A R K I V F O R M A T E M A T I K B a n d 3 n r 35
R e a d 13 F e b r u a r y 1957
O n l i n e a r r e c u r r e n c e s w i t h c o n s t a n t coefficients
B y TRYGVE NAGELL
1 . - - A n arithmetical function A ( n ) = An of n m a y be defined b y recursion in the following w a y : The value of An is defined for n = O , 1, 2, .... m - 1, a n d there is given a rule indicating h o w the value of Am+n m a y be determined when the values of A , are known for # = n , n + l , n + 2 . . . n + m - 2 , n + m - 1 , n being a n integer ~ 0.
T h e infinite sequence
A0, A1, A~, A3 ... An ....
thus defined is said to be a recurrent sequence. We denote it b y {A~}. The rule of recursion has often the shape of a recursive /ormula.
F o r instance, t h e function An = n ! satisfies the reeursive formula
A~+I = (n + 1)As (1)
a n d the initial condition A0 = 1.
T h e general solution of t h e reeursive formula
A , + I = 2 A~ (2)
is obviously A~ = 2 ~ A0.
T h e arithmetical function An satisfying the reeursive formula
A~÷~ = V-X~+~ A~
(3)
is a function of n, A0 a n d A1.
Another example is the function An defined b y the recursive formula
A~+2 = An+l + A~ (4)
a n d the initial conditions A 0 = A 1 = 1. I n this case we get the following series:
1, 1, 2, 3, 5, 8, 13, 21 . . . the so-called Fibonacci numbers.
2 . - - I n Algebra and in N u m b e r T h e o r y we often h a v e to do with linear recur- renees, t h a t is to say reeursive formulae of the t y p e
A,~+~ = al Am+n-1 + a2Am+~_e + "" + am A ~ + b, (5) 395
T. :N'AGELL, On linear recurrences with constant coefficients
where t h e coefficients al, as ... am a n d b are functions of n. I n t h e sequel we shall o n l y consider t h e case in which t h e coefficients are constants. W e shall de- velop a n e l e m e n t a r y t h e o r y of this c a t e g o r y of linear recurrences.
W h e n am # 0 , t h e recurrence (5) is said to be of t h e ruth order. W h e n am =
= a m - l = . . . = a v + l = 0 a n d a ~ # 0 , t h e order of t h e recurrence is/z. I t suffices t o consider t h e case with am # 0 .
T h e r e c u r r e n t sequence
A0, A1, As, As . . . . , An . . . . (6) is said t o be of t h e ruth order if t h e n u m b e r s A , satisfy a linear r e c u r r e n c e of order m b u t no recurrence of a lower order. A r e c u r r e n t sequence of t h e m t h order satisfies e x a c t l y one recurrence of t h e t y p e (5). I n fact, if it satisfied an- o t h e r r e c u r r e n c e
Am+, = cl Am+,-~ + c~Am+n-z + "'" + cm A n + d, we s h o u l d h a v e b y elimination of Am+,
( a l - cl) Am+,-1 + ( a 2 - c2) Am+n-~ + " " + (am - Cm) An + ( b - d) = O.
B u t this recurrence is a t m o s t of order m - 1 .
T h e f o r m u l a (2) is of t h e first order. T h e f o r m u l a (4) is of t h e second order.
W h e n b = 0, t h e recurrence (5) is homogeneous. Tl~e h o m o g e n e o u s recurrence Xm+n = al Xm+n-1. + a2 Xm+n-2 + "'" + am X n (7) is said t o h a v e t h e seals [al, az . . . a m ] .
As a direct consequence of t h e a b o v e definition we h a v e
T h e o r e m t . I ] { A , } a n d {Bn} are two recurrent sequences satislying the recur- rence (7), then { A , + B n } is also a recurrent sequence satisfying (7).
3 . - - W e shall p r o v e
T h e o r e m 2. Suppose that the numbers al, az, as . . . am are given, a m # O.
I f { A n } is a recurrent sequence such that the numbers An satisfy the honmgeneous
r e c u r r e n c e
Am+n = al A m + . - 1 + a2 Am+n-~ + "'" + am An (8) o/ order m, we have the relation
~ A n bo + bl z + b2z 2 + ... + bm_lz m-1
z "
n - o 1 - - a l z - - a 2 z 2 . . . a m Z m ' (9)
where bo, bl, b2 . . . bm-1 are constants which are uniquely determined by A0, A1 . . . Am-l, a l , a ~ , . . . , a m .
Conversely, w h e n bo, bl . . . bin-1 are arbitrarily given constants, the coeHicient8 A , i n (9) satisfy the recurrence (8).
Proof. Given t h e r e c u r r e n t sequence { A , } satisfying (8) it is easy t o see t h a t t h e infinite series
¢o
~ A n z n
(10)
n = 0
ARKIV FOR MATEMATIK. Bd 3 nr 35 has a certain circle of convergence. I n fact, we will s h o w b y i n d u c t i o n t h a t , for all N_-_O,
I AN I < QN Q~, (11)
where Q = l + l a l [ + l a s l + ... + ] a m ] a n d Qx = m a x ( ] A o [ , I A I ] . . . ]Am_l ]).
T h e relation (11) is clearly t r u e for N = 0 , 1, 2 . . . . , m - 1. I t follows f r o m (8) t h a t (11) is t r u e for N = m . I f we suppose t h a t (11) is t r u e for N = m , m + 1, m + 2 . . . r e + v , we g e t f r o m (8)
[Am+,+x[~_ Q . m a x ( [ A m ÷ , [ . . . [A,+~[) a n d , since (11) is t r u e for all N < = m + v ,
] Am+~+x ] g Q QI" m a x (Qm+,, Qm+,-1 . . . Q~+I) =< Qm+~ Q1.
This proves t h a t (11) is t r u e for all N=>0. H e n c e t h e circle of convergence of t h e series (10) has a radius which is
= l i m s u p ~ 1 _ _ ~ l i m ~ 1 _ 1
Thus, m u l t i p l y i n g t h e series (10) b y t h e p o l y n o m i a l
1 - al z - as z s . . . am z m, (12)
we get, since t h e c o n v e r g e n c e is absolute in t h e inner of t h e circle, t h e following p r o d u c t
m-1
÷n~o a A ~z n+m (13)
~ o b h Z h (A~+n - a l A m + , ~ - l - a 2 A , ~ + m - 2 . . . m ,~, ,
where t h e coefficients ba are u n i q u e l y d e t e r m i n e d b y t h e relations
b o = Ao, ]
L
b 1 = A 1 - a 1 A0,
b s = A s - a 1 A1 - a 2 A 0,
(14)
... j
b i n - 1 = A m - 1 - al A m - s . . . a m - 1 A o . I n v i r t u e of (8) t h e p r o d u c t (13) is equal to t h e p o l y n o m i a l
bo + bl z + bs z ~ + "" + bin-1 Z m -1.
This proves t h e first p a r t of T h e o r e m 2.
Suppose n e x t t h a t t h e n u m b e r s bo, bl . . . bin-1 are a r b i t r a r i l y given a n d e x p a n d t h e rational f u n c t i o n
bo + bl z + b2zg + ... + bm_l Z 'n-1 (15)
1 - - a l z - a s z ~ . . . am z m
T. NAG]ELL, On linear recurrences with constant coefficients
in a p o w e r series. I f this series is given b y (10), a n d if we m u l t i p l y it b y t h e p o l y n o m i a l (12), we find as a b o v e t h a t t h e coefficients A0, A1, As . . . A a - 1 are d e t e r m i n e d b y t h e s y s t e m (14) a n d f u r t h e r t h a t t h e coefficients A~+~, for all n _>- 0, satisfy t h e r e c u r r e n c e (8).
T h e r a t i o n a l f u n c t i o n (15) is t h e generating /unction of t h e r e c u r r e n t sequence {A=}. This f u n c t i o n m a y be w r i t t e n in t h e f o r m
~o+~, ~+ f l ~ + . . . + ~._~ ~-~
1 - ~ z - o c s z s . . . ~,,z ~ "
where t h e n u m e r a t o r a n d t h e denominafx)r h a v e no c o m m o n divisor z - 0 , a n d where ~ 0 . T h e n t h e order of t h e sequence (An} i s - - / ~ . I n fact, suppose t h a t it was of t h e o r d e r )L<p. T h e n , it would satisfy a h o m o g e n e o u s recurrence w i t h t h e scale [el, cz . . . cA] where e~¢ 0.
H e n c e we s h o u l d have, in v i r t u e of T h e o r e m 2,
flo + f l l z + f l 2 z 2 + . . . + f l , , _ l z ~ - i = e o + e l z + ~ zS + . . . + e x _ i z ~ - 1 1 - ~1 z - ~2 z 2 . . . a ~ z " 1 - - e 1 z - - C 2 Z 2 . . . C A Z ~ '
where Co, e~, ..., ez_~ are constants. T h u s (/~0 +/~1 z + --. + / ~ - 1 z "-~ ) (1 - c l z . . . e ~ ~)
: ( e O " k e l z q - " " q - ex-1 z;t-1) (1 - 0c1z . . . 0~ z~).
H e n c e 1 - cl z . . . ca z a would be divisible b y 1 - ~1 z . . . ~ z ~. B u t this is im- possible since ~ < ~u.
There is of course a n infinity of r e c u r r e n t sequences of a given order m a n d satisfying t h e h o m o g e n e o u s recurrence with t h e given scale [al, a2 . . . am], a~ ¢ 0.
4 . - - W e a d d t h e foUowing result:
T h e o r e m 3. Denote by 01, 02, 0 a, etc. the distinct roots o/ the algebraic equation
Z m - - a l z m - 1 . . . a m - 1 z - am = 0, (16)
where am ~ 0 . T h e n we obtain all the recurrent sequences {An} satis/ying the recur- rence with the scale [al, as . . . am] by the /oUowing formula
o~ t~ v i - 1 + d , ~ - l . ~ \ v ~ _ 2 2 ) + ' " + d 2 . ~ 1
where the s u m is extended over all the distinct roots Ot and where v, is the mul$i- plicity o/ Oz. T h e eoeHicients dl.,, dz., . . . d~, ~ are arbitrary constants.
F o r t h e p r o o f it suffices t o observe t h a t t h e f u n c t i o n (15) m a y be w r i t t e n
[ dvt, i dvt-l.t dl.i ],
~ [ ( 1 - 0~ z) ~ + (1 - 0t z) ~i-1 + " " + 1 -- 0~ zJ a n d t h a t we h a v e t h e expansion
ARKIV FOR MATEMATIK. Bd 3 Jar 35
( 1 - 0 ~ z ) q . = o O~z ~.
T h e n u m b e r of coefficients dk.~ in f o r m u l a (17) is equal t o m. I f t h e initial values A0, AI . . . Am-1 are given, a n d if the roots of e q u a t i o n (16) are k n o w n , we o b t a i n f r o m (17) a set of m linear equations for t h e d e t e r m i n a t i o n of t h e coefficients dk. ~.
A corollary of T h e o r e m 3 is t h e following p r o p o s i t i o n :
L e t v 1 d e n o t e t h e m n l t i p h c i t y of t h e r o o t 01 of (16), a n d let A= be given b y (17). T h e n t h e difference
B n = A n - d " l \ V1--1 satisfies t h e recurrence
B m + ~ - i = bl Bm+~-2 + b2 Bm+n-3 + "'" + bm-x B . , where t h e coefficients bl, b2 . . . bm_l are d e t e r m i n e d b y t h e i d e n t i t y
Z m - - a i z m - 1 . . . a m : Zm_ 1 _ 51 z m - 2 . . . b i n - 1 . z - 01
W e n o w t u r n t o t h e i n h o m o g e n e o u s recurrences. Consider t h e recurrence o f order m
Am+, = al Am+,-1 + a 2 A , n + , - s + " " + am A n + b, (19) where am a n d b are c 0 . Suppose first t h a t
h = 1 - a l - a 2 . . . a m ¢ O ,
a n d p u t c = b / h a n d Bn = A . - c.
T h e n it is easily seen t h a t B~ satisfies t h e h o m o g e n e o u s recurrence
B m + n = a l B i n + n - 1 + a2 B i n + n - 2 + "'" + a m B n . (20) Suppose n e x t t h a t h = 0. T h e n t h e e q u a t i o n (16) has t h e r o o t z = 1. D e n o t e b y
# t h e multiplicity of this root. W e m a y eliminate b b e t w e e n (19) a n d t h e f o r m u l a
A m + n + l = a l A m + n + a 2 A m + n - 1 + "'" + a m A n + t + b.
T h e n
A m + n + l = (al + 1) Am+. + (as - al) A m + n - 1 + (as - as) A m + n - 2 + " " + + (am -- a m - i ) A n + I - am A n . This recurrence is h o m o g e n e o u s a n d its scale is
[ a l + 1 , a s - - a l , a s - - a2 . . . a m - - a m - l , - - a m ] .
P l a i n l y
Z m + l - - (a i + 1) z m - - ( a s - - al) z m-1 ..... (am -- a m - i ) z + a m
= (z-- 1) (z m -- a l z m-1 -- a ~ z m-2 . . . am).
T. NAGELL, On linear recurrences w i t h constant coefficients
Hence, in virtue of the corollary of Theorem 3, we h a v e (for all n >_-0)
A , = B , + ( n ~ / z ) d ,
(21)where B . satifies (20), a n d where d is a certain constant (i. e. independent of n).
T o determine d we h a v e the relations
Am+, = Bm+, + (m + n + /zl d
a n dai Am+,~-t = ~ a~ Bm+,,-i + t as d.
t f f i l i f f i l
Since Am a n d B . satisfy (19) a n d (20) respectively, we get b y subtraction
b= m~-n+/z --_as #
Since b a n d d are constants, we m a y take n = 0 . Hence
d=
b# / i-1 #
where /z denotes the multiplicity of the root z = 1 in equation (16).
I n this w a y the inhomogeneous case has been reduced to t h e homogeneous case.
5 . - - A general t h e o r y of recurrences is developed in N. E. NSrlund,
Vorl~ungen giber Dif/erenzrechnung,
Berlin 1924 (Verl. Springer). Linear recurrences are t r e a t e d in K a p i t e l 14, § 2. B u t the m e t h o d employed is quite different f r o m the simple one a d o p t e d in this note.6 . - - W e finish with a few examples:
(1) The sequence of t h e
Fibonar~i numbers F,~
1, 1, 2, 3, 5, 8, 13, 21 . . . . satisfy the recurrence of the second order
F . + 2 =
F.+I
+F~
a n d the initial conditions F 1 = F 2 = 1. One finds easily
(2) The general solution of the recurrence of the first order A , + l = a A n + b
400
ARKIV FOR MATEMATIK. B d 3 n r 3 5
n " a n - 1 .
is easily found to be A . = a A0 + ~ 0, valid also for a = 1.
(3) The coefficients A . in the expansion
1 ~ A
1 - 7 z ~ - 6 z a .=o satisfy the recurrence of the third order
A,~+s= 7 A,~+x + 6 An,
a n d the initiM conditions A 0 = l , A I = 0 , A ~ = 7 . Further we find A . = - 1 ( - 1)n + ~ ( - 2)" +~o3".
I n fact the equation z 3 - 7 z - 6 = 0 has the roots z = - 1 , - 2 , and 3.
(4) I f we p u t in the reeursive formula (3), for all n, B . = log A . ,
we get the homogeneous linear recurrence B,~+2=I B.+I + ~ B . .
Hence B . = (x + fl ( _ ~)1 n,
where a a n d /3 arc determined b y the relations
B0= ~+/3, B1=~--~/3.
Bn=a gBo + gBl + ~ ( B o - B1) ( - ~ ) n 2 Thus
a n d finally
A 3 _ A 2 ( - ½ ) n + l / ] 2 - 2 ( - ½)n
T r y c k t d e n 15 m a r s 1957
U p p s a l a 1957. Almqvist & Wiksslls B o k t r y c k e r i A B