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Core Definitions ​

This page collects the formal definitions used throughout the site. Each definition includes a worked example. The notation follows Paper 1 (dropping) conventions; see the Terminology Map for equivalences with Paper 2 (stopping). None of these objects is new: the dropping time is the stopping time of Terras (1976) counted in single steps, a dropping set is the set of integers with a given stopping time, and the Syracuse map is classical. Only the names differ.

Collatz Step ​

Definition 1 (Collatz Step). The Collatz function f:Z+→Z+f : \mathbb{Z}^+ \to \mathbb{Z}^+ is defined by

f(n)={n/2if n is even3n+1if n is oddf(n) = \begin{cases} n/2 & \text{if } n \text{ is even} \\ 3n+1 & \text{if } n \text{ is odd} \end{cases}

Example. f(7)=3(7)+1=22f(7) = 3(7)+1 = 22 (odd rule), then f(22)=22/2=11f(22) = 22/2 = 11 (even rule).

Orbit ​

Definition 2 (Orbit). The orbit of nn is the sequence n,  f(n),  f2(n),  …n,\; f(n),\; f^2(n),\; \ldots obtained by iterating ff. From 1 it repeats 1→4→2→11 \to 4 \to 2 \to 1 forever, so an orbit that reaches 1 is written up to its first 1:

orbit(n)=[n,  f(n),  f2(n),  …,  1]\text{orbit}(n) = [n,\; f(n),\; f^2(n),\; \ldots,\; 1]

Whether every orbit reaches 1 is the Collatz conjecture.

Example. The orbit of 6 is [6,3,10,5,16,8,4,2,1][6, 3, 10, 5, 16, 8, 4, 2, 1].

Dropping Time ​

Definition 3 (Dropping Time). The dropping time of n>1n > 1 is the smallest k≥1k \geq 1 such that fk(n)<nf^k(n) < n, if there is one:

drop(n)=min⁡{k≥1:fk(n)<n}\text{drop}(n) = \min\{k \geq 1 : f^k(n) < n\}

This is the stopping time of the literature (Terras, 1976, who counts an odd step and the halving after it as one step) and of Paper 2. The OEIS also calls it the dropping time (A060445).

Example. For n=5n = 5: f(5)=16f(5) = 16, f2(5)=8f^2(5) = 8, f3(5)=4f^3(5) = 4. Since 4<54 < 5 and this is the first value below 5, the dropping time is 3.

Dropping Destination ​

Definition 4 (Dropping Destination). The dropping destination of nn is the first iterate that falls below nn:

dest(n)=fdrop(n)(n)\text{dest}(n) = f^{\text{drop}(n)}(n)

Example. The dropping destination of 5 is f3(5)=4f^3(5) = 4.

Dropping Orbit ​

Definition 5 (Dropping Orbit). The dropping orbit of nn is the segment of the orbit from nn up to (but not including) the dropping destination dd:

dorbit(n)=[n,  f(n),  …,  2d]\text{dorbit}(n) = [n,\; f(n),\; \ldots,\; 2d]

It includes nn and excludes the destination. Its length equals the dropping time.

Example. The dropping orbit of 5 is [5,16,8][5, 16, 8] (length 3). The destination 44 is excluded.

For n=3n = 3: orbit proceeds 3→10→5→16→8→43 \to 10 \to 5 \to 16 \to 8 \to 4, and the destination is 2<32 < 3. The dropping orbit is [3,10,5,16,8,4][3, 10, 5, 16, 8, 4] (length 6).

Dropping Set ​

Definition 6 (Dropping Set). For each positive integer kk, the dropping set Dsetk\text{Dset}_k is defined as:

Dsetk={n∈Z:n>1,  fk(n)<n, and fj(n)≥n for all 0<j<k}\text{Dset}_k = \{n \in \mathbb{Z} : n > 1,\; f^k(n) < n, \text{ and } f^j(n) \geq n \text{ for all } 0 < j < k\}

That is: the set of all integers whose first drop below themselves occurs at exactly step kk.

Note on existence

This is a definition, not a claim. We do not assert that every integer greater than 1 belongs to some Dsetk\text{Dset}_k — that assertion would be equivalent to the Collatz conjecture. We define Dsetk\text{Dset}_k as the set of integers with dropping time kk, and the results on this site describe properties of these sets. An integer with no finite dropping time would simply not belong to any Dsetk\text{Dset}_k.

Example.

  • Dset1\text{Dset}_1 = all even numbers {2,4,6,8,…}\{2, 4, 6, 8, \ldots\}, since one halving gives n/2<nn/2 < n.
  • Dset3={5,9,13,17,…}={n>1:n≡1(mod4)}\text{Dset}_3 = \{5, 9, 13, 17, \ldots\} = \{n > 1 : n \equiv 1 \pmod{4}\}.

Both examples are arithmetic progressions. For a general dropping set the Affine Orbit Structure gives this much: the pattern of odd and even steps in the first kk steps depends only on nn modulo a power of 2, and when a pattern's multiplier 3s/2k−s3^s/2^{k-s} (with ss odd steps) first falls below 1 at step kk, every large enough number with that pattern has dropping time exactly kk. That every such number greater than 1 does, so that each Dsetk\text{Dset}_k is exactly a union of arithmetic progressions, is Terras's coefficient stopping time conjecture (so named by Lagarias, 1985). It is open, it would rule out nontrivial cycles, and it has been verified for all nn from 2 to 2.8×10192.8 \times 10^{19} (Rozier and Terracol, 2026).

Orbital Oddity ​

Definition 7 (Orbital Oddity). The orbital oddity of nn is the count of odd numbers in its dropping orbit. Within the verified range above, all members of Dsetk\text{Dset}_k share the same orbital oddity ss; that they always do would follow from the coefficient stopping time conjecture.

Example. The orbital oddity of 3 is 2. Its dropping orbit is [3,10,5,16,8,4][3, 10, 5, 16, 8, 4], which contains two odd values: 3 and 5.

Syracuse Map ​

Definition 8 (Syracuse Map). The Syracuse map S:Zodd+→Zodd+S : \mathbb{Z}_{\text{odd} }^+ \to \mathbb{Z}_{\text{odd} }^+ compresses each 3n+13n+1 step and all subsequent halvings into a single operation:

S(n)=3n+12v2(3n+1)S(n) = \frac{3n+1}{2^{v_2(3n+1)} }

where v2(m)v_2(m) is the 2-adic valuation of mm (the largest power of 2 dividing mm).

Example. S(7)=3(7)+12v2(22)=2221=11S(7) = \frac{3(7)+1}{2^{v_2(22)} } = \frac{22}{2^1} = 11, since 22=2⋅1122 = 2 \cdot 11.

Alpha Value ​

Definition 9 (Alpha Value). For odd nn, the alpha value is the 2-adic valuation of 3n+13n+1:

α(n)=v2(3n+1)\alpha(n) = v_2(3n+1)

This counts how many halvings follow the 3n+13n+1 step in the Syracuse map.

Example. α(5)=v2(16)=4\alpha(5) = v_2(16) = 4, since 3(5)+1=16=243(5)+1 = 16 = 2^4.

Alpha Sequence ​

Definition 10 (Alpha Sequence). The alpha sequence of an odd number nn is the sequence of alpha values along its Syracuse orbit until reaching 1:

α-seq(n)=[α(n),  α(S(n)),  α(S2(n)),  …]\alpha\text{-seq}(n) = [\alpha(n),\; \alpha(S(n)),\; \alpha(S^2(n)),\; \ldots]

The sequence terminates if and when the Syracuse orbit reaches 1.

Example. The alpha sequence of 3 is [1,4][1, 4]:

  • S(3)=(3⋅3+1)/21=5S(3) = (3 \cdot 3 + 1)/2^1 = 5, so α(3)=1\alpha(3) = 1
  • S(5)=(3⋅5+1)/24=1S(5) = (3 \cdot 5 + 1)/2^4 = 1, so α(5)=4\alpha(5) = 4

The alpha sequence of 7 is [1,1,2,3,4][1, 1, 2, 3, 4].