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I am trying to find a way to express this sequence,

$$ 1, \frac{1}{7}, \frac{7}{17}, \frac{7}{23}, \frac{17}{31}, \frac{1}{41}, \frac{23}{47}, \frac{31}{49}, \frac{17}{73}, \frac{49}{71}, \frac{23}{89}, \frac{7}{103}, ... $$

It showed up while searching for solutions to a circle. So far I have found that for any m or n, that is $\frac{m}{n}$, m and n must be apart of the OEIS sequence here.

  1. Numbers whose prime factors are all congruent to +1 or -1 modulo 8.
  2. Numbers of the form $x^2 - 2*y^2$, where x is odd and x and y are relatively prime.

So is there a way to generate this sequence? Are there any interesting properties you noticed looking at it?

(Note: if necessary 1 can be excluded)

Edit: To add some better context the way I generated the sequence was by taking the integer solutions to $x^2+y^2=2(n)^2$ and adding term $\frac{y}{x}$ in reduced form to the sequence if it is new $\forall n$ that came before (thus no two terms in the sequence are the same).

Edit2: My apologies one of the terms was wrongly recorded in the series.

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  • $\begingroup$ @sil sorry if it's unclear I understand your point, I'll try to add more context. $\endgroup$
    – PMaynard
    Oct 24, 2019 at 7:58
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    $\begingroup$ I've updated the question to show how it was generated. It goes up to around n=100 $\endgroup$
    – PMaynard
    Oct 24, 2019 at 8:17
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    $\begingroup$ Why $\frac1{41}$ twice? (Position 6 and position 10) $\endgroup$ Oct 24, 2019 at 8:39

1 Answer 1

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It is possible to generate almost this sequence, except for the value of $1/49$ and the positions of the $1/41$ and $79/119$:

  • find primitive Pythagorean triples, i.e. $a^2+b^2=c^2$ where $a$ and $b$ are coprime and $a \lt b$, and order them by $c$ (and ties perhaps by $a$)
  • calculate $\dfrac{b-a}{b+a}$

You would then get

a   b   c   (b-a)/(b+a)
0   1   1   1/1
3   4   5   1/7
5   12  13  7/17
8   15  17  7/23
7   24  25  17/31
20  21  29  1/41
12  35  37  23/47
9   40  41  31/49
28  45  53  17/73
11  60  61  49/71
16  63  65  47/79
33  56  65  23/89
48  55  73  7/103
13  84  85  71/97
36  77  85  41/113
39  80  89  41/119
65  72  97  7/137
20  99  101 79/119
60  91  109 31/151 
...

which is close to your sequence.

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  • $\begingroup$ Thanks, for this I'm currently looking at the sequence further to see if it can be another form of what you just mentioned $\endgroup$
    – PMaynard
    Oct 24, 2019 at 8:05
  • $\begingroup$ I've updated the question to show how it was generated. If you find a way to connect the two please let me know. Once again thanks $\endgroup$
    – PMaynard
    Oct 24, 2019 at 8:17
  • $\begingroup$ My series was wrongly recorded for the values you mentioned your answer was correct please let me know if you can think of another way to define the sequence or any interesting properties of it. Thanks! $\endgroup$
    – PMaynard
    Oct 24, 2019 at 8:35
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    $\begingroup$ The reason behind this being that $(a-b)^2+(a+b)^2=2a^2+2b^2=2c^2$ $\endgroup$ Oct 24, 2019 at 8:40

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