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Can you work out the mathematical rule for these number sequences? It is the same rule for both sequences

2, 4, 16, 7, 49, 25, 9, 81, 9, 81....

8, 64, 22, 6, 36, 15, 6, 36, 15....

Hint 1

this is a mathematical puzzle

From time to time hints will be updated.

Note: this sequence cannot be found in The On-Line Encyclopedia of Integer Sequences

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2 Answers 2

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Is it...

Finding the smallest digit, squaring it, then adding with other digits.

$2$ -> $2^2$ = $4$.
$4$ -> $4^2$ = $16$.
$16$ -> $1^2 + 6$ = $7$.
$7$ -> $7^2$ = $49$.
$49$ -> $4^2 + 9$ = $25$.
$25$ -> $2^2+5$ = $9$.
...

And that's why $64$ -> $4^2 + 6$ = $22$.
Trivia: The numbers will not exceed 99 btw.

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  • $\begingroup$ Very nice, this seems to be the cleanest fit! $\endgroup$
    – Phylyp
    Commented Mar 16, 2018 at 4:38
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    $\begingroup$ Even more trivia: the numbers won't exceed 81. (For any 2-digit number, the next number will be strictly smaller) $\endgroup$
    – Bass
    Commented Mar 16, 2018 at 9:25
  • $\begingroup$ Great, Well done! $\endgroup$
    – tom
    Commented Mar 16, 2018 at 10:31
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    $\begingroup$ @Bass good point, though note that 98 --> 73 and 9 --> 81, but 99-->90 (compare with 22-->6 in question). Having said that you cannot get to 99 starting from a single digit. $\endgroup$
    – tom
    Commented Mar 16, 2018 at 13:05
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I am not sure if this is the intended pattern, but it could be:

The sum of all digits, where the lowest (non zero) digit is squared. For numbers less than ten there is only one digit, so it will simply be the square of that number.
Some (non-consecutive) examples from the question:
4 -> 16 because 4 is the lowest (and only) digit, so it is squared
16 -> 7 because 1 is the lowest digit, is squared, and added to 6
36 -> 15 because 3 is the lowest digit, is squared, and added to 6

EDIT: Same as athin's answer, just a few minutes slower (and without the interesting trivia)

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    $\begingroup$ Ninja'd by athin $\endgroup$
    – somebody
    Commented Mar 16, 2018 at 4:37
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    $\begingroup$ Great, well done, +1! $\endgroup$
    – tom
    Commented Mar 16, 2018 at 10:31

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