Seeking series formula for: $3, 2, 4, 7, 11, 16, 26, 39, 63, 94, 152, 227, 367, 548, 886, 1323, 2139, 3194$

Question

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.

Answer 1

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.

Answer 2

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:

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

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

Answer 3

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.

Answer 4

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.