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I have a sequence of numbers here, which follow a very simple rule that can be expressed using a single fraction:

1, 2, 10, 11, 13, 15, 20, 22, 25, 28, ...

What is this rule (in particular, what is the fraction), and what are the next few numbers in the sequence?


Since nobody's gotten it yet, I'll add a few hints over the next little while.

(06-16 13:22) 1. The numbers in the sequence never get more than 2 digits long.

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  • $\begingroup$ Shall we consider 1, 2 as 01, 02? $\endgroup$ – newzad Jun 11 '16 at 10:16
  • $\begingroup$ @nikamed What would that imply? $\endgroup$ – user88 Jun 11 '16 at 23:41
  • $\begingroup$ because if it's like this one can find infinitely many solutions. $\endgroup$ – newzad Jun 12 '16 at 18:34
  • $\begingroup$ @nikamed Can you give me a few examples? $\endgroup$ – user88 Jun 12 '16 at 18:41
  • $\begingroup$ You can make a number and take every two digits after the point to create your sequence. For example: 0.0102101113152022252831.. leads to a fraction $680486027/66648247604$. $\endgroup$ – newzad Jun 12 '16 at 18:47
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The items in the sequence should be numbered from 2 onwards, and then item $b$ is

the base-$b$ representation of $\displaystyle \left \lfloor \frac{b^2}{4} \right \rfloor$, which can also be written as just $\displaystyle \left \lfloor \frac{100}{4} \right \rfloor$, since $b$ in base $b$ is always $10$.

So the fraction that defines the sequence is

100/4

and the next few entries in the sequence are:

30, 33, 37, 3B, 40, 44, 49, 4E, 50.

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    $\begingroup$ (My original answer gave a different fraction and a slightly different -- but equivalent -- explanation. Joe Z has adjusted what I wrote to match his intention.) $\endgroup$ – Gareth McCaughan Jun 17 '16 at 16:50
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1, 2, 10, 11, 13, 15, 20, 22, 25, 28, ...

1, 2, 10, 11, 13, 15, 20, 22, 25, 28, 31, 100, 103, 107, 111, 115, 119, 110, ...

The pattern is explained as follows:

Add (1) to first number (1) time, then add a zero to first number

1, 2, 10

Add (1) to new number once

11

Add (1+1) to new number (1+1) times, then add a zero to second number

13, 15, 20

Add (1+1) to new number once

22

Add (1+1+1) to new number (1+1+1) times, then add a zero to third number

25, 28, 31, 100

Add (1+1+1) to new number once

103

Add (1+1+1+1) to new number (1+1+1+1) times, then add a zero to forth number

107, 111, 115, 119, 110

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  • $\begingroup$ I don't see any fractions in your definition. Also, you got the next few numbers wrong. $\endgroup$ – user88 Jun 11 '16 at 3:05

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