A SIMPLE RECURRENCE RELATION IN FINITE ABELIAN GROUPS

A SIMPLE RECURRENCE RELATION IN FINITE ABELIAN GROUPS H. P. YAP University of Singapore, Singapore 10 A finite abelian group G is said to have a simp...
Author: Bonnie Houston
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A SIMPLE RECURRENCE RELATION IN FINITE ABELIAN GROUPS H. P. YAP University of Singapore, Singapore 10

A finite abelian group G is said to have a simple recurrence relation of length n if there exist distinct nonzero elements a l 9 a 2 , • • • , a of that a* + a2 = a3? a2 + a3 = a 4 , • • • , a +a 1 = a , a + a. = n—a II—x n n—± n a n + a l = a 2 4 tt i s P r o v e d t i i a t if n = 6m or n = 2^3^m fe3), where = 1, ^ = 0, 2, or 3 and /3 = 0, 1, or 2, then there exists a finite group which has a simple recurrence relation of length n. Let G be a finite abelian group written additively and a l9 a 2 ,

G such & and i (6,m) abelian ,8S

, a

be distinct nonzero elements of G. If a-i + a2 = a 3s a2 + a3 = a 4 , • • • , a n _ 2 + a ^ a . - + a = an-1 n 1

and

a + a- = a n n 1 2

= an , 9

then we say that the ordered set A = (a l 9 a 2 , • •• , a n } has a simple recurrence relation (SRR). If G has an ordered subset A such that the cardinal of A is n ^ 3 ) and A has a SRR, then we say that G has a SRR of length n. We use the notation 1(G) = n to mean that G has a SRR of length n. Suppose A = (a*, a 2 , •• • , a n } has a SRR; then we have a3 = a^ + a 2s a 4 = a t + 2a 2 , a5 = 2aj + 3a 2 ,

.

Let U0 = 0, Ui = 1, U2 = 1, U3 = 2, U4 = 3, U5 = 5 , - , U i+ 2 255

= U

i+Ui+l5""s

256

A SIMPLE RECURRENCE RELATION

be a Fibonacci sequence ([1], p. 148). Then

U. =

(1)

1

— N/5

l^j - (^)']

i >0

Thus

a2+i = U . a i + U.+1a2 ,

(2)

From a

(3)

i > 0

, + a = a, and a„ + an = a„, we have n-l n 1 n x ^

< U n-l "

1)a

+ U

l

na2 = °

and



0< i < n - 2 .

Now we prove that A = {a l 9 a 2 , • • • , a^}, where a4 is chosen such that (a l5 f(n)) = 1 and a 2 + . f

0 < i < n - 2 is given by (11), has a SRR.

We need to verify (1) a i j a 2 5 # 0 ' s a

considered as elements in CL, y are distinct and nonzero;

(II) ai + a2 = a39 a2 + a3 = a4s

e

• • , an_2 + a ^

= an?

an_1 + an = alt

and

==

a ~r a., a^o n 1 2 First we prove (n): For this part* we need only to verify that a a2.

- +a

= a- and a + a- =

In fact,

V l

+ a

n - CrUn(1 " U n - 1 >+

U

n-lK

=

C (1 " U n - 1 > + — a^ |

U

n - A

< *

+ U

X

0< i < j < n - 2 ,

i =

0 < i < j < n - 2 rU

j+l(1 "

U

n-1>

9

+ U

j '

then r

< V - Ui+l^ 1 " Un-1>

+



1970]

IN FINITE ABELIAN GROUPS

from which it follows that (

Vl-Ui-M)(1-Un-l)+Un(Uj-Ui)

= °'

i . e, j

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