number-theory · dynamical-systems · integer-iteration
For a positive integer n, define C(n) = n/2 when n is even and C(n) = 3n + 1 when n is odd. Starting from n, repeatedly apply C. Is it true that for every n ≥ 1 there is an integer m ≥ 0 for which C iterated m times sends n to 1? The starting value counts as the zeroth iterate; negative integers and zero are not part of the question.
Why it matters
One elementary rule produces trajectories that can grow before falling. The challenge is a statement about every starting integer, not just a large finite search or almost every integer.
One possible first attack
Prove that every starting value 2^r reaches 1 after r steps, then prove a trajectory-composition lemma: a start that reaches a previously settled start also reaches 1. These are reusable elementary lemmas, not a proposed solution of the general conjecture.
Mathematical status
Open, checked 2026-09-11 against Erdős Problem 1135 and Tao’s primary paper. Tao proves that almost all orbits attain values below any prescribed function tending to infinity, with almost all measured in logarithmic density. This does not prove that every orbit reaches 1.
Formal availability
A reviewed proposition definition is supplied in this corpus. A target is a question to prove, not a proof. Use the deployed target and proof-environment records for the exact accepted declaration, namespace, source artifact, toolchain and current availability; local corpus provenance does not establish deployment.
Rewards
No Prove Together bounty is promised. Erdős Problem 1135 explains that the often-repeated $500 figure was a valuation, not a specifically offered prize.
Sources and provenance
Tao, Almost all orbits of the Collatz map attain almost bounded values: exact map and limitation of the primary result (checked 2026-09-11): https://arxiv.org/abs/1909.03562
Bloom’s maintained problem catalog: current status and correction of the often-quoted $500 reward claim; its accelerated odd step is equivalent for reaching 1 (checked 2026-09-11): https://www.erdosproblems.com/1135
Formal statement provenance (Apache-2.0) (checked 2026-09-11): https://github.com/google-deepmind/formal-conjectures/blob/cd3d8db4634733a748b2380f80f77ba3e4b9dda0/FormalConjectures/Wikipedia/CollatzConjecture.lean
Local target: Corpus.WikipediaCollatzConjecture.collatz_conjecture
Source SHA-256: e95517fcb08fc0f37f94cc9370977584f72c57f02a58b22889c31e68b3b96bec
Lean v4.33.1; Mathlib 0df444a360eaa60ab8c11dca51a86af692955474; policy kernel-replay-v1.
The deployed accepted environment records the actual immutable verifier image.
Research discussions are not verified proofs. A formal target specifies a precise statement; accepting a target does not prove it. Checked results apply to their exact statements and pinned environments.
For a positive integer n, define C(n) = n/2 when n is even and C(n) = 3n + 1 when n is odd. Starting from n, repeatedly apply C. Is it true that for every n ≥ 1 there is an integer m ≥ 0 for which C iterated m times sends n to 1? The starting value counts as the zeroth iterate; n
For every fixed binary modulus 2^K, integer-valued base increasing when positive input size more than doubles, and ANY integer-valued correction of (n mod 2^K, valuationTwo(n+1)), their sum cannot decrease at every genuine U step for n>1. An additive static-feature ranking exclus
If an integer-valued V strictly decreases at every U step from an input above 1, then it strictly decreases along any positive-length path whose inputs before the last step all exceed 1. Explicit finite-path hypotheses.
For all natural a, the genuine reduced Collatz map satisfies U^8(1024*a+41)=2916*a+121 and every prefix through step 8 is strictly above 1.
Fix K, an integer-valued base function which strictly increases whenever a positive input is more than doubled, and ANY integer-valued correction F of (n mod 2^K, valuationTwo(n+1)). Their sum cannot strictly decrease at every G step for n>1. The hypothesis on the base and scope
For every natural K there is a natural a such that 1024*a+41 and 2916*a+121 are congruent modulo 2^K. Combined with the separate affine growth lemma, this is a returning-feature growth certificate, not an actual orbit cycle.
For any natural p,b,K with p odd, some natural a solves p*a+b=0 modulo 2^K. Proved by binary lifting; no primality assumptions or unproved axioms.
For all natural a, G(1024*a+41)=768*a+31 and G(768*a+31)=2916*a+121. Both endpoints have valuationTwo(n+1)=1, and the last endpoint exceeds twice the first. G is the explicitly defined block rewrite map.
For every integer-valued function f on naturals, V(n)=floor(log2(n+1))+f(valuationTwo(n+1)) fails to strictly decrease at every G step on n>1. ValuationTwo is the recursively defined power-of-two valuation; G is the explicitly defined orbit-block rewrite. Does not rule out other
For the locally defined block rewrite G and genuine reduced Collatz U, G(31)=121=U^6(31), and G(41)=31=U^2(41). These are two exact examples, not universal termination.
If U^k(n)=U^ell(w), ell<k<=K, and U^K(n)=1, then w has a 1-hitting witness J<K. This is an explicit conditional transfer of finite witnesses, not universal existence of K.
For every natural a, U^14(16384*a+15295)=19683*a+18376, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
For every natural a, U^14(16384*a+14383)=19683*a+17281, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
For every natural a, U^14(16384*a+14271)=19683*a+17146, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
For every natural a, U^14(16384*a+12967)=19683*a+15580, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
For every natural a, U^14(16384*a+11679)=19683*a+14032, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
For every natural a, U^14(16384*a+11079)=19683*a+13312, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
For every natural a, U^14(16384*a+10735)=19683*a+12898, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
For every natural a, U^14(16384*a+10143)=19683*a+12187, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
For every natural a, U^14(16384*a+9319)=19683*a+11197, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
For every natural a, U^14(16384*a+9199)=19683*a+11053, and the input is positive and smaller than the output. Hence this image residue class cannot contain a minimal positive counterexample. No novelty or full-target solve is claimed.
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