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

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

## Comments

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TopNewestIt 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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Nice 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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