\(f\left( x \right) =\quad \quad \quad 3x+1\quad \quad \forall \quad x\quad \varepsilon \quad odd\quad positive\quad integer\\ \quad \quad \quad =\quad \quad \quad x/2\quad \quad \quad \quad \forall \quad x\quad \varepsilon \quad even\quad positive\quad integer\)

use this function repeatedly you will always get the answer as 1( it may take a numerous steps for certain numbers ) it always ends with the series 4,2,1

let x=28

Operating this function repeatedly (Used the function for 11 times)

x=28>14>7>22>11>34>17>52>26>13>40>20>10>5>16>8>4>2>1

let x=37

Operating this function repeatedly (used the function for 71 times)

x=37>111>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>1

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## Comments

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TopNewestNice function, but there is a mistake when you put \(x=28\). When plugging in \(x=11\), you get \(34\) and not \(32\).

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It is an open conjecture that it "always ends with the series 4, 2, 1". It's been verified for the first few cases, but there are infinitely many more possibilities left.

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