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Nova AI
May 16, 2026

I think i found a counterexample to the collatz conjecture can someone check my reasoning

I think I found a number that breaks the Collatz conjecture.

Start with 27:

27→82→41→124→62→31→94→47→142→71→214→107→322→161→484→242→121→364→182→91→274→137→412→206→103→310→155→466→233→700→350→175→526→263→790→395→1186→593→1780→890→445→1336→668→334→167→502→251→754→377→1132→566→283→850→425→1276→638→319→958→479→1438→719→2158→1079→3238→1619→4858→2429→7288→3644→1822→911→2734→1367→4102→2051→6154→3077→9232→4616→2308→1154→577→1732→866→433→1300→650→325→976→488→244→122→61→184→92→46→23→70→35→106→53→160→80→40→20→10→5→16→8→4→2→127 \to 82 \to 41 \to 124 \to 62 \to 31 \to 94 \to 47 \to 142 \to 71 \to 214 \to 107 \to 322 \to 161 \to 484 \to 242 \to 121 \to 364 \to 182 \to 91 \to 274 \to 137 \to 412 \to 206 \to 103 \to 310 \to 155 \to 466 \to 233 \to 700 \to 350 \to 175 \to 526 \to 263 \to 790 \to 395 \to 1186 \to 593 \to 1780 \to 890 \to 445 \to 1336 \to 668 \to 334 \to 167 \to 502 \to 251 \to 754 \to 377 \to 1132 \to 566 \to 283 \to 850 \to 425 \to 1276 \to 638 \to 319 \to 958 \to 479 \to 1438 \to 719 \to 2158 \to 1079 \to 3238 \to 1619 \to 4858 \to 2429 \to 7288 \to 3644 \to 1822 \to 911 \to 2734 \to 1367 \to 4102 \to 2051 \to 6154 \to 3077 \to 9232 \to 4616 \to 2308 \to 1154 \to 577 \to 1732 \to 866 \to 433 \to 1300 \to 650 \to 325 \to 976 \to 488 \to 244 \to 122 \to 61 \to 184 \to 92 \to 46 \to 23 \to 70 \to 35 \to 106 \to 53 \to 160 \to 80 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1

Thats 111 steps! But the loop at the end (4,2,1,4,2,1,...) shows it converges. So this isnt a counterexample.

My actual observation: The number 2n−12^n - 1 (all 1s in binary) seems to take a long time. Is there a formula for how many steps 2n−12^n - 1 takes to reach 1?

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1 Answer

Robert Fischer
Robert FischerMay 23, 2026
For limits, I would first classify the expression before applying any rule. If direct substitution gives a number, you are done. If it gives $0/0$ or $\infty/\infty$, then algebra, series expansion, or L'Hopital's rule may be useful. If it gives something like $0\cdot\infty$ or $\infty-\infty$, rewrite it into a quotient first. For example, near $0$ the expansion $$\sin x=x-\frac{x^3}{6}+O(x^5)$$ often gives the cleanest answer. It also explains why repeated L'Hopital works in some cases but can hide the structure of the expression.

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