Centered hexagonal number
A centered hexagonal number, also known as a hex number, is a figurate number representing the total number of dots arranged in a hexagonal pattern with a central dot surrounded by successive layers of dots forming regular hexagons.[1] These numbers arise in the context of centered polygonal numbers for a hexagon, where the nth term counts the points in a figure with n layers around the center.[2] The sequence of centered hexagonal numbers begins 1, 7, 19, 37, 61, 91, 127, 169, ... and is given by the explicit formula H_n = 3n^2 - 3n + 1 for n \geq 1, or equivalently H_n = 3n(n-1) + 1, which simplifies to the difference of consecutive cubes: H_n = n^3 - (n-1)^3.[1][2] Alternatively, using zero-based indexing starting at n=0, the formula is H_n = 3n^2 + 3n + 1, or H_n = (n+1)^3 - n^3.[3] A key property is that the sum of the first [n+1](/page/N+1) centered hexagonal numbers equals (n+1)^3, highlighting their intimate connection to cubic numbers.[1] They also satisfy the recurrence relation H_n = 2H_{n-1} - H_{n-2} + 6 for n \geq 2, with initial terms H_0 = 1 and H_1 = 7.[1] Centered hexagonal numbers appear in various mathematical contexts, such as the crystal ball sequence for a hexagonal lattice, where they count the number of lattice points up to a given shell in a 2D hexagonal grid, each point having six neighbors.[3] This lattice structure relates to applications in crystallography and discrete geometry, modeling atomic arrangements in hexagonal crystals.[2] Additionally, they enumerate the number of ordered integer triples (a, b, c) with -n \leq a, b, c \leq n such that a + b + c = 0, providing insights into combinatorial problems on integer lattices.[3] Notable intersections with other figurate numbers include triangular centered hexagonal numbers (1, 91, 8911, ...) and square ones (1, 169, 32761, ...), though only 1 is both square and triangular among them.[1]Definition and Geometry
Geometric Interpretation
Centered hexagonal numbers are a type of figurate number that can be visualized as a hexagon constructed from dots arranged in a hexagonal lattice, beginning with a single central dot and expanding outward through successive concentric layers. Each layer forms a regular hexagonal ring around the previous structure, with the dots positioned at the vertices and along the edges of the hexagon in a symmetric pattern. This geometric arrangement emphasizes the centered nature of the figure, where the central dot serves as the nucleus, and the surrounding layers build a cohesive, star-like or honeycomb pattern that approximates a filled hexagon as the number of layers increases.[4][1] The layering process starts with layer 0 consisting of just the central dot. The first layer (k=1) adds 6 dots, forming the innermost hexagon around the center. Subsequent layers add 6k dots for the kth layer, where k increases outward: for example, the second layer (k=2) adds 12 dots, creating a larger hexagonal perimeter. This incremental addition ensures that each new layer maintains the hexagonal symmetry and lattice alignment, with dots spaced equidistantly to preserve the overall geometric integrity.[1][4] Visually, the progression of these figures illustrates the cumulative growth: for n=1, there is a single central dot; for n=2, this expands to 7 dots by adding the first layer of 6; and for n=3, it reaches 19 dots with the addition of the second layer's 12 dots, forming a more pronounced hexagonal outline. This step-by-step layering provides an intuitive foundation for understanding the spatial distribution of dots in a hexagonal grid, where each complete figure up to the nth layer encapsulates all preceding layers in a compact, self-similar form.[1][4]Sequence and Examples
The sequence of centered hexagonal numbers begins as 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, and continues onward (OEIS A003215).[3] These numbers represent the total count of dots in successively larger centered hexagonal figures. For n=1, the value is 1, a single central dot with no surrounding layers. For n=2, it is 7, formed by adding one layer of 6 dots around the center. The progression continues with n=3 yielding 19 (two layers, adding 12 more dots), n=4 giving 37 (three layers, adding 18 dots to form a complete hexagon), and n=5 producing 61 (four layers, adding 24 dots). This incremental layering ties directly to the geometric buildup of hexagonal patterns, where each new layer encircles the previous figure with 6 times the layer index in additional dots.[1] The following table lists the first 10 centered hexagonal numbers, along with the corresponding n and the number of surrounding layers (equal to n-1):| n | H_n | Number of layers |
|---|---|---|
| 1 | 1 | 0 |
| 2 | 7 | 1 |
| 3 | 19 | 2 |
| 4 | 37 | 3 |
| 5 | 61 | 4 |
| 6 | 91 | 5 |
| 7 | 127 | 6 |
| 8 | 169 | 7 |
| 9 | 217 | 8 |
| 10 | 271 | 9 |
Mathematical Formulation
Explicit Formula
The explicit formula for the nth centered hexagonal number H(n), where n \geq 1, is given byH(n) = 3n(n-1) + 1.
This expands to the quadratic form
H(n) = 3n^2 - 3n + 1. [3][4] To verify the equivalence, expand the first expression:
$3n(n-1) + 1 = 3n^2 - 3n + 1,
which matches the second form directly by algebraic distribution.[1] This formula derives from the geometric construction of centered hexagonal numbers, where the figure consists of a central point surrounded by n-1 concentric hexagonal layers. The central point contributes 1, and the kth layer (for k = 1 to n-1) adds $6k points, forming the six sides of the hexagon. Thus,
H(n) = 1 + \sum_{k=1}^{n-1} 6k = 1 + 6 \sum_{k=1}^{n-1} k = 1 + 6 \cdot \frac{(n-1)n}{2} = 1 + 3n(n-1),
confirming the explicit expression.[4] To determine if a given positive integer m is a centered hexagonal number, solve for n in $3n^2 - 3n + 1 = m:
$3n^2 - 3n + (1 - m) = 0.
Applying the quadratic formula yields
n = \frac{3 \pm \sqrt{9 - 12(1 - m)}}{6} = \frac{3 \pm \sqrt{12m - 3}}{6}.
The positive root must be an integer for m to be centered hexagonal; the discriminant $12m - 3 must be a perfect square.[1][3]
Recurrence Relation
The centered hexagonal numbers satisfy the first-order linear recurrence relation H(1) = 1 and H(n+1) = H(n) + 6n for n \geq 1.[3] This relation allows sequential computation of the sequence by iteratively adding terms that reflect the incremental growth of the figurate pattern. Geometrically, this recurrence arises from the construction of centered hexagonal figures, where each successive layer forms a hexagonal ring around the previous structure. The central point counts as H(1) = 1. The first ring adds 6 points, one at each vertex of the hexagon. For the second ring, 12 additional points are placed (6*2), forming the next concentric layer with points along each extended side, excluding overlaps at vertices. In general, the n-th ring contributes exactly $6n points, as the hexagon has six sides and the n-th layer requires n points per side beyond the inner connections, leading to the additive term $6n in the recurrence.[1] To illustrate, compute the first few terms: H(2) = H(1) + 6 \cdot 1 = 1 + 6 = 7, representing the center plus the initial ring; H(3) = H(2) + 6 \cdot 2 = 7 + 12 = 19, incorporating the second ring. This iterative approach is particularly suited for building the sequence layer by layer, contrasting with direct evaluation via the closed-form expression H(n) = 3n(n-1) + 1.[3]Generating Function
The ordinary generating function for the centered hexagonal numbers H_n = 3n^2 - 3n + 1 (for n \geq 1) is given by \sum_{n=1}^{\infty} H_n x^n = \frac{x(1 + 4x + x^2)}{(1 - x)^3}. [3] This expression can be derived from the explicit formula for H_n using the known generating functions for the powers of n. Specifically, the series expansions \sum_{n=1}^{\infty} n x^n = \frac{x}{(1 - x)^2}, \quad \sum_{n=1}^{\infty} n^2 x^n = \frac{x(1 + x)}{(1 - x)^3}, \quad \sum_{n=1}^{\infty} x^n = \frac{x}{1 - x} are standard results for quadratic sequences.[5] Substituting H_n = 3n^2 - 3n + 1 yields \sum_{n=1}^{\infty} H_n x^n = 3 \cdot \frac{x(1 + x)}{(1 - x)^3} - 3 \cdot \frac{x}{(1 - x)^2} + \frac{x}{1 - x}, which simplifies to the given rational function after combining terms over the common denominator (1 - x)^3.[5] The denominator (1 - x)^3 arises from the quadratic nature of H_n, reflecting the general form for generating functions of second-degree polynomials in n. This can be expressed in relation to binomial expansions, as \frac{1}{(1 - x)^3} = \sum_{n=0}^{\infty} \binom{n+2}{2} x^n, so the full generating function is x(1 + 4x + x^2) \sum_{n=0}^{\infty} \binom{n+2}{2} x^n, where the numerator polynomial adjusts the coefficients to match the centered hexagonal sequence.[5] This generating function facilitates analytic techniques, such as deriving asymptotic behaviors or partial sums of the sequence, and has been used to establish identities involving centered hexagonal numbers, including connections to other figurate number series.[3]Properties
Algebraic Properties
The nth centered hexagonal number is given by the explicit formulaH(n) = 3n^2 - 3n + 1
for positive integers n.[3] This expression reveals an intrinsic algebraic structure, as it can be transformed to show a direct relation to triangular numbers:
H(n) - 1 = 6 T_{n-1},
where T_m = \frac{m(m+1)}{2} denotes the mth triangular number.[1] To verify this, substitute the triangular formula:
$6 T_{n-1} = 6 \cdot \frac{(n-1)n}{2} = 3n(n-1),
so
$1 + 3n(n-1) = 3n^2 - 3n + 1,
confirming the equivalence algebraically. This transformation highlights how each centered hexagonal number builds upon layers of triangular arrangements, with the central unit plus six times the preceding triangular number. Centered hexagonal numbers possess the property that all terms are odd integers. From the formula, n(n−1) is the product of two consecutive integers and hence even, making 3n(n−1) even; adding 1 then yields an odd result.[3] A notable summation identity is
\sum_{k=1}^n H(k) = n^3. [6]
This can be established by mathematical induction on n. For the base case n=1, H(1) = 1 = 1^3. Assume the statement holds for some positive integer n, so \sum_{k=1}^n H(k) = n^3. For n+1,
\sum_{k=1}^{n+1} H(k) = n^3 + H(n+1) = n^3 + [3(n+1)n + 1] = n^3 + 3n^2 + 3n + 1 = (n+1)^3.
By the principle of mathematical induction, the identity holds for all positive integers n.[7]