Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Pell number
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
== Computations and connections == The following table gives the first few powers of the [[silver ratio]] ''Ξ΄'' = ''Ξ΄''<sub>S</sub> = 1 + {{sqrt|2}} and its [[conjugate (square roots)|conjugate]] {{overline|''Ξ΄''}} = 1 β {{sqrt|2}}. :{| class="wikitable" style="text-align:center" |- ! ''n'' ! (1 + {{sqrt|2}})<sup>''n''</sup> ! (1 β {{sqrt|2}})<sup>''n''</sup> |- ! 0 | 1 + 0{{sqrt|2}} = 1 | 1 β 0{{sqrt|2}} = 1 |- ! 1 | 1 + 1{{sqrt|2}} = 2.41421β¦ | 1 β 1{{sqrt|2}} = β0.41421β¦ |- ! 2 | 3 + 2{{sqrt|2}} = 5.82842β¦ | 3 β 2{{sqrt|2}} = 0.17157β¦ |- ! 3 | 7 + 5{{sqrt|2}} = 14.07106β¦ | 7 β 5{{sqrt|2}} = β0.07106β¦ |- ! 4 | 17 + 12{{sqrt|2}} = 33.97056β¦ | 17 β 12{{sqrt|2}} = 0.02943β¦ |- ! 5 | 41 + 29{{sqrt|2}} = 82.01219β¦ | 41 β 29{{sqrt|2}} = β0.01219β¦ |- ! 6 | 99 + 70{{sqrt|2}} = 197.9949β¦ | 99 β 70{{sqrt|2}} = 0.0050β¦ |- ! 7 | 239 + 169{{sqrt|2}} = 478.00209β¦ | 239 β 169{{sqrt|2}} = β0.00209β¦ |- ! 8 | 577 + 408{{sqrt|2}} = 1153.99913β¦ | 577 β 408{{sqrt|2}} = 0.00086β¦ |- ! 9 | 1393 + 985{{sqrt|2}} = 2786.00035β¦ | 1393 β 985{{sqrt|2}} = β0.00035β¦ |- ! 10 | 3363 + 2378{{sqrt|2}} = 6725.99985β¦ | 3363 β 2378{{sqrt|2}} = 0.00014β¦ |- ! 11 | 8119 + 5741{{sqrt|2}} = 16238.00006β¦ | 8119 β 5741{{sqrt|2}} = β0.00006β¦ |- ! 12 | 19601 + 13860{{sqrt|2}} = 39201.99997β¦ | 19601 β 13860{{sqrt|2}} = 0.00002β¦ |} The [[coefficient]]s are the half-companion Pell numbers ''H<sub>n</sub>'' and the Pell numbers ''P<sub>n</sub>'' which are the (non-negative) solutions to {{nowrap|1=''H''<sup>2</sup> β 2''P''<sup>2</sup> = Β±1}}. A [[square triangular number]] is a number :<math>N = \frac{t(t+1)}{2} = s^2,</math> which is both the ''t''-th triangular number and the ''s''-th square number. A ''near-isosceles Pythagorean triple'' is an integer solution to {{nowrap|1=''a''<sup>2</sup> + ''b''<sup>2</sup> = ''c''<sup>2</sup>}} where {{nowrap|1=''a'' + 1 = ''b''}}. The next table shows that splitting the [[parity (mathematics)|odd]] number ''H<sub>n</sub>'' into nearly equal halves gives a square triangular number when ''n'' is even and a near isosceles Pythagorean triple when ''n'' is odd. All solutions arise in this manner. :{| class="wikitable" style="text-align:center" |- !''n'' !''H<sub>n</sub>'' !''P<sub>n</sub>'' !''t'' !''t'' + 1 !''s'' !''a'' !''b'' !''c'' |- !0 |1 |0 |0 |1 |0 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !1 |1 |1 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |0 |1 |1 |- !2 |3 |2 |1 |2 |1 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !3 |7 |5 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |3 |4 |5 |- !4 |17 |12 |8 |9 |6 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !5 |41 |29 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |20 |21 |29 |- !6 |99 |70 |49 |50 |35 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !7 |239 |169 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |119 |120 |169 |- !8 |577 |408 |288 |289 |204 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !9 |1393 |985 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |696 |697 |985 |- !10 |3363 |2378 |1681 |1682 |1189 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !11 |8119 |5741 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |4059 |4060 |5741 |- !12 |19601 |13860 |9800 |9801 |6930 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |} === Definitions === The half-companion Pell numbers ''H<sub>n</sub>'' and the Pell numbers ''P<sub>n</sub>'' can be derived in a number of easily equivalent ways. ==== Raising to powers ==== :<math>\left(1+\sqrt2\right)^n = H_n+P_n\sqrt{2}</math> :<math>\left(1-\sqrt2\right)^n = H_n-P_n\sqrt{2}.</math> From this it follows that there are ''closed forms'': :<math>H_n = \frac{\left(1+\sqrt2\right)^n+\left(1-\sqrt2\right)^n}{2}.</math> and :<math>P_n\sqrt2 = \frac{\left(1+\sqrt2\right)^n-\left(1-\sqrt2\right)^n}{2}.</math> ==== Paired recurrences ==== :<math>H_n = \begin{cases}1&\mbox{if }n=0;\\H_{n-1}+2P_{n-1}&\mbox{otherwise.}\end{cases}</math> :<math>P_n = \begin{cases}0&\mbox{if }n=0;\\H_{n-1}+P_{n-1}&\mbox{otherwise.}\end{cases}</math> ==== Reciprocal recurrence formulas ==== Let ''n'' be at least 2. :<math>H_n = (3P_n-P_{n-2})/2 = 3P_{n-1}+P_{n-2};</math> :<math>P_n = (3H_n-H_{n-2})/4 = (3H_{n-1}+H_{n-2})/2.</math> ==== Matrix formulations ==== :<math>\begin{pmatrix} H_n \\ P_n \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} H_{n-1} \\ P_{n-1} \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}.</math> So :<math>\begin{pmatrix} H_n & 2P_n \\ P_n & H_n \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}^n .</math> === Approximations === The difference between ''H<sub>n</sub>'' and ''P<sub>n</sub>''{{sqrt|2}} is :<math>\left(1-\sqrt2\right)^n \approx (-0.41421)^n,</math> which goes rapidly to zero. So :<math>\left(1+\sqrt2\right)^n = H_n+P_n\sqrt2</math> is extremely close to 2''H<sub>n</sub>''. From this last observation it follows that the integer ratios ''{{sfrac|H<sub>n</sub>|P<sub>n</sub>}}'' rapidly approach {{sqrt|2}}; and {{sfrac|''H<sub>n</sub>''|''H''<sub>''n''β1</sub>}} and {{sfrac|''P<sub>n</sub>''|''P''<sub>''n''β1</sub>}} rapidly approach 1 + {{sqrt|2}}. === ''H''<sup>2</sup> β 2''P''<sup>2</sup> = Β±1 === Since {{sqrt|2}} is irrational, we cannot have ''{{sfrac|H|P}}'' = {{sqrt|2}}, i.e., :<math>\frac{H^2}{P^2} = \frac{2P^2}{P^2}.</math> The best we can achieve is either :<math>\frac{H^2}{P^2} = \frac{2P^2-1}{P^2}\quad \mbox{or} \quad \frac{H^2}{P^2} = \frac{2P^2+1}{P^2}.</math> The (non-negative) solutions to {{nowrap|1=''H''<sup>2</sup> β 2''P''<sup>2</sup> = 1}} are exactly the pairs {{nowrap|(''H<sub>n</sub>'', ''P<sub>n</sub>'')}} with ''n'' even, and the solutions to {{nowrap|1=''H''<sup>2</sup> β 2''P''<sup>2</sup> = β1}} are exactly the pairs {{nowrap|(''H<sub>n</sub>'', ''P<sub>n</sub>'')}} with ''n'' odd. To see this, note first that :<math>H_{n+1}^2-2P_{n+1}^2 = \left(H_n+2P_n\right)^2-2\left(H_n+P_n\right)^2 = -\left(H_n^2-2P_n^2\right),</math> so that these differences, starting with {{nowrap|1=''H''{{su|b=0|p=2}} β 2''P''{{su|b=0|p=2}} = 1}}, are alternately 1 and β1. Then note that every positive solution comes in this way from a solution with smaller integers since :<math>(2P-H)^2-2(H-P)^2 = -\left(H^2-2P^2\right).</math> The smaller solution also has positive integers, with the one exception: {{nowrap|1=''H'' = ''P'' = 1}} which comes from ''H''<sub>0</sub> = 1 and ''P''<sub>0</sub> = 0. === Square triangular numbers === {{main|Square triangular number}} The required equation :<math>\frac{t(t+1)}{2}=s^2</math> is equivalent to <math>4t^2+4t+1 = 8s^2+1,</math> which becomes {{nowrap|1=''H''<sup>2</sup> = 2''P''<sup>2</sup> + 1}} with the substitutions ''H'' = 2''t'' + 1 and ''P'' = 2''s''. Hence the ''n''-th solution is :<math>t_n = \frac{H_{2n}-1}{2} \quad\mbox{and}\quad s_n = \frac{P_{2n}}{2}.</math> Observe that ''t'' and ''t'' + 1 are relatively prime, so that {{sfrac|''t''(''t'' + 1)|2}} = ''s''<sup>2</sup> happens exactly when they are adjacent integers, one a square ''H''<sup>2</sup> and the other twice a square 2''P''<sup>2</sup>. Since we know all solutions of that equation, we also have :<math>t_n=\begin{cases}2P_n^2&\mbox{if }n\mbox{ is even};\\H_{n}^2&\mbox{if }n\mbox{ is odd.}\end{cases}</math> and <math>s_n=H_nP_n.</math> This alternate expression is seen in the next table. :{| class="wikitable" style="text-align:center" |- !''n'' !''H<sub>n</sub>'' !''P<sub>n</sub>'' !''t'' !''t'' + 1 !''s'' !''a'' !''b'' !''c'' |- !0 |1 |0 |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |style="background: grey;"| |- !1 |1 |1 |1 |2 |1 |3 |4 |5 |- !2 |3 |2 |8 |9 |6 |20 |21 |29 |- !3 |7 |5 |49 |50 |35 |119 |120 |169 |- !4 |17 |12 |288 |289 |204 |696 |697 |985 |- !5 |41 |29 |1681 |1682 |1189 |4059 |4060 |5741 |- !6 |99 |70 |9800 |9801 |6930 |23660 |23661 |33461 |} === Pythagorean triples === The equality {{nowrap|1=''c''<sup>2</sup> = ''a''<sup>2</sup> + (''a'' + 1)<sup>2</sup> = 2''a''<sup>2</sup> + 2''a'' + 1}} occurs exactly when {{nowrap|1=2''c''<sup>2</sup> = 4''a''<sup>2</sup> + 4''a'' + 2}} which becomes {{nowrap|1=2''P''<sup>2</sup> = ''H''<sup>2</sup> + 1}} with the substitutions {{nowrap|1=''H'' = 2''a'' + 1}} and {{nowrap|1=''P'' = ''c''}}. Hence the ''n''-th solution is {{nowrap|1=''a<sub>n</sub>'' = {{sfrac|''H''<sub>2''n''+1</sub> β 1|2}}}} and {{nowrap|1=''c<sub>n</sub>'' = ''P''<sub>2''n''+1</sub>}}. The table above shows that, in one order or the other, ''a<sub>n</sub>'' and {{nowrap|1=''b<sub>n</sub>'' = ''a<sub>n</sub>'' + 1}} are {{nowrap|1=''H<sub>n</sub>H''<sub>''n''+1</sub>}} and {{nowrap|1=2''P<sub>n</sub>P''<sub>''n''+1</sub>}} while {{nowrap|1=''c<sub>n</sub>'' = ''H''<sub>''n''+1</sub>''P<sub>n</sub>'' + ''P''<sub>''n''+1</sub>''H<sub>n</sub>''}}.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)