A tetrahedral number counts dots arranged in a tetrahedron — a triangular pyramid. The pyramid is built from stacked triangular layers: the bottom layer contains Tn dots, the next layer up contains Tn−1 dots, and so on to the apex with one dot.
Te1=1,Te2=4,Te3=10,Te4=20,Te5=35,Te6=56,…
The closed form is
Ten=6n(n+1)(n+2)
As a running sum of triangulars.Ten=T1+T2+⋯+Tn. The n-th tetrahedral number is the partial sum of the first n triangular numbers.
As a binomial coefficient.Ten=(3n+2). Tetrahedral numbers form the fourth diagonal of Pascal's triangle (counting from 0).
As a 3D figurate number. Tetrahedral numbers are the 3D analog of triangular numbers, the simplest 3-dimensional pyramidal sequence. They count dots in a triangular pyramid the way triangulars count dots in a triangle.
For deeper coverage see tetrahedral number, pyramidal numbers, and triangular number.
The Closed Form and Its Derivations
Three distinct ways to arrive at Ten=n(n+1)(n+2)/6.
Sum of triangulars. From Ten=∑k=1nTk=∑k=1n2k(k+1):
Binomial coefficient.(3n+2)=3!(n−1)!(n+2)!=6(n+2)(n+1)n=Ten. The hockey-stick identity in Pascal's triangle says that summing entries along a column equals the next column's entry — and the triangular numbers in column 2 sum to tetrahedral numbers in column 3.
Why the formula is always an integer. Of three consecutive integers n, n+1, n+2, one is divisible by 3, and at least one is even. So the product is divisible by 3⋅2=6.
Asymptotic behavior.Ten∼n3/6. Tetrahedral numbers grow cubically, in contrast to the quadratic growth of polygonal numbers.
The Membership Test — Why It's Harder
Unlike triangulars (test via 8m+1) or squares (test via square root), tetrahedral numbers have no clean closed-form inverse. The cubic equation n(n+1)(n+2)/6=m, or equivalently n3+3n2+2n−6m=0, can be solved by the cubic formula but the discriminant is messy.
Practical approach: estimate then verify. Since Ten≈n3/6 for large n, the inverse is approximately n≈36m. The algorithm:
1. Compute nest=round(36m). 2. Verify Tek=m for k∈{nest−1,nest,nest+1}. 3. If any match, m is tetrahedral. Otherwise, no.
Why a window of three? The estimate 36m is accurate to within a small constant for all m, so verifying nearby integers handles rounding errors and edge cases.
Example. Is 120 tetrahedral? Estimate 36⋅120=3720≈8.96. Round to 9. Check Te8=120, Te9=165, Te7=84. Match at k=8: 120=Te8. Verified: 8⋅9⋅10/6=120.
Non-example. Is 100 tetrahedral? Estimate 3600≈8.43. Round to 8. Check Te7=84, Te8=120, Te9=165. None equals 100. So 100 is not tetrahedral.
Trade-off. The estimate-then-verify approach is fast (constant time) but slightly less elegant than the closed-form perfect-square tests of polygonal numbers.
Properties and Identities
• Recurrence: Ten=Ten−1+Tn with Te1=1. Each tetrahedral number adds the next triangular layer. • Binomial coefficient: Ten=(3n+2). The number of unordered triples from n+2 objects. • Hockey-stick identity: ∑k=1n(2k+1)=(3n+2). The sum of column 2 of Pascal's triangle (the triangular numbers) gives column 3 (the tetrahedrals). • Sum of first $n$ tetrahedrals: ∑k=1nTek=(4n+3) — the 4D pentatope numbers. • Connection to squares: Ten+Ten−1=∑k=1nk2, the sum of the first n squares. • Generating function: ∑n=0∞Tenxn=(1−x)4x. • Asymptotic growth: Ten∼n3/6. • The only tetrahedral squares: 1, 4, 19600. Just three tetrahedral numbers are perfect squares (proved by Anglin in 1985, though Mordell-style arguments suggested this much earlier).
Tetrahedral Numbers and the Cannonball Problem
A famous classical problem: in how many ways can cannonballs be stacked in a tetrahedral pile, such that the total number of cannonballs is also a perfect square?
The question asks: for which n is Ten a perfect square?
No other tetrahedral number is a perfect square. This was a conjecture by Eduard Lucas in 1875 and proved by W.S. Anglin in 1985.
The geometric meaning. Cannonballs piled in a tetrahedron with 48 layers contain exactly 140×140=19600 balls — the same total as cannonballs arranged in a flat square of side 140. The trivial cases n=1 and n=2 are uninteresting; n=48 is the only nontrivial solution.
Why so few? The equation Ten=m2 is a Diophantine equation, n(n+1)(n+2)/6=m2. After clearing denominators and substituting, it reduces to a problem about integer points on a cubic curve. The finiteness of solutions follows from Mordell's theorem (specific case) and the analysis is delicate.
This is one of the most striking finite-solution results in elementary number theory.
Pascal's Triangle and Hockey-Stick Identity
Tetrahedral numbers form the fourth diagonal of Pascal's triangle (counting diagonals from 0).
Pascal's triangle diagonals (each one a figurate-number sequence):
The pattern continues: each diagonal is the cumulative-sum sequence of the previous diagonal.
The hockey-stick identity. Summing entries down a diagonal (a "hockey stick") gives the next diagonal's entry where the stick "hooks":
k=r∑n(rk)=(r+1n+1)
For r=2:
(22)+(23)+⋯+(2n+1)=(3n+2)
The left side is T1+T2+⋯+Tn (triangular numbers); the right side is Ten. This is the cleanest proof that the sum of the first n triangulars is the n-th tetrahedral.
Geometric meaning. A tetrahedron of side n is built from triangular layers of sides n,n−1,…,1. The total dot count is the sum of layer dot counts, which is the sum of triangulars.
Common Mistakes
• Confusing tetrahedral with triangular numbers. Triangular: Tn=n(n+1)/2, 2D. Tetrahedral: Ten=n(n+1)(n+2)/6, 3D. Tetrahedrals are partial sums of triangulars.
• Forgetting the divide by 6. The formula has a denominator of 6, not 2 or 4. Skipping it overshoots by a factor of 6.
• Misremembering the membership test. Unlike triangulars and squares, tetrahedrals do not have a perfect-square discriminant test. Use estimate-by-cube-root then verify.
• Trying to use a perfect-square test. Discriminant tests work for polygonal numbers (which are quadratic in n) but not for tetrahedrals (cubic in n). The cubic-root approximation is the standard practical approach.
• Confusing with pyramidal numbers in general. Tetrahedral numbers are the simplest pyramidal numbers — the triangular-base pyramid case. Square pyramidal numbers Pn□=n(n+1)(2n+1)/6 count balls in a square-base pyramid; these are different.
• Off-by-one indexing.Te1=1, not Te0=0. Some sources start at n=0; this explorer uses n=1.
• Confusing the cannonball problem variants. The tetrahedral cannonball problem has three solutions (n = 1, 2, 48). The square pyramidal cannonball problem has just two non-trivial solutions (n=1 and n=24, where the latter gives 1+4+⋯+576=4900=702).
Related Sequences and Concepts
Triangular Numbers — Tn=n(n+1)/2. The 2D base case; tetrahedrals are running sums of these.
Square Pyramidal Numbers — Pn□=n(n+1)(2n+1)/6. Count dots in a square-base pyramid; different from tetrahedral.
Pentatope Numbers — (4n+3)=n(n+1)(n+2)(n+3)/24. The 4D analog of tetrahedral; the next Pascal's triangle diagonal.
Pascal's Triangle — the source of binomial coefficients. Tetrahedrals form its fourth diagonal.
Hockey-Stick Identity — the summation identity that connects column k of Pascal's triangle to column k+1.
Cannonball Problem — the classical question of when tetrahedral numbers are also perfect squares. Three solutions only: 1, 4, 19600.
Pyramidal Numbers — the umbrella term for 3D figurate numbers in pyramidal arrangements. Tetrahedral and square pyramidal are the two most common.
Figurate Numbers — all polygonal, pyramidal, and higher-dimensional analogs. Tetrahedrals are the 3D, 3-sided base case.
Binomial Coefficient — (3n+2)=Ten. The number of unordered triples from n+2 items.
Triangular Pyramid — the geometric solid whose dot count is a tetrahedral number. Four triangular faces, all equilateral if regular.
3-Simplex — the technical name for a tetrahedron in higher-dimensional terminology. The k-simplex generalizes to k dimensions; the k-th simplex numbers are (kn+k−1).
Mordell's Theorem — the deep result enabling the proof that only finitely many tetrahedral numbers are squares.