Pell number should not be confused with Bell number.
In mathematics, the Pell numbers are an infinite sequence of integers, known since ancient times, that comprise the denominators of the closest rational approximations to the square root of 2. This sequence of approximations begins,,,, and, so the sequence of Pell numbers begins with 1, 2, 5, 12, and 29. The numerators of the same sequence of approximations are half the companion Pell numbers or Pell–Lucas numbers; these numbers form a second infinite sequence that begins with 2, 6, 14, 34, and 82.
Both the Pell numbers and the companion Pell numbers may be calculated by means of a recurrence relation similar to that for the Fibonacci numbers, and both sequences of numbers grow exponentially, proportionally to powers of the silver ratio 1 + . As well as being used to approximate the square root of two, Pell numbers can be used to find square triangular numbers, to construct integer approximations to the right isosceles triangle, and to solve certain combinatorial enumeration problems.[1]
As with Pell's equation, the name of the Pell numbers stems from Leonhard Euler's mistaken attribution of the equation and the numbers derived from it to John Pell. The Pell–Lucas numbers are also named after Édouard Lucas, who studied sequences defined by recurrences of this type; the Pell and companion Pell numbers are Lucas sequences.
The Pell numbers are defined by the recurrence relation:
Pn=\begin{cases}0&ifn=0;\\1&ifn=1;\\2Pn-1+Pn-2&otherwise.\end{cases}
In words, the sequence of Pell numbers starts with 0 and 1, and then each Pell number is the sum of twice the previous Pell number, plus the Pell number before that. The first few terms of the sequence are
0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, … .
Analogously to the Binet formula, the Pell numbers can also be expressed by the closed form formula
P | ||||
|
.
A third definition is possible, from the matrix formula
\begin{pmatrix}Pn+1&Pn\ Pn&Pn-1\end{pmatrix}=\begin{pmatrix}2&1\ 1&0\end{pmatrix}n.
Many identities can be derived or proven from these definitions; for instance an identity analogous to Cassini's identity for Fibonacci numbers,
Pn+1Pn-1
2 | |
-P | |
n |
=(-1)n,
Pell numbers arise historically and most notably in the rational approximation to . If two large integers x and y form a solution to the Pell equation
x2-2y2=\pm1,
11, | |||||||||
|
,
41 | |
29 |
,
99 | |
70 |
,...
Pn-1+Pn | |
Pn |
.
\sqrt2 ≈
577 | |
408 |
These approximations can be derived from the continued fraction expansion of
\sqrt2
\sqrt2=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}}}.
577 | |
408 |
=1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2}}}}}}}.
As Knuth (1994) describes, the fact that Pell numbers approximate allows them to be used for accurate rational approximations to a regular octagon with vertex coordinates and . All vertices are equally distant from the origin, and form nearly uniform angles around the origin. Alternatively, the points
(\pm(Pi+Pi-1),0)
(0,\pm(Pi+Pi-1))
(\pmPi,\pmPi)
A Pell prime is a Pell number that is prime. The first few Pell primes are
2, 5, 29, 5741, 33461, 44560482149, 1746860020068409, 68480406462161287469, ... .The indices of these primes within the sequence of all Pell numbers are
2, 3, 5, 11, 13, 29, 41, 53, 59, 89, 97, 101, 167, 181, 191, 523, 929, 1217, 1301, 1361, 2087, 2273, 2393, 8093, ... These indices are all themselves prime. As with the Fibonacci numbers, a Pell number Pn can only be prime if n itself is prime, because if d is a divisor of n then Pd is a divisor of Pn.
The only Pell numbers that are squares, cubes, or any higher power of an integer are 0, 1, and 169 = 132.[7]
However, despite having so few squares or other powers, Pell numbers have a close connection to square triangular numbers.[8] Specifically, these numbers arise from the following identity of Pell numbers:
l(\left(Pk-1+Pk\right) ⋅
2 | |
P | |
kr) |
=
| |||||||||||||||||||
2 |
.
Falcón and Díaz-Barrero (2006) proved another identity relating Pell numbers to squares and showing that the sum of the Pell numbers up to P4n +1 is always a square:
4n+1 | |
\sum | |
i=0 |
Pi=
n | |
\left(\sum | |
r=0 |
2r{2n+1\choose2r}\right)2=\left(P2n+P2n+1\right)2.
1, 7, 41, 239, 1393, 8119, 47321, …,are known as the Newman–Shanks–Williams (NSW) numbers.
If a right triangle has integer side lengths a, b, c (necessarily satisfying the Pythagorean theorem), then (a,b,c) is known as a Pythagorean triple. As Martin (1875) describes, the Pell numbers can be used to form Pythagorean triples in which a and b are one unit apart, corresponding to right triangles that are nearly isosceles. Each such triple has the form
\left(2PnPn+1,
2 | |
P | |
n+1 |
-
2, | |
P | |
n |
2 | |
P | |
n+1 |
+
2=P | |
P | |
2n+1 |
\right).
(4,3,5), (20,21,29), (120,119,169), (696,697,985), …
The companion Pell numbers or Pell–Lucas numbers are defined by the recurrence relation
Qn=\begin{cases}2&ifn=0;\\2&ifn=1;\\2Qn-1+Qn-2&otherwise.\end{cases}
In words: the first two numbers in the sequence are both 2, and each successive number is formed by adding twice the previous Pell–Lucas number to the Pell–Lucas number before that, or equivalently, by adding the next Pell number to the previous Pell number: thus, 82 is the companion to 29, and The first few terms of the sequence are : 2, 2, 6, 14, 34, 82, 198, 478, …
Like the relationship between Fibonacci numbers and Lucas numbers,
Q | ||||
|
The companion Pell numbers can be expressed by the closed form formula
Qn=\left(1+\sqrt2\right)n+\left(1-\sqrt2\right)n.
These numbers are all even; each such number is twice the numerator in one of the rational approximations to
\sqrt2
Like the Lucas sequence, if a Pell–Lucas number Qn is prime, it is necessary that n be either prime or a power of 2. The Pell–Lucas primes are
3, 7, 17, 41, 239, 577, … .
For these n are
2, 3, 4, 5, 7, 8, 16, 19, 29, 47, 59, 163, 257, 421, … .
The following table gives the first few powers of the silver ratio δ = δ S = 1 + and its conjugate = 1 − .
n | (1 +)n | (1 −)n | |
---|---|---|---|
0 | 1 + 0 = 1 | 1 − 0 = 1 | |
1 | 1 + 1 = 2.41421… | 1 − 1 = −0.41421… | |
2 | 3 + 2 = 5.82842… | 3 − 2 = 0.17157… | |
3 | 7 + 5 = 14.07106… | 7 − 5 = −0.07106… | |
4 | 17 + 12 = 33.97056… | 17 − 12 = 0.02943… | |
5 | 41 + 29 = 82.01219… | 41 − 29 = −0.01219… | |
6 | 99 + 70 = 197.9949… | 99 − 70 = 0.0050… | |
7 | 239 + 169 = 478.00209… | 239 − 169 = −0.00209… | |
8 | 577 + 408 = 1153.99913… | 577 − 408 = 0.00086… | |
9 | 1393 + 985 = 2786.00035… | 1393 − 985 = −0.00035… | |
10 | 3363 + 2378 = 6725.99985… | 3363 − 2378 = 0.00014… | |
11 | 8119 + 5741 = 16238.00006… | 8119 − 5741 = −0.00006… | |
12 | 19601 + 13860 = 39201.99997… | 19601 − 13860 = 0.00002… |
The coefficients are the half-companion Pell numbers Hn and the Pell numbers Pn which are the (non-negative) solutions to .A square triangular number is a number
N=
t(t+1) | |
2 |
=s2,
which is both the t-th triangular number and the s-th square number. A near-isosceles Pythagorean triple is an integer solution to where .
The next table shows that splitting the odd number Hn 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.
n | Hn | Pn | t | t + 1 | s | a | b | c | |
---|---|---|---|---|---|---|---|---|---|
0 | 1 | 0 | 0 | 1 | 0 | ||||
1 | 1 | 1 | 0 | 1 | 1 | ||||
2 | 3 | 2 | 1 | 2 | 1 | ||||
3 | 7 | 5 | 3 | 4 | 5 | ||||
4 | 17 | 12 | 8 | 9 | 6 | ||||
5 | 41 | 29 | 20 | 21 | 29 | ||||
6 | 99 | 70 | 49 | 50 | 35 | ||||
7 | 239 | 169 | 119 | 120 | 169 | ||||
8 | 577 | 408 | 288 | 289 | 204 | ||||
9 | 1393 | 985 | 696 | 697 | 985 | ||||
10 | 3363 | 2378 | 1681 | 1682 | 1189 | ||||
11 | 8119 | 5741 | 4059 | 4060 | 5741 | ||||
12 | 19601 | 13860 | 9800 | 9801 | 6930 |
The half-companion Pell numbers Hn and the Pell numbers Pn can be derived in a number of easily equivalent ways.
\left(1+\sqrt2\right)n=Hn+Pn\sqrt{2}
\left(1-\sqrt2\right)n=Hn-Pn\sqrt{2}.
From this it follows that there are closed forms:
Hn=
\left(1+\sqrt2\right)n+\left(1-\sqrt2\right)n | |
2 |
.
Pn\sqrt2=
\left(1+\sqrt2\right)n-\left(1-\sqrt2\right)n | |
2 |
.
Hn=\begin{cases}1&ifn=0;\\Hn-1+2Pn-1&otherwise.\end{cases}
Pn=\begin{cases}0&ifn=0;\\Hn-1+Pn-1&otherwise.\end{cases}
Let n be at least 2.
Hn=(3Pn-Pn-2)/2=3Pn-1+Pn-2;
Pn=(3Hn-Hn-2)/4=(3Hn-1+Hn-2)/2.
\begin{pmatrix}Hn\ Pn\end{pmatrix}=\begin{pmatrix}1&2\ 1&1\end{pmatrix}\begin{pmatrix}Hn-1\ Pn-1\end{pmatrix}=\begin{pmatrix}1&2\ 1&1\end{pmatrix}n\begin{pmatrix}1\ 0\end{pmatrix}.
So
\begin{pmatrix}Hn&2Pn\ Pn&Hn\end{pmatrix}=\begin{pmatrix}1&2\ 1&1\end{pmatrix}n.
The difference between Hn and Pn is
\left(1-\sqrt2\right)n ≈ (-0.41421)n,
which goes rapidly to zero. So
\left(1+\sqrt2\right)n=Hn+Pn\sqrt2
is extremely close to 2Hn.
From this last observation it follows that the integer ratios rapidly approach ; and and rapidly approach 1 + .
Since is irrational, we cannot have = , i.e.,
H2 | |
P2 |
=
2P2 | |
P2 |
.
The best we can achieve is either
H2 | |
P2 |
=
2P2-1 | |
P2 |
or
H2 | |
P2 |
=
2P2+1 | |
P2 |
.
The (non-negative) solutions to are exactly the pairs with n even, and the solutions to are exactly the pairs with n odd. To see this, note first that
2 | |
H | |
n+1 |
=\left(Hn+2P
2-2\left(H | |
n+P |
2 | |
n\right) |
=
2\right), | |
-\left(H | |
n |
so that these differences, starting with, are alternately 1 and −1. Then note that every positive solution comes in this way from a solution with smaller integers since
(2P-H)2-2(H-P)2=-\left(H2-2P2\right).
The smaller solution also has positive integers, with the one exception: which comes from H0 = 1 and P0 = 0.
See main article: Square triangular number. The required equation
t(t+1) | |
2 |
=s2
is equivalent to
4t2+4t+1=8s2+1,
tn=
H2n-1 | |
2 |
and sn=
P2n | |
2 |
.
Observe that t and t + 1 are relatively prime, so that = s 2 happens exactly when they are adjacent integers, one a square H  2 and the other twice a square 2P  2. Since we know all solutions of that equation, we also have
tn=\begin{cases}2P
2&ifnisodd.\end{cases} | |
n |
sn=HnPn.
This alternate expression is seen in the next table.
n | Hn | Pn | t | t + 1 | s | a | b | c | |
---|---|---|---|---|---|---|---|---|---|
0 | 1 | 0 | |||||||
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 |
The equality occurs exactly when which becomes with the substitutions and . Hence the n-th solution is and .
The table above shows that, in one order or the other, an and are and while .