5
$\begingroup$

I have a series: $$3, 2, 4, 7, 11, 16, 26, 39, 63, 94, 152, 227, 367, 548, 886, 1323, 2139, 3194, ...$$ which represents values of $m,n$ in Euclid's formula to generate Pythagorean triples where $A-B=\pm17.$

I have found a number of series on MSE and oeis.org/A001333 or oeis.org/A266504

but not this one.

I also tried WolframAlpha and it properly continued the series as:

$$3, 2, 4, 7, 11, 16, 26, 39, 63, 94, 152, 227, 367, 548, 886, 1323, 2139, 3194, 5164, 7711, 12467, 18616, 30098, 44943, 72663, 108502, 175424, ...$$

but the formula it gave me: $$G_n(a_n)(z) = (-3 z^3 + 2 z^2 - 2 z - 3)/(z^4 + 2 z^2 - 1)$$ generates $$-3 -1, -0.734693878, -0.595818815, -0.50148368, -0.432333577, -0.379503603, -0.337911437, -0.304373698, -0.276791842$$ instead.

Since WolframAlpha correctly extend the series from what I entered, it seems that the formula it offered should work so I must be interpreting something incorrectly. Can anyone point me in the right direction for this series or tell me what I'm doing wrong with Wolfram?

WolframAlpha extrapolated correctly for $A-B=\pm17$ when I included the first elements but not when I omitted the first two.

$\endgroup$
  • 6
    $\begingroup$ Presumably that's meant to be the generating function. That is, $G(z)=\sum_{n=0}^{\infty}a_nz^n$. See this $\endgroup$ – lulu Nov 8 '19 at 19:06
  • $\begingroup$ I would like to be able to generate the $n^{th}$ term in the series directly. Can this summation formula be turned into a polynomial as in the one to generate Pell numbers? $\endgroup$ – poetasis Nov 8 '19 at 19:10
  • 1
    $\begingroup$ $$\frac{i^{-n} \left(2 i^{-n} \left(\left(519 \sqrt{2}+734\right) \left((-1)^n-1\right)+\sqrt{703666 \sqrt{2}+995134} \left((-1)^n+1\right)\right) \left(\sqrt{2}+1\right)^{n/2}+4 i^n e^{-\frac{1}{2} n \operatorname{arc sinh}(1)} \left(\left(775 \sqrt{2}+1096\right) \sin \left(\frac{\pi n}{2}\right)+\left(519 \sqrt{2}+734\right) \sqrt{\sqrt{2}+1} \cos \left(\frac{\pi n}{2}\right)\right)\right)}{16 \left(\sqrt{2}+1\right)^{13/2}}$$ $\endgroup$ – AccidentalFourierTransform Nov 8 '19 at 19:13
  • 1
    $\begingroup$ The quartic denominator strongly suggests that no closed formula will be pleasant. $\endgroup$ – lulu Nov 8 '19 at 19:14
  • 1
    $\begingroup$ @ Ross Millikan The first $6$ triples where $A-B=\pm17$ are: $f(4,3)=7,24,25\qquad f(7,2)=45,28,53\qquad f(11,4)=105,88,137\qquad f(16,7)=207,224,305\qquad f(26,11)=555,572,797\qquad f(39,16)=1265,1248,1777$ $\endgroup$ – poetasis Nov 9 '19 at 12:03
4
$\begingroup$

Once we find the fundamental solutions $m=7,n=2$, giving the $45,28,53$ triangle and $m=4,n=3$ giving the $7,24,25$ triangle we can note that you want $(m^2-n^2)-2mn=\pm 17$, which we can rewrite as $(m-n)^2-2n^2=\pm 17$ We can then use the Brahmagupta identity to take one $m,n$ pair and say the next pair is $(2m+n,m)$. This recurrence gives the even terms and odd terms in your sequence separately.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I came up with the same method (though I found it was not original) to generate sequential pairs for differences of $1, 2, 4, 8, 9, 16, 18, 25, 32, 36, 50, 64, 72, 81$ where $m_x=n_{x+1}$. Here, I see it would generate every other triple. Perhaps I can explore those numbers because the pattern of $m+x=n_{x+2}$ holds for differences of $7, 14, 17, 23, 28, 31, 34, $… then gets confusing for $41$. $\endgroup$ – poetasis Nov 9 '19 at 13:18
9
$\begingroup$

There is a recursive formula given by $a_{n+4} = 2 a_{n+2} + a_{n}$ . This suggests separating the formula into even and odd components. I'll work on getting an explicit formula.

Edit: the formula I obtained is:

$$a_{2n} = \frac{1}{4}((6 - \sqrt{2} )(1+\sqrt{2})^n + (6 + \sqrt{2} )(1-\sqrt{2})^n)$$

$$a_{2n+1} = \frac{1}{4}((6 - \sqrt{2} )(1-\sqrt{2})^n + (6 + \sqrt{2} )(1+\sqrt{2})^n)$$

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ And since $|1-\sqrt{2}|<1$, we have $a_{2n}=\operatorname{round}((6 - \sqrt{2} )(1+\sqrt{2})^n/4)$ and $a_{2n+1} = \operatorname{round}((6 + \sqrt{2} )(1+\sqrt{2})^n/4)$ for sufficiently large $n$ (in fact only first few values differ). $\endgroup$ – Sil Nov 8 '19 at 21:09
  • $\begingroup$ In my exploration of $a-b=\pm1$ about a year ago, I came up with two equations, one for $A-B=1$ and another for $B-A=1$ but I didn't like it so I found the equation to generate Pell numbers. Now, here, I have one equation in terms of $n$ an I found the formula to generate $n$. $\endgroup$ – poetasis Nov 9 '19 at 19:45
  • $\begingroup$ @user3257842 I tried your first equation with negative $n$ and it appears to generate the complete set of $out$ values I needed as did the formula I found by myself. I don't have time, with my work schedule, to explore it much now but I upvoted your answer and I will add my findings to $my$ answer one I complete them. Question: How did you come up with the formula? $\endgroup$ – poetasis Nov 11 '19 at 18:41
  • 1
    $\begingroup$ Both the odd and even terms of the series obey the recurrence relationship given by $t_{n+2} = 2 t_{n+1} + t_{n}$. This is linear, that is, if two series obey this recurrence, so does their sum or linear combination. Assuming $t_{n}$ is of form $r^n$, and setting $n = 0$ we would get $r^{2} = 2r^{1} + 1$. This the characteristic polynomial of the recurrence. It can be written as $(r-(1+\sqrt{2}))(r-(1-\sqrt{2})) = 0$. From this we get $r$ can be one of those two roots, so the general formula for $t_{n} = c_{1}(1+\sqrt{2})^{n} +c_{2} (1-\sqrt{2})^{n}$ $\endgroup$ – user3257842 Nov 12 '19 at 7:43
  • 1
    $\begingroup$ For the odd series we have $t_{0} = 3$ and $t_{1} = 4$ giving $c_1 + c_2 = 3 $ and $c_1 (1+ \sqrt{2}) + c_2 (1- \sqrt{2}) = 3$ . This gives $c_1 = \frac{6 + \sqrt{2}}{4}$ and $c_2 = \frac{6 - \sqrt{2}}{4}$. For the even series we have $t_{1} = 2$ and $t_{2} = 7$, giving $c_1 (1+ \sqrt{2}) + c_2 (1- \sqrt{2}) = 2$ and $c_1 (1+ \sqrt{2})^{2} + c_2 (1- \sqrt{2})^{2} = 3$. This gives $c_1 = \frac{6 - \sqrt{2}}{4}$ and $c_2 = \frac{6 + \sqrt{2}}{4}$. These are the two formulas that describe the series. $\endgroup$ – user3257842 Nov 12 '19 at 7:49
2
$\begingroup$

You misunderstood the output of Wolfram. Perhaps the words "generating function" were in there somwhere? What it means is $$ {\frac {-3\,{x}^{3}+2\,{x}^{2}-2\,x-3}{{x}^{4}+2\,{x}^{2}-1}} =3+2\,x+ 4\,{x}^{2}+7\,{x}^{3}+11\,{x}^{4}+16\,{x}^{5}+26\,{x}^{6}+39\,{x}^{7}+ 63\,{x}^{8}+94\,{x}^{9}+152\,{x}^{10}+227\,{x}^{11}+367\,{x}^{12}+548 \,{x}^{13}+886\,{x}^{14}+1323\,{x}^{15}+2139\,{x}^{16}+3194\,{x}^{17}+ 5164\,{x}^{18}+7711\,{x}^{19}+\dots $$ when you expand as a power series. This is the same as the recursion in user's answer.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I am answering my own question only because there is more room here to explain my findings. Ross Millikan's solution for the $next$ pair yielded only every other term in the series but it got me thinking and.... oeis.org/A221172 had a series for every other number, i.e. $(3,4,11,26,63,152,367,886)$ and a formula:

$$\frac{(5\sqrt{2}((1+\sqrt{2})^n-(1-\sqrt{2})^n )-4((1+\sqrt{2})^n+(1-\sqrt{2})^n ))}{4}$$

that generates them but oeis did not find a series for the other set of every other term in the series. On a hunch, seeing that the series began with $-2$, I used $n=0$ and got $-2$; I used $n=-1$ and got $7$; I used $n=-2$ and got $-16$, etc.

$$\begin{array}{c|c|c|c|} series \# &n & out & desired \\ \hline 1&1 & 3 & 3\\ \hline 2&0 & -2 & 2\\ \hline 3&2 & 4 & 4\\ \hline 4&-1 & 7 &7 \\ \hline 5&3 & 11 & 11 \\ \hline 6&-2 & -16 & 16 \\ \hline 7&4 & 26 & 26 \\ \hline 8&-3 & 39 & 39\\ \hline 9&5 & 63 & 63\\ \hline 10&-4 & -94 & 94 \\ \hline \end{array}$$

Update: from http://oeis.org/A097140

$$n=\frac{3+(2(x+1)-3) (-1)^{x+1}}{4}$$

where $x$ is the series element number.

Update 2: I put user3257842's first equation to the test with negative $n$ as I did with the equation I found above.

$$a_{2n} = \frac{1}{4}((6 - \sqrt{2} )(1+\sqrt{2})^n + (6 + \sqrt{2} )(1-\sqrt{2})^n)$$

It does indeed generate all the values I need using sequence identified in http://oeis.org/A001057

$$n=\frac{1-(2(x-1)+1)(-1)^{x-1}}{4}$$

where $x$ is the series element number as in the first example above.

I will find both useful in my study of how such functions may be generalized.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ The formula in my answer also works and is more compact. We get $a_1 = 3 , a_2 = 2, a_3 = 4, a_4 = 7$, etc. $\endgroup$ – user3257842 Nov 9 '19 at 15:30

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.