jon E. Schoenfield, Feb 13 2023

The asymptotic density of this sequence is Product_{p prime} (1 - 3/p^2) = A206256 = 0.125486980905... (Tsang, 1985). - Amiram Eldar, Feb 26 2024

The numbers of terms that do not exceed 10^k, for k = 1, 2, ..., are 8, 53, 504, 5029, 50187, 501925, 5019527, 50194688, 501948054, 5019478733, ... . Conjecture: The asymptotic density of this sequence is 4 * Product_{p prime} (1 - 3/p^2) = 4 * A206256 = 0.50194792... . - Amiram Eldar, Sep 24 2024

For numbers a and c, see A172186 and A179017. Numbers b are this sequence.

These numbers c, with distribution a+b=c such that a=(c-1)/2 and b=(c+1)/2, have minimal possible values with function L(a,b,c) = log(c)/log(N[a,b,c]) = log(c)/log((c^2-1)c/4).

There exist no numbers or distributions for which L < log(c)/log((c^2-1)c/4). - Artur Jasinski

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