Inmathematics, thePell numbers are an infinitesequence of integers, known since ancient times, that comprise thedenominators of theclosest rational approximations to thesquare root of 2. Thissequence of approximations begins⁠1/1⁠,⁠3/2⁠,⁠7/5⁠,⁠17/12⁠, and⁠41/29⁠, so the sequence of Pell numbers begins with 1, 2, 5, 12, and 29. The numerators of the same sequence of approximations are half thecompanion Pell numbers orPell–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 arecurrence relation similar to that for theFibonacci numbers, and both sequences of numbersgrow exponentially, proportionally to powers of thesilver ratio 1 + √2. As well as being used to approximate the square root of two, Pell numbers can be used to findsquare triangular numbers, to constructinteger approximations to theright isosceles triangle, and to solve certaincombinatorial enumeration problems.^[1]

As withPell's equation, the name of the Pell numbers stems fromLeonhard Euler's mistaken attribution of the equation and the numbers derived from it toJohn Pell. The Pell–Lucas numbers are also named afterÉdouard Lucas, who studied sequences defined by recurrences of this type; the Pell and companion Pell numbers areLucas sequences.

The Pell numbers are defined by therecurrence relation:

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, … (sequenceA000129 in theOEIS).

Analogously to theBinet formula, the Pell numbers can also be expressed by the closed form formula

${\displaystyle P_n=\frac{{\left(1+\sqrt{2}\right)}^n-{\left(1-\sqrt{2}\right)}^n}{2\sqrt{2}}.}$For large values ofn, the(1 +√2)ⁿ term dominates this expression, so the Pell numbers are approximately proportional to powers of thesilver ratio1 +√2, analogous to the growth rate of Fibonacci numbers as powers of thegolden ratio.

A third definition is possible, from thematrix formula

Manyidentities can be derived orproven from these definitions; for instance an identity analogous toCassini's identity for Fibonacci numbers,

${\displaystyle P_{n+1}{P}_{n-1}-P_n^2=(-1{)}^n,}$

is an immediate consequence of the matrix formula (found by considering thedeterminants of the matrices on the left and right sides of the matrix formula).^[2]

Pell numbers arise historically and most notably in therational approximation to√2. If two large integersx andy form a solution to thePell equation

${\displaystyle x^2-2y^2=\pm 1,}$

then their ratio⁠x/y⁠ provides a close approximation to√2. The sequence of approximations of this form is

${\displaystyle \frac{1}{1},\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29},\frac{99}{70},\dots }$

where the denominator of eachfraction is a Pell number and the numerator is the sum of a Pell number and its predecessor in the sequence. That is, the solutions have the form

${\displaystyle \frac{P_{n-1}+P_n}{P_n}.}$

The approximation

${\displaystyle \sqrt{2}\approx \frac{577}{408}}$

of this type was known to Indian mathematicians in the third or fourth century BCE.^[3] The Greek mathematicians of the fifth century BCE also knew of this sequence of approximations:^[4] Plato refers to the numerators asrational diameters.^[5] In the second century CETheon of Smyrna used the term theside and diameter numbers to describe the denominators and numerators of this sequence.^[6]

These approximations can be derived from thecontinued fraction expansion of${\displaystyle \sqrt{2}}$:

${\displaystyle \sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\ddots }}}}}.}$

Truncating this expansion to any number of terms produces one of the Pell-number-based approximations in this sequence; for instance,

${\displaystyle \frac{577}{408}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}}}}.}$

AsKnuth (1994) describes, the fact that Pell numbers approximate√2 allows them to be used for accurate rational approximations to a regular octagon with vertex coordinates(± P_i, ± P_i +1) and(± P_i +1, ± P_i ). All vertices are equally distant from theorigin, and form nearly uniformangles around the origin. Alternatively, the points${\displaystyle (\pm (P_i+P_{i-1}),0)}$,${\displaystyle (0,\pm (P_i+P_{i-1}))}$, and${\displaystyle (\pm P_i,\pm P_i)}$ form approximate octagons in which the vertices are nearly equally distant from the origin and form uniform angles.

APell prime is a Pell number that isprime. The first few Pell primes are

2, 5, 29, 5741, 33461, 44560482149, 1746860020068409, 68480406462161287469, ... (sequenceA086383 in theOEIS).

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, ... (sequenceA096650 in theOEIS)

These indices are all themselves prime. As with the Fibonacci numbers, a Pell numberP_n can only be prime ifn itself is prime, because ifd is adivisor ofn thenP_d is a divisor ofP_n.

The only Pell numbers that aresquares,cubes, or any higherpower 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 tosquare triangular numbers.^[8] Specifically, these numbers arise from the following identity of Pell numbers:

${\displaystyle (\left(P_{k-1}+P_k\right)\cdot P_k{)}^2=\frac{{\left(P_{k-1}+P_k\right)}^2\cdot \left({\left(P_{k-1}+P_k\right)}^2-(-1{)}^k\right)}{2}.}$

The left side of this identity describes a square number, while the right side describes atriangular number, so the result is a square triangular number.

Falcón and Díaz-Barrero (2006) proved another identity relating Pell numbers to squares and showing that the sum of the Pell numbers up toP_4n +1 is always a square:

${\displaystyle \sum _{i=0}^{4n+1}{P}_i={\left(\sum _{r=0}^n{2}^r(\genfrac{}{}{0ex}{}{2n+1}{2r})\right)}^2={\left(P_{2n}+P_{2n+1}\right)}^2.}$

For instance, the sum of the Pell numbers up toP₅,0 + 1 + 2 + 5 + 12 + 29 = 49, is the square ofP₂ +P₃ = 2 + 5 = 7. The numbersP_2n +P_2n +1 forming the square roots of these sums,

1, 7, 41, 239, 1393, 8119, 47321, … (sequenceA002315 in theOEIS),

are known as theNewman–Shanks–Williams (NSW) numbers.

If aright triangle has integer side lengthsa,b,c (necessarily satisfying thePythagorean theorema² +b² =c²), then (a,b,c) is known as aPythagorean triple. As Martin (1875) describes, the Pell numbers can be used to form Pythagorean triples in whicha andb are one unit apart, corresponding to right triangles that are nearly isosceles. Each such triple has the form

${\displaystyle \left(2P_n{P}_{n+1},P_{n+1}^2-P_n^2,P_{n+1}^2+P_n^2=P_{2n+1}\right).}$

The sequence of Pythagorean triples formed in this way is

(4,3,5), (20,21,29), (120,119,169), (696,697,985), …

Thecompanion Pell numbers orPell–Lucas numbers are defined by therecurrence relation

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, and82 = 2 × 34 + 14 = 70 + 12. The first few terms of the sequence are (sequenceA002203 in theOEIS):2, 2,6,14,34,82, 198,478, …

Like the relationship betweenFibonacci numbers andLucas numbers,

${\displaystyle Q_n=\frac{P_{2n}}{P_n}}$

for allnatural numbersn.

The companion Pell numbers can be expressed by the closed form formula

${\displaystyle Q_n={\left(1+\sqrt{2}\right)}^n+{\left(1-\sqrt{2}\right)}^n.}$

These numbers are alleven; each such number is twice the numerator in one of the rational approximations to${\displaystyle \sqrt{2}}$ discussed above.

Like the Lucas sequence, if a Pell–Lucas number⁠1/2⁠Q_n is prime, it is necessary thatn be either prime or apower of 2. The Pell–Lucas primes are

3, 7, 17, 41, 239, 577, … (sequenceA086395 in theOEIS).

For thesen are

2, 3, 4, 5, 7, 8, 16, 19, 29, 47, 59, 163, 257, 421, … (sequenceA099088 in theOEIS).

The following table gives the first few powers of thesilver ratioδ =δ_S = 1 + √2 and itsconjugateδ = 1 − √2.

n	(1 +√2)n	(1 −√2)n
0	1 + 0√2 = 1	1 − 0√2 = 1
1	1 + 1√2 = 2.41421…	1 − 1√2 = −0.41421…
2	3 + 2√2 = 5.82842…	3 − 2√2 = 0.17157…
3	7 + 5√2 = 14.07106…	7 − 5√2 = −0.07106…
4	17 + 12√2 = 33.97056…	17 − 12√2 = 0.02943…
5	41 + 29√2 = 82.01219…	41 − 29√2 = −0.01219…
6	99 + 70√2 = 197.9949…	99 − 70√2 = 0.0050…
7	239 + 169√2 = 478.00209…	239 − 169√2 = −0.00209…
8	577 + 408√2 = 1153.99913…	577 − 408√2 = 0.00086…
9	1393 + 985√2 = 2786.00035…	1393 − 985√2 = −0.00035…
10	3363 + 2378√2 = 6725.99985…	3363 − 2378√2 = 0.00014…
11	8119 + 5741√2 = 16238.00006…	8119 − 5741√2 = −0.00006…
12	19601 + 13860√2 = 39201.99997…	19601 − 13860√2 = 0.00002…

Thecoefficients are the half-companion Pell numbersH_n and the Pell numbersP_n which are the (non-negative) solutions toH² − 2P² = ±1.Asquare triangular number is a number

${\displaystyle N=\frac{t(t+1)}{2}=s^2,}$

which is both thet-th triangular number and thes-th square number. Anear-isosceles Pythagorean triple is an integer solution toa² +b² =c² wherea + 1 =b.

The next table shows that splitting theodd numberH_n into nearly equal halves gives a square triangular number whenn is even and a near isosceles Pythagorean triple whenn 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 numbersH_n and the Pell numbersP_n can be derived in a number of easily equivalent ways.

${\displaystyle {\left(1+\sqrt{2}\right)}^n=H_n+P_n\sqrt{2}}$

${\displaystyle {\left(1-\sqrt{2}\right)}^n=H_n-P_n\sqrt{2}.}$

From this it follows that there areclosed forms:

${\displaystyle H_n=\frac{{\left(1+\sqrt{2}\right)}^n+{\left(1-\sqrt{2}\right)}^n}{2}.}$

and

${\displaystyle P_n\sqrt{2}=\frac{{\left(1+\sqrt{2}\right)}^n-{\left(1-\sqrt{2}\right)}^n}{2}.}$

Letn be at least 2.

${\displaystyle H_n=(3P_n-P_{n-2})/2=3P_{n-1}+P_{n-2};}$

${\displaystyle P_n=(3H_n-H_{n-2})/4=(3H_{n-1}+H_{n-2})/2.}$

So

The difference betweenH_n andP_n√2 is

${\displaystyle {\left(1-\sqrt{2}\right)}^n\approx (-0.41421{)}^n,}$

which goes rapidly to zero. So

${\displaystyle {\left(1+\sqrt{2}\right)}^n=H_n+P_n\sqrt{2}}$

is extremely close to 2H_n.

From this last observation it follows that the integer ratios⁠H_n/P_n⁠ rapidly approach√2; and⁠H_n/H_n −1⁠ and⁠P_n/P_n −1⁠ rapidly approach 1 + √2.

Since√2 is irrational, we cannot have⁠H/P⁠ = √2, i.e.,

${\displaystyle \frac{H^2}{P^2}=\frac{2P^2}{P^2}.}$

The best we can achieve is either

${\displaystyle \frac{H^2}{P^2}=\frac{2P^2-1}{P^2}\text{or}\frac{H^2}{P^2}=\frac{2P^2+1}{P^2}.}$

The (non-negative) solutions toH² − 2P² = 1 are exactly the pairs(H_n,P_n) withn even, and the solutions toH² − 2P² = −1 are exactly the pairs(H_n,P_n) withn odd. To see this, note first that

${\displaystyle H_{n+1}^2-2P_{n+1}^2={\left(H_n+2P_n\right)}^2-2{\left(H_n+P_n\right)}^2=-\left(H_n^2-2P_n^2\right),}$

so that these differences, starting withH²
₀ − 2P²
₀ = 1, are alternately 1 and −1. Then note that every positive solution comes in this way from a solution with smaller integers since

${\displaystyle (2P-H{)}^2-2(H-P{)}^2=-\left(H^2-2P^2\right).}$

The smaller solution also has positive integers, with the one exception:H =P = 1 which comes fromH₀ = 1 andP₀ = 0.

The required equation

${\displaystyle \frac{t(t+1)}{2}=s^2}$

is equivalent to${\displaystyle 4t^2+4t+1=8s^2+1,}$which becomesH² = 2P² + 1 with the substitutionsH = 2t + 1 andP = 2s. Hence then-th solution is

${\displaystyle t_n=\frac{H_{2n}-1}{2}\text{and}{s}_n=\frac{P_{2n}}{2}.}$

Observe thatt andt + 1 are relatively prime, so that⁠t (t + 1)/2⁠ = s² happens exactly when they are adjacent integers, one a squareH² and the other twice a square 2P². Since we know all solutions of that equation, we also have

and${\displaystyle s_n=H_n{P}_n.}$

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 equalityc² =a² + (a + 1)² = 2a² + 2a + 1 occurs exactly when2c² = 4a² + 4a + 2 which becomes2P² =H² + 1 with the substitutionsH = 2a + 1 andP =c. Hence then-th solution isa_n =⁠H_2n +1 − 1/2⁠ andc_n =P_2n +1.

The table above shows that, in one order or the other,a_n andb_n =a_n + 1 areH_n H_n +1 and2P_n P_n +1 whilec_n =H_n +1 P_n +P_n +1 H_n.

• Weisstein, Eric W."Pell Number".MathWorld.
• OEISsequence A001333 (Numerators of continued fraction convergents to sqrt(2))—The numerators of the same sequence of approximations

• Integer sequences
• Recurrence relations
• Unsolved problems in mathematics

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