Generalized Pell Numbers Tony Reix ([email protected]) 2005, 27th of March - v0.4 This paper presents a generalization of Pell numbers. With no proof.

1 1.1

General Properties Definitions

Let define: Gn [m, r] =

X

0≤k≤n k ≡ r (mod m)

  n−r   m X  n  k−r n mi m m = mi + r k i=0

We have the recursion: m−1 X m Gn [m, r] = (−1)i+1 Gn−i [m, r] + (m + (−1)m+1 )Gn−m [m, r] i i=1

and the limit:

√ Gn+1 [m, r] =1+ mm n→∞ Gn [m, r] lim

with: 0 ≤ n ≤ m , 0 ≤ r < m , Gn [m, r] =

  n , except: : Gm [m, 0] = m + 1 r

Let define the following polynomial: m−1 X m (−1)i X m−i − (m + (−1)m+1 ) = (X − 1)m − m Pm (X) = i i=0

Its discriminant is: (−1)(m−(m mod 2))/2 m2m−1 and: 1 +

1.2

√

m

m is a root.

Gn [m, 0] modulo a prime

With: M = 2m , Gn [M, 0] mod (M + 1) gives interesting results: For m = 2, 4, 8 , but NOT for m = 3, 5, 6, 7, 9 , we have: Gn [M, 0] mod (M + 1) = 1 for all n except for n = M, 2M, 2M + 1, 3M, 3M + 1, 3M + 2, ..., kM + (0..k − 1) . (Hereafter we note: Gn [m, 0] = Gn [m] .) G1×4+0 [4] mod 5 = G1×16+0 [16] mod 17 = G1×256+0 [256] mod 257 = 0 G2×4+0 [4] mod 5 = G2×16+0 [16] mod 17 = G2×256+0 [256] mod 257 = 2 G2×4+1 [4] mod 5 = G2×16+1 [16] mod 17 = G2×256+1 [256] mod 257 = −1 G3×4+0 [4] mod 5 = G3×16+0 [16] mod 17 = G3×256+0 [256] mod 257 = 0 G3×4+1 [4] mod 5 = G3×16+1 [16] mod 17 = G3×256+1 [256] mod 257 = 5 1

G3×4+2 [4] mod 5 = G3×16+2 [16] mod 17 = G3×256+2 [256] mod 257 = −3 G4×4+0 [4] mod 5 = G4×16+0 [16] mod 17 = G4×256+0 [256] mod 257 = 2 G4×4+1 [4] mod 5 = G4×16+1 [16] mod 17 = G4×256+1 [256] mod 257 = −5 G4×4+2 [4] mod 5 = G4×16+2 [16] mod 17 = G4×256+2 [256] mod 257 = 13 G4×4+3 [4] mod 5 = G4×16+3 [16] mod 17 = G4×256+3 [256] mod 257 = −7 The same properties seem to appear when M is any number such that M +1 is prime. Checked for M = 6, 10, 12, 18, 22.

2 2.1

Examples m=1

With m = 1 and r = 0, Gn [1, 0] gives the powers of 2: 2Gn [1, 0] = P Gn−1 [1, 0] = 2n , with: P = 2 The polynomial is: P(x) = x − P = x − 2 = (x√− 1)1 − 1 , which discriminant is: D = 1 = 12×1−1 and has 1 root: α = 1 + 1 1 = 2. And we have: α = P .

2.2

m=2

With m = 2 and r = 0 or 1, Gn [2, r] gives the Pell numbers: (P, Q) = (2, −1) Gn [2, r] = P Gn−1 [2, r] − QGn−2 [2, r] = 2Gn−1 [2, r] + Gn−2 [2, r] 2Gn [2, 0] = Vn (P, Q) = P Vn (P, Q) − QVn (P, Q) Gn [2, 1] = Un (P, Q) = P Un (P, Q) − QUn (P, Q)

The polynomial is: P(x) = x2 − P x + Q = x2 − 2x − 1 = (x − 1)2 − 2 , which √ 2 2 discriminant √ is: D = P 2 − 4Q = 8 = 22×2−1 and has 2 roots:√α = 1 + √ 2 and β = 1 − 2. And we have: α + β = P , αβ = Q, α − β = D = 2 2. Un (2, −1) = (αn − β n )/(α − β) and Vn (2, −1) = αn + β n .

2.3

m=3

With m = 3 and r = 0, 1 or 2, Gn [3, r] gives: (P, Q, R) = (3, 3, 4) Gn [3, r] = P Gn−1 [3, r] − QGn−2 [3, r] + RGn−3 [3, r] Gn [3, r] = 3Gn−1 [3, r] − 3Gn−2 [3, r] + 4Gn−3 [3, r]

The polynomial is: P(x) = x3 −P x2 +Qx−R = x3 −3x2 +3x−4 = (x−1)3 −3 , which discriminant is: D = P 2 Q2 + 18P QR − 4Q3 − 4P 3 R − 27R2 = √ √ √ 6 3 3 2×3−1 and has 3 roots: α = 1 + 3 , β = 1 − ( 3 − i 35 )/2 , and −243 = −3 √ √ 6 3 γ = 1 − ( 3 + i 35 )/2 . And we have: α + β + γ = P , αβ + βγ + γα = Q, and αβγ = R. Let define: An = αn + β n + γ n . We have: 3Gn [3, 0] = An .

2

m 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8

r 0 0 1 0 1 2 0 1 2 3 0 1 2 3 4 0 1 2 3 4 5 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7

0 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0

1 2 1 1 1 1 0 1 1 0 0 1 1 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0

2 4 3 2 1 2 1 1 2 1 0 1 2 1 0 0 1 2 1 0 0 0 1 2 1 0 0 0 0 1 2 1 0 0 0 0 0

3 8 7 5 4 3 3 1 3 3 1 1 3 3 1 0 1 3 3 1 0 0 1 3 3 1 0 0 0 1 3 3 1 0 0 0 0

4 16 17 12 13 7 6 5 4 6 4 1 4 6 4 1 1 4 6 4 1 0 1 4 6 4 1 0 0 1 4 6 4 1 0 0 0

5 32 41 29 31 20 13 21 9 10 10 6 5 10 10 5 1 5 10 10 5 1 1 5 10 10 5 1 0 1 5 10 10 5 1 0 0

6 64 99 70 70 51 33 61 30 19 20 31 11 15 20 15 7 6 15 20 15 6 1 6 15 20 15 6 1 1 6 15 20 15 6 1 0

7 128 239 169 169 121 84 141 91 49 39 106 42 26 35 35 43 13 21 35 35 21 8 7 21 35 35 21 7 1 7 21 35 35 21 7 1

8 256 577 408 421 290 205 297 232 140 88 281 148 68 61 70 169 56 34 56 70 56 57 15 28 56 70 56 28 9 8 28 56 70 56 28 8

9 512 1393 985 1036 711 495 649 529 372 228 631 429 216 129 131 505 225 90 90 126 126 253 72 43 84 126 126 84 73 17 36 84 126 126 84 36

10 1024 3363 2378 2521 1747 1206 1561 1178 901 600 1286 1060 645 345 260 1261 730 315 180 216 252 841 325 115 127 210 252 210 361 90 53 120 210 252 210 120

11 2048 8119 5741 6139 4268 2953 3961 2739 2079 1501 2586 2346 1705 990 605 2773 1991 1045 495 396 468 2311 1166 440 242 337 462 462 1321 451 143 173 330 462 462 330

Table 1: First values of Gn [m, r], n = 1..12, r = 0..m − 1, m = 1..8

3

12 4096 19601 13860 14998 10407 7221 9965 6700 4818 3580 5611 4932 4051 2695 1595 5581 4764 3036 1540 891 864 5545 3477 1606 682 579 799 924 3961 1772 594 316 503 792 924 792

Values of Gn [m, r] for m = 1..8

2.4

Table 1 provides the first 12 values for m = 1 to m.

3

A proof of the recursion by Zhi-Wei SUN

As the characteristic function of k = r (mod m) equals: m−1

m−1 X

e2πij(k−r)/m ,

j=0

we have: Gn [m, r] = m−1

m−1 X

m−r/m e−2πijr/m (1 + m1/m e2πij/m )n .

j=0

Thus Gn [m, r] is a recurrent sequence (with respect to n) of order m with the characteristic polynomial: P (x) =

m−1 Y j=0

(x − 1 − m1/m e2πij/m ) = (x − 1)m − m .

So Gn [m, r] satisfies the recursion you described. A more detailed explanation: If P (x) = (x − α1 )...(x − αm ) = xm − a1 xm−1 − ... − am , then un =

m X j=1

satisfies the recursion un =

m X i=1

cj ∗ αjn

ai un−i (for n ≥ m).

In fact, m X

ai un−i =

i=1

m X i=1

=

m X j=1

ai

m X

cj αjn−i =

j=1

m X j=1

cj αjn−m

m X i=1

cj αjn−m (αjm − P (αj )) = un .

4

ai αjm−i