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I have a series I can generate sequentially if I pair elements with Pell numbers. If I pair them with $1,2,5,12,29,...$ (Pell numbers, $p_n$) and call elements of this series $q_n$, then $q_n=q_{n-1}+p_{n-1}$ where $q_0=1$.

$$1\quad 2\quad 4\quad 9\quad 21\quad 50\quad 120\quad 289\quad 697\quad 1682\quad 4060\quad 9801\quad 23661\quad 57122$$

I would like to generate any $q_n$ directly. I can get $p_n$ directly but so far my formula depends on knowing $q_{n-1}$ beforehand. Can the $n^{th} q$ be generated directly?

poetasis
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    Yes, $q_n=\dfrac{2+(1+\sqrt2)^{n+1}+(1-\sqrt2)^{n+1}}4$ – J. W. Tanner Oct 24 '19 at 23:46
  • @J. W. Tanner Your formula works. How did you find it? – poetasis Oct 25 '19 at 11:34
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    I will try to answer in a few hours. Are you interested in how I found the formula or mathematical justification of it? In your definition of Pell numbers, is $p_0=1$? (Usually $p_0=0$, but I can work with your definition) – J. W. Tanner Oct 25 '19 at 12:56
  • @J. W. Tanner Your answer let me generate the pairs: $$1,1\quad 2,2\quad 4,5\quad 9,12\quad 21,29\quad 50,70\quad 120,169\quad 289,408\quad 697,985\quad 1682,2378\quad 4060,5741\quad 9801,13860\quad 23661,33461\quad 57122,80782\quad 137904,195025\quad 137904,195025\quad 332929,470832\quad $$ for input to a formula I developed: $$A=(2n-1)^2+2(2n-2)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+k^2$$ to make it generate Pythagorean triples where $B-A=\pm1$ – poetasis Oct 25 '19 at 19:42
  • There is some discussion of Pythagorean triples and Pell numbers on the Wikipedia page – J. W. Tanner Oct 25 '19 at 19:49
  • I'm familiar with the page. I was referred to it and found an answer to the same sort of question for Euclid's formula in which all the inputs to find $B-A=\pm1$ are Pell numbers. – poetasis Oct 25 '19 at 19:55

2 Answers2

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One way to find a formula is to convert the system of difference equations into first order difference equations - the Pell numbers $p_n=2p_{n-1}+p_{n-2}$ is a second order difference equation. To do this, we introduce a dummy sequence $$a_n=p_{n-1}$$ for $n\ge1$; this lets us write $$\begin{align*} a_n&=p_{n-1}\\ p_n&=2p_{n-1}+a_{n-1}\\ q_n&=q_{n-1}+p_{n-1}. \end{align*}$$ Now the reason we like this first order system is that we can define $$x_n=\begin{pmatrix}a_n\\p_n\\q_n\end{pmatrix}$$ so that for $n\ge 2$ $$x_n=\begin{pmatrix}p_{n-1}\\2p_{n-1}+a_{n-1}\\q_{n-1}+p_{n-1}\end{pmatrix}=\begin{pmatrix}0&1&0\\1&2&0\\0&1&1\end{pmatrix}x_{n-1}.$$ Letting $A=\begin{pmatrix}0&1&0\\1&2&0\\0&1&1\end{pmatrix}$, we get $$x_n=Ax_{n-1}=A^2x_{n-2}=\cdots=A^{n-1}x_1.$$ Now all that remains is to compute $A^{n-1}$ by writing as Jordan Normal Form, i.e. finding the eigenvalues and eigenvectors of $A$.

Heath Winning
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  • I'm not familiar with Jordan Normal Form or eigenvalues. Your final equation also seems to make reference to a prior value of $x$ which is what I do not need. – poetasis Oct 25 '19 at 11:29
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The Pell numbers (OEIS A000129 $ 0, 1, 2, 5, 12, 29, 70, ...)$ are usually given by

$P_0=0, P_1=1, P_n=2P_{n-1}+P_{n-2}=\dfrac{(1+\sqrt2)^n-(1-\sqrt2)^n}{2\sqrt2}$.

The Pell-Lucas numbers (OEIS A002203 $2,2,6,14,34,82,198,...)$ are usually given by

$Q_0=2, Q_1=2, Q_n=2Q_{n-1}+Q_{n-2}=(1+\sqrt2)^n+(1-\sqrt2)^n.$

Your $q_n$ (OEIS A024537 $1, 2, 4, 9, 21, 50,...)$ is $\dfrac{Q_{n+1}+2}4.$

J. W. Tanner
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  • I know how to generate pell numbers. I never paid attention to Pell-Lucas numbers. Your final equation involving $Q$ to obtain $q$ is the same as the comment above and it works, so you get the prize. Thank you. – poetasis Oct 25 '19 at 18:02
  • Did you realize this answer was given by the same J.W. Tanner as above? Fractions with half-Pell-Lucas numbers as numerators and Pell numbers as denominators approach $\sqrt2: \frac11, \frac32, \frac75, \frac{17}{12}, \frac{41}{29}, \frac{99}{70}, ...$ – J. W. Tanner Oct 25 '19 at 18:13