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Give the first few terms of a geometric[1] sequence such that:

  • the sequence is increasing and infinite
  • only one digit is used throughout the sequence
  • the terms are written as base ten decimals and are integers

[1]For a geometric sequence, each subsequent term is found by multiplying the previous one by a fixed non-zero number.

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  • $\begingroup$ Do you have a particular sequence in mind that you know exists? $\endgroup$
    – AHKieran
    Dec 20, 2019 at 9:37
  • $\begingroup$ @AHKieran, yes, one, and replies may be with its first few terms, instead of later terms of the same seq. $\endgroup$
    – Tom
    Dec 20, 2019 at 9:41
  • $\begingroup$ Are you asking for the first few terms because the pattern of only one digit breaks at some point? Because a geometric sequence is uniquely determined by its first term and its common ratio. $\endgroup$
    – Taladris
    Dec 21, 2019 at 7:16

2 Answers 2

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How about this sequence

$9.999\ldots$
$99.999\ldots$
$999.999\ldots$
$9999.999\ldots$

as each term is

an integer equal to $10, 100, 1000, 10000$ etc

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  • 1
    $\begingroup$ appears to fail in the requirement that the terms are integers $\endgroup$
    – Penguino
    Dec 20, 2019 at 10:04
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    $\begingroup$ @Penguino these are integers $\endgroup$
    – hexomino
    Dec 20, 2019 at 10:04
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    $\begingroup$ Nice. Is that lateral-thinking? $\endgroup$
    – Jay
    Dec 20, 2019 at 11:07
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    $\begingroup$ @MatthewWells This isn't rounding, those are real equal signs. $\endgroup$ Dec 20, 2019 at 12:47
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    $\begingroup$ Are you using 10 as the fixed non-zero integer? $\endgroup$
    – Smock
    Dec 20, 2019 at 13:52
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There are infinite of these sequences

All have their geometric factor (not sure about this term) of

1x

Example:

0, 0, 0, ...
5, 5, 5, ...
11, 11, 11, ...
888, 888, 888, ...
...

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    $\begingroup$ Maybe this is strictly correct but I meant strictly increasing :P $\endgroup$
    – Tom
    Dec 20, 2019 at 9:20
  • $\begingroup$ The "geometric factor" is usually called the common ratio of the sequence, since it is the ratio that any two consecutive terms share. $\endgroup$
    – Taladris
    Dec 21, 2019 at 7:25

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