5
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Can you fill in the holes in this pattern?

361
722
726
[A]
323
650
223
49729
99458
[B]
19024
2538
2240
98
102
302
22805
[C]

EDIT: As the answer has been found, the clues would have been:
1) An edit to the title to be "A Prime Example..."
2) Adding 19 to the beginning of the sequence
3) A link to an online calculator for finding prime factorials
4) The statement that the last clue just about summed it up


Bonus Question:
What can you choose as the starting number such that the pattern repeats itself forever?

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4
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I think I see a pattern in there, it only works in case one of your calculations is incorrect though.

The solution:

A = 255, B = 99462, C = 20802746

The incorrect transition:

99462 -> 2538, you forgot to mention 19024 in between.

The pattern:

Take a number, split it into its prime factors, add the square of each prime factor to arrive at the next number.
Example: 2538 is 2*3*3*3*47, adding the squares is 4+9+9+9+2209, next number is 2240.

Bonus answer:

Start with 1, you'll stay at that value.

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  • $\begingroup$ Amazingly well done. You're right that I missed a step when making my table and dropped 19,024 from the list. I'll put it back in for completeness sake. You missed the bonus question, though, because 1 is not a prime number and the only factor of 1 is 1 so it has no prime factors. Figure out a different answer for that and you can have an upvote to go along with the answer acceptance. $\endgroup$ – Engineer Toast Mar 13 '15 at 15:37
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    $\begingroup$ Fair enough, in that case I'll pick 16. :-) $\endgroup$ – Moghwyn Mar 13 '15 at 15:44
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    $\begingroup$ Spot on. I also would have accepted 27. There aren't any more below 1,000 and I wonder if there are any more at all. $\endgroup$ – Engineer Toast Mar 13 '15 at 15:49
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    $\begingroup$ @EngineerToast For the bonus, I've brute-forced for any cycles (max length 2048, max value 10^50) with starting values from 2 to 100k without finding any other than 16 → 16 and 27 → 27. $\endgroup$ – LegionMammal978 Oct 27 '16 at 22:41
  • $\begingroup$ Points for completeness, @LegionMammal978 ! $\endgroup$ – Engineer Toast Oct 27 '16 at 23:31

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