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In other words, if a prime p divides n, then p == 1 mod 3.

Equivalently, products of primes == 1 (mod 6), products of elements of A002476.

Positive integers n such that n+d+1 is divisible by 3 for all divisors d of n. For example, a(13)=91 since 91=7*13, 91+1+1=93=3*31, 91+7+1=99=9*11, 91+13+1=105=3*7*5, 91+91+1=183=3*61. The only prime p such that x+d+1 is divisible by p for all divisors d of x is p=3. The sequence consists of 1 and all integers whose prime divisors are of the form 6k+1. - Walter Kehowski, Aug 09 2006

Also z such that z^2 = x^2 + x*y + y^2 and gcd(x,y,z) = 1. - Frank M Jackson, Jul 30 2013

From Jean-Christophe Hervé, Nov 24 2013: (Start)

Apart from the first term (for all in this comment), this sequence is the analog of A008846 (hypotenuses of primitive Pythagorean triangles) for triangles with integer sides and a 120-degree angle: a(n), n>1, is the sequence of lengths of the longest side of the primitive triangles.

Not only the square of these numbers is equal to x^2 + xy + y^2 with x and y > 0, but the numbers themselves also are; the sequence starting at n=2 is then a subsequence of A024606.

(End)

Numbers n such that 3/n cannot be written as the sum of 2 unit fractions. - Carl Schildkraut, Jul 19 2016

a(n), n>1, is the sequence of lengths of the middle side b of the primitive triangles such that A < B < C with an angle B = 60 degrees (A335895). Compare with comment of Nov 24 2013 where a(n), n>1, is the sequence of lengths of the longest side of the primitive triangles that have an angle = 120 degrees. - Bernard Schott, Mar 29 2021

Equivalently, sequence A024612 with duplicates removed; i.e., numbers of the form i^2 - i*j + j^2, where 1 <= i < j.

A subsequence of A135412, which consists of multiples (by squarefree factors) of the numbers listed here. It appears that this lists numbers > 1 which have in their factorization: (a) no even power of 3 unless there is a factor == 1 (mod 6); (b) no odd power of 2 or of a prime == 5 (mod 6) and no even power unless there is a factor 3 or == 1 (mod 6). - M. F. Hasler, Aug 17 2016

If we regroup the entries in a triangle with row lengths A004526

3,

7,

12, 13,

19, 21,

27, 28, 31,

37, 39, 43,

... it seems that the j-th row of the triangle contains the numbers i^2+j^2-i*j in row j>=2 and column i = floor((j+1)/2) .. j-1. - R. J. Mathar, Aug 21 2016

Proof of the above characterization: the sequence is the union of 3*(the squares A000290) and A024606 (numbers x^2+xy+y^2 with x > y > 0). For the latter it is known that these are the numbers with a factor p==1 (mod 6) and any prime factor == 2 (mod 3) occurring to an even power. The former (3*n^2) are the same as (odd power of 3)*(even power of any other prime factor). The union of the two cases yields the earlier characterization. - M. F. Hasler, Mar 04 2018

Least term that can be written in exactly n ways is A300419(n). - Altug Alkan, Mar 04 2018

For the general theory see the Fouvry et al. reference and A296095. Bounds used in the Julia program are based on the theorems in this paper. - Peter Luschny, Mar 10 2018

Original definition: Solutions c of cot(2*Pi/3)*(-(a+b+c)*(-a+b+c)*(-a+b-c)*(a+b-c))^(1/2)=a^2+b^2-c^2, c>a,b integers.

Note cot(2*Pi/3) = -1/sqrt(3).

Also the c-values for solutions to c^2 = a^2 + ab + b^2 in positive integers. Also the numbers which occur as the longest side of some triangle with integer sides and a 120-degree angle. - Paul Boddington, Nov 05 2007

The sequence can also be defined as the numbers w which are Heronian means of two distinct positive integers u and v, i.e., w = [u+sqrt(uv)+v]/3. E.g., 28 is the Heronian mean of 4 and 64 (and also of 12 and 48). - Pahikkala Jussi, Feb 16 2008

From Jean-Christophe Hervé, Nov 24 2013: (Start)

This sequence is the analog of hypotenuse numbers A009003 for triangles with integer sides and a 120-degree angle. There are two integers a and b > 0 such that a(n)^2 = a^2 + ab + b^2, and a, b and a(n) are the sides of the triangle: a(n) is the sequence of lengths of the longest side of these triangles. A004611 is the same for primitive triangles.

a and b cannot be equal because sqrt(3) is not rational. Then the values a(n) are such that a(n)^2 is in A024606. It follows that a(n) is the sequence of multiples of primes of form 6k+1 A002476.

The sequence is closed under multiplication. The primitive elements are those with exactly one prime divisor of the form 6k+1 with multiplicity one, which are also those for which there exists a unique 120-degree integer triangle with its longest side equals to a(n).

(End)

Conjecture: Numbers m such that abs(Sum_{k=1..m} [k|m]*A008683(k)*(-1)^(2*k/3)) = 0. - Mats Granvik, Jul 06 2024

Squares of distances between two points in the triangular lattice in two or more nontrivially different ways.

Numbers whose prime factorization contains at least two (not necessarily distinct) primes congruent to 1 mod 6 and any prime factor congruent to 2 mod 3 has even multiplicity. Products of two elements of A024606.

If k is in the sequence then so is k * m^2 for m > 0. - David A. Corneth, Jun 21 2018

These are the primitive elements of A024606, the integers which are expressible as x^2 + xy + y^2 with distinct nonzero x and y.

As a subsequence of A003136 (Loeschian numbers), the sequence is related with the triangular lattice: circles with radius sqrt(a(n)) centered at a grid point in this lattice hit exactly 12 points, cf. A004016.

Numbers with exactly one prime factor of form 6k+1 with multiplicity one and no prime factor of form 3k+2 with odd multiplicity, that is a(n) is of form 3^a*p*q^2, with a>=0, p a prime of form 6k+1, and q an integer with all its prime factors of form 3k+2. There is thus no square in the sequence.

From a(n) = 3^a*p*q^2, it is easily seen that sigma(a(n)) = 2 mod 6,

thus this sequence is a subsequence of A074628: the two sequences are equal up to a(308) = 1723; then A074628(309)= 1729 = a(1)*a(2)*a(3), the famous Ramanujan's taxi number, and a(309) = A074628(310) = 1731.

The square of these numbers is also uniquely decomposable into the form x^2 + xy + y^2 with x and y > 0, thus this sequence is a subsequence of A232437.

For the numbers with representations as a sum of three distinct nonzero squares see A004432. For their multiplicity see A025442.

Here only even numbers are considered, and a representation 2*m = a^2 + b^2 + c^2, a > b > c > 0 denoted by the triple (a,b,c), is called 'balanced' if a = b + c. E.g., 62 is represented by (6, 5, 1) and (7, 3, 2) but only (6, 5, 1) is balanced because 6 = 5 + 1.

The multiplicities are given in A240228.

These numbers a(n) play a role in the problem proposed in A236300: Find all numbers which are of the form (x + y + z)*(u^2 + v^2 + w^2)/2, x >= y >= z >= 0, where u = x-y, v = x-z, w = y-z, with u >= 0, v >=0, w >= 0, u - v + w = 0 and even u^2 + v^2 + w^2 >= 4. The special case (called in a comment on A236300 case (iib)) with distinct u, v, and w, each >=1, needs the numbers a(n) of the present sequence. If the triple is taken as (u, u+w, w) with u < w then the [x, y, z] values are [2*u+w, u+w, u] and the number from A236300 is (2*u+w)*(u^2 + w^2 + u*w) =(2*u+w)*a(n). If this number from A236300 has multiplicity A240228(n) >=2 then there are A240228(n) different values for [x, y, z] and corresponding different A236300 numbers.

Numbers without prime factor of form 6k+1 and without prime factor of form 3k+2 to an odd multiplicity.

The sequence is closed under multiplication. Primitive elements are 3 and square of primes of form 3k+2 (A003627). Sequence A003136 (Loeschian numbers) is the union of {0}, this sequence and A024606 (numbers of form x^2+xy+y^2 with y > x > 0). These 4 sequences are all closed under multiplication.