From the simple continued fraction of $\pi$, one gets the convergents,

$$p_n = \frac{3}{1}, \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \frac{103993}{33102}, \frac{104348}{33215}, \frac{208341}{66317}, \frac{312689}{99532}, \frac{833719}{265381}, \frac{1146408}{364913}, \dots$$

starting with $n=1$, where the numerators and denominators are A002485 and A002486, respectively. If you stare at it hard enough, a pattern will emerge between three consecutive convergents. Define,

$$a_n,\,b_n,\,c_n = p_{n}-3,\;\; p_{n+1}-3,\;\; p_{n+2}-3$$


then for even $n \ge 2$,

$$F(n) = \sqrt{\frac{a_n c_n}{a_n-c_n}-v_n}=\text{Integer}\, (often)$$

For example, for $n = 2$,

$$a_2,\,b_2,\,c_2, = \frac{22}{7}-3,\; \frac{333}{106}-3,\; \frac{355}{113}-3$$

$$F(2) = 1$$

More generally,

$$\begin{array}{cc} n&F(n) \\ 2&1 \\ 4&16\\ 6&4703\\ 8&14093\\ 10&51669\\ 12&122126\sqrt{2}\\ 14&7468474\\ 16&\frac{18549059}{\sqrt{2}}\\ \end{array}$$

and so on. For even $n<100$, I found half of the $F(n)$ were either integer or half-integer. (And all the non-integers were of form $N\sqrt{d}$ for some very small d.)

Some questions:

  1. For $n<500$, $n<1000$, etc, how many $F(n)$ are integers or half-integers?
  2. More importantly, why is $F(n)$ often an integer?
  • 2
    $\begingroup$ Is $\pi$ special? Does a similar pattern hold for $\sqrt{2}$ or $e$? $\endgroup$ – Calvin Lin Sep 28 '13 at 22:07
  • $\begingroup$ Now why didn't I think of that? I checked and a similar pattern exists for $e$, as well as for $\sqrt{2}$ (though I have to check if there's a bug in Mathematica). But it seems the special one is $\sqrt{5}$ since apparently all the $F(n)$ are integers and a subset of the Fibonacci numbers. $\endgroup$ – Tito Piezas III Sep 28 '13 at 23:55
  • $\begingroup$ Is there any correlation between the partial quotients and the (near-)integrality of $F$? Say, if the $n$th partial quotient is 1, then $F(n)$ is an integer? $\endgroup$ – Gerry Myerson Sep 29 '13 at 0:13
  • $\begingroup$ That's a good point. I'll have to check that, too. $\endgroup$ – Tito Piezas III Sep 29 '13 at 0:23

The $q_n = p_n-3$ are the convergent fractions of $\pi-3$ (it really doesn't matter to do this change by the way, you could have started straight from $\pi$, only by picking $n \ge 3$ odd instead of $n \ge 2$ even)

3 consecutive convergent fractions are of the form $\frac ab, \frac cd, \frac{a+kc}{b+kd}$ for some integers $a,b,c,d,k$ and $ad-bc=1$ (because we picked $n$ even).

$F(n) = \sqrt{\frac {a(a+kc)}{a(b+kd)-b(a+kc)}-ac} = \sqrt{\frac{a^2+kac}k-ac} = a/\sqrt k$

From the wikipedia page of $\pi$ I can only see the first $3$ relevant $k$, and they are all $1$, so $F(n) = numerator(a_n)$ for $n=2,4,6$ at least.

  • 1
    $\begingroup$ And here $k$ is a partial quotient, and it is believed that about 41% of the partial quotients for $\pi$ are 1. $\endgroup$ – Gerry Myerson Sep 29 '13 at 7:57
  • $\begingroup$ Ah, so that explains why about half the $F(n)$ were integers. $\endgroup$ – Tito Piezas III Sep 29 '13 at 15:43
  • $\begingroup$ @mercio, can you kindly look at this post on the asymptotics of $a^3+b^3 = c^3\pm 1$, since you were able to answer the one on $a^2+b^2 = c^2\pm 1$. $\endgroup$ – Tito Piezas III Sep 29 '13 at 20:56
  • $\begingroup$ @GerryMyerson: You may be interested in this post. It seems that 41% of the partial quotients of other constants are also 1. $\endgroup$ – Tito Piezas III Jul 14 '15 at 7:34
  • 1
    $\begingroup$ @Watson, nobody knows whether the partial quotients for $\pi$ are bounded, but no one has put forward any reason for thinking they are bounded. The consensus is that there's nothing very special going on with the partial quotients for $\pi$. $\endgroup$ – Gerry Myerson Aug 22 '16 at 23:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.