- Email: [email protected]
On the packing sums of pairs
Microelearon. Reliab., Vol. 33, No. 2, pp. 169-173, 1993. ih'inted in Great Britain.
0026-271419356.00+.00 © 1993 Pergamon Press Ltd
ON THE PACKING SUMS OF PAIRS G. R. HOFMEISTER Johannes Gutenberg University, 6500 Malnz, Saarstr. 21, F.R.G. A. M. IBRAHIM Department of Mathematics, Faculty of Science, Alexandria University, EI-Shatby, Alexandria, Egypt and
G. A. SHADIA Suez Canal University, Ismallia, Egypt
(Receivedfor publication 30 November 1991)
~cl:. This paper is concerned with the determination of the length of the largest interval of consecutive integers of the set hA~, where A k is a sequence of integers which is a Bh-sequenca.
We shall recall the following definitions (cf. [1]).
DeflnitiontlL The sequence of all distinct integers of the form a+b (a ~ A, b e B) will be denoted by A + B, and referred to as the sum of the sequences A and B.
The sum of more than two sequences is defined similarly. It is convenient to write A+A as 2A and, more generally, the h.fold sum A+A+...+A as hA.
P..~[QJJLqD.(~. A sequence {a~,a2..... ar} of integers is called a B= -sequence if the sums al + aj, 1 < i < j <__r, are all different.
Deflnitioni3L Let g, h be natural numbers, where h>2. We denote by Bh [g] the class of all (finite or infinite) sequences A such that every integer n has at most g representations of the form n = a °) + a ~) + ... + a (h), (a 0), a R) ..... a (h) ¢ A) subject to
a0)<_ a (2).<. . . < a(h).
D e f l n l t l o n f 4 L If A lies in the class B h [g], w e say that "a is a B h [9]-sequence". We shall
refer to the class Bh [1] simply as the class Bh and accordingly, we refer to sequences in the class Bh [1] simply as Bh-saquence. In other words, a Bh-sequence (h>._2) may be defined as a sequence al,a2..... a r of integers in ascending order of magnitude such that the sums ]69
170
G . R . HOFMEiSTERet aL
a~ + a~ . ... + a~,
(/1
'= ~ '~ ... < /,)
are all different.
Let A k be a B h -sequence, Ih (Ak) = [Ik, mk] be the largest consecutive interval in the set hA k, and d e n o t e by r h (Ak) the length of [I k, ink], i.e. r h (Ak) = m k - Ik + 1, and let r. (k) =
max. r h (Ak). A k ~ Bh
Note(1), If A k is a B h -sequence, then also A k - a 1 = {a i - a 1 I a I e Ak} c: N O is a B h -sequence and A k - a 1 has the same length of the interval in h(Ak-al), and O~ (A k - al).
Note(2) For arbitrary h, h > 2 (1) {0,1} is a B h -sequence, r h (2) _> h + l (2) {0,1,h + 1} is a B h -sequence, r h (3) > 2 h
+ 1
From this, it is clear that for fixed positive integers k > 2, we can select a positive integer h such that we get a set A k which has k elements and A~ is a B h -sequence and the length of an interval lies in hA k > I ( I arbitrary integer ) i.e. for any I ~ N and any k >_ 2 there exists h ~ N, A K SUCh that i)
A k is a B h -sequence
ii)
h A k contains I consecutive integers.
T h e o r e m . For arbitrary positive integers h, k > 2. r h ( k + 2 ) _ > r h(k) + 1 Proof. Without loss of generality, let (1)
A k = {a 1, a2..... a k} , 0 = a l < a = < . . . < a k ; k > 2
be a B h -sequence such that
(2)
r h (Ak) = r h (k).
We c h o o s e ak+l, ak+2 as
(3)
ak. 1 ;----- (h+2) a k , ak. 2 := m k + 1 - (h-l) ak. 1
i.e.
(4)
aK~.2 + (h-l) ak. 1 = m k + 1
and
ak+ 2 = m k + 1- (h-l) (h+2) a k
= m k + 1 - (h2+h-2) a k Since 0 <__mk< h a k , we have
(s)
- (h2+h-2) a k + 1 < ak+2 <__-(h2-2) a k + 1. Let
Packing sums of pairs
171
~'k+2 ' = {ak+l, ak+2} U A k , and Ak+ 2 := Ak+ 2 + I ak+ 2 I ~: N o , with 0 • Ak. 2 , then [I k , m k + 1] C h/t,k. 2. It suffices to show that Ak+ 2 is a B h -sequence, because we then get r h (k+2) > r h (k) + 1.
We note that any element in h/L.k+ 2 can be written in the following from (6)
S= = a 1 aK+ 1 + G2 ak+ 2 + S, ; a 1 + a 2 3
S=s= S ( a , ,
"i" G 3 --'-- h;
at > 0
a, 2..... a,,~ = a l + a , + ... + a , , : a . A k
i.e
(7)
0 <__S~j<__a 3 a k.
Let s.,
-
.
.
Sp, " 8A + 8j= + . . .
If
...
.
,
...
+ 8/, = , /1 < J'z < " "
S=s= Sp3, then a 3 = /33 and a=,= all for 1 < l < a
;
< jp, •
s.
In this case we shall write S%= Spsi.e.
s
..... a,.,) - s (a,, a/, ..... %)
Where %"
ps a n d a& - a h ,
a~ - a k . . . . .
al, = - a/~=.
If 0 ~ A k , then S = ~ Sp3gives S=3= S~ and a 1 < ... < a=.
Now we shall prove, if S= = Sp are two elements in h/t.k+ 2 , then a t = #1 , a 2 =/32, and a 3 = /33 , i.e h/~'k+2 is a B h -sequence.
I. Let a 2 = ~2
11. If
al = /~1,
S= = S B gives S=3 = S~s (since A k is a B h -sequence) and this is possible if and only if S=3 = S~(i.e. a 3 = /~3)I=) I f a 1,, / 3 1 , s a y a 1 > ~ 1 , S= = S n gives al ak+l + S=s= ~1 ak+l + SJa, ( a I - /31)
ak+l = S t ; S=s
172
O. R. HOFP~IS'rH eta/.
but (=1 " #1) ak+l --> (h+2) a=, (from (3) and since =1 - #I .P- 1). From (7)
and then we get contradiction.
II. Let a 2 ,, #2 say a 2 > #2 and this gives =2 > 1, (~1 + == < h-l.
II 1. If =1 = h-1 ; gives =3 = O, =2=1 , # 2 = 0 S= = Sp gives (h-l) ak+ I + ak+ 2 = #1 ak+l + Sjl a and from (3) m k + 1 = Pl ax+l + ,-qp= and this is impossible, for (1) If #1 > 1, then #1 ak+l + Sns> mk + 1 ( because m k < h a k , ak+ 1 = (h+2) a k ) (2) I f # l = 0 , t h e n S n a "
m k + 1, (since m k + 1 ~ h A k).
112" If =1 < h-2 S= = S n gives a 1 ak+ 1 + (Z2 ak+ 2 + S=a= #1 ak+1 + #2 ak+2 + SB~ (8)
(=2" #2) ak+2 = #1 ak+l + Sn; (orl ak+l + S= ), 3 Since ak+ 2 < 0 , 22 " #2 > 1, from (4) we have
(9)
LH.S. of (8) < ak+ 2 < - (h 2 - 2) a k + 1
(1) If =1 = h-2, i.e. a= < 1
R.H.S.of (8) > - ( ¢1 mk+l + 8= I ), ( ,ffh?oo Pl ink.1 + Sill = 0 ) - -[(h
-
2) ,,.,
+ S.]
- - [(h - 2) (h + 2) =, + S,,]
- -p,-
4) . , + s.]
- [(h = - 4) =, + =,],
- - (h = - 3) a, i.e.
(10)
R.H.S. of (8) > - (h 2 - 3) a k . From (9) and (10)
(=~
=,,
1)
Packing sums of pairs
173
R.H.S. of (8) > LH.S. of (8), (since a= :; 1) and we get contradiction. (2)
ff cr1 < h - 3 ,
i.e. a 3
<__h-1
R.H.S.of (S) > - [(h - 3)(h + 2) a~ + S,,] - - [(h' - h-
6) ,,+
S,,]
> - [ ( h = - h - 6 ) ak+ ( h - 1) ak], (sincees =; h - 1) -
-
(h = -
7)
a,
i.e°
(11)
R.H.S of (8) > - (h2- 7) a,. From (9) and (11) R.H.S. of (8) > L.H.S. of (8)
and we get also contradiction, and the proof of the theorem is completed.
References [1] H.Halberstam and K.F.Roth, sequences. Vol.l, Oxford Univ. Press, Oxford, 1966.
0026-271419356.00+.00 © 1993 Pergamon Press Ltd
ON THE PACKING SUMS OF PAIRS G. R. HOFMEISTER Johannes Gutenberg University, 6500 Malnz, Saarstr. 21, F.R.G. A. M. IBRAHIM Department of Mathematics, Faculty of Science, Alexandria University, EI-Shatby, Alexandria, Egypt and
G. A. SHADIA Suez Canal University, Ismallia, Egypt
(Receivedfor publication 30 November 1991)
~cl:. This paper is concerned with the determination of the length of the largest interval of consecutive integers of the set hA~, where A k is a sequence of integers which is a Bh-sequenca.
We shall recall the following definitions (cf. [1]).
DeflnitiontlL The sequence of all distinct integers of the form a+b (a ~ A, b e B) will be denoted by A + B, and referred to as the sum of the sequences A and B.
The sum of more than two sequences is defined similarly. It is convenient to write A+A as 2A and, more generally, the h.fold sum A+A+...+A as hA.
P..~[QJJLqD.(~. A sequence {a~,a2..... ar} of integers is called a B= -sequence if the sums al + aj, 1 < i < j <__r, are all different.
Deflnitioni3L Let g, h be natural numbers, where h>2. We denote by Bh [g] the class of all (finite or infinite) sequences A such that every integer n has at most g representations of the form n = a °) + a ~) + ... + a (h), (a 0), a R) ..... a (h) ¢ A) subject to
a0)<_ a (2).<. . . < a(h).
D e f l n l t l o n f 4 L If A lies in the class B h [g], w e say that "a is a B h [9]-sequence". We shall
refer to the class Bh [1] simply as the class Bh and accordingly, we refer to sequences in the class Bh [1] simply as Bh-saquence. In other words, a Bh-sequence (h>._2) may be defined as a sequence al,a2..... a r of integers in ascending order of magnitude such that the sums ]69
170
G . R . HOFMEiSTERet aL
a~ + a~ . ... + a~,
(/1
'= ~ '~ ... < /,)
are all different.
Let A k be a B h -sequence, Ih (Ak) = [Ik, mk] be the largest consecutive interval in the set hA k, and d e n o t e by r h (Ak) the length of [I k, ink], i.e. r h (Ak) = m k - Ik + 1, and let r. (k) =
max. r h (Ak). A k ~ Bh
Note(1), If A k is a B h -sequence, then also A k - a 1 = {a i - a 1 I a I e Ak} c: N O is a B h -sequence and A k - a 1 has the same length of the interval in h(Ak-al), and O~ (A k - al).
Note(2) For arbitrary h, h > 2 (1) {0,1} is a B h -sequence, r h (2) _> h + l (2) {0,1,h + 1} is a B h -sequence, r h (3) > 2 h
+ 1
From this, it is clear that for fixed positive integers k > 2, we can select a positive integer h such that we get a set A k which has k elements and A~ is a B h -sequence and the length of an interval lies in hA k > I ( I arbitrary integer ) i.e. for any I ~ N and any k >_ 2 there exists h ~ N, A K SUCh that i)
A k is a B h -sequence
ii)
h A k contains I consecutive integers.
T h e o r e m . For arbitrary positive integers h, k > 2. r h ( k + 2 ) _ > r h(k) + 1 Proof. Without loss of generality, let (1)
A k = {a 1, a2..... a k} , 0 = a l < a = < . . . < a k ; k > 2
be a B h -sequence such that
(2)
r h (Ak) = r h (k).
We c h o o s e ak+l, ak+2 as
(3)
ak. 1 ;----- (h+2) a k , ak. 2 := m k + 1 - (h-l) ak. 1
i.e.
(4)
aK~.2 + (h-l) ak. 1 = m k + 1
and
ak+ 2 = m k + 1- (h-l) (h+2) a k
= m k + 1 - (h2+h-2) a k Since 0 <__mk< h a k , we have
(s)
- (h2+h-2) a k + 1 < ak+2 <__-(h2-2) a k + 1. Let
Packing sums of pairs
171
~'k+2 ' = {ak+l, ak+2} U A k , and Ak+ 2 := Ak+ 2 + I ak+ 2 I ~: N o , with 0 • Ak. 2 , then [I k , m k + 1] C h/t,k. 2. It suffices to show that Ak+ 2 is a B h -sequence, because we then get r h (k+2) > r h (k) + 1.
We note that any element in h/L.k+ 2 can be written in the following from (6)
S= = a 1 aK+ 1 + G2 ak+ 2 + S, ; a 1 + a 2 3
S=s= S ( a , ,
"i" G 3 --'-- h;
at > 0
a, 2..... a,,~ = a l + a , + ... + a , , : a . A k
i.e
(7)
0 <__S~j<__a 3 a k.
Let s.,
-
.
.
Sp, " 8A + 8j= + . . .
If
...
.
,
...
+ 8/, = , /1 < J'z < " "
S=s= Sp3, then a 3 = /33 and a=,= all for 1 < l < a
;
< jp, •
s.
In this case we shall write S%= Spsi.e.
s
..... a,.,) - s (a,, a/, ..... %)
Where %"
ps a n d a& - a h ,
a~ - a k . . . . .
al, = - a/~=.
If 0 ~ A k , then S = ~ Sp3gives S=3= S~ and a 1 < ... < a=.
Now we shall prove, if S= = Sp are two elements in h/t.k+ 2 , then a t = #1 , a 2 =/32, and a 3 = /33 , i.e h/~'k+2 is a B h -sequence.
I. Let a 2 = ~2
11. If
al = /~1,
S= = S B gives S=3 = S~s (since A k is a B h -sequence) and this is possible if and only if S=3 = S~(i.e. a 3 = /~3)I=) I f a 1,, / 3 1 , s a y a 1 > ~ 1 , S= = S n gives al ak+l + S=s= ~1 ak+l + SJa, ( a I - /31)
ak+l = S t ; S=s
172
O. R. HOFP~IS'rH eta/.
but (=1 " #1) ak+l --> (h+2) a=, (from (3) and since =1 - #I .P- 1). From (7)
and then we get contradiction.
II. Let a 2 ,, #2 say a 2 > #2 and this gives =2 > 1, (~1 + == < h-l.
II 1. If =1 = h-1 ; gives =3 = O, =2=1 , # 2 = 0 S= = Sp gives (h-l) ak+ I + ak+ 2 = #1 ak+l + Sjl a and from (3) m k + 1 = Pl ax+l + ,-qp= and this is impossible, for (1) If #1 > 1, then #1 ak+l + Sns> mk + 1 ( because m k < h a k , ak+ 1 = (h+2) a k ) (2) I f # l = 0 , t h e n S n a "
m k + 1, (since m k + 1 ~ h A k).
112" If =1 < h-2 S= = S n gives a 1 ak+ 1 + (Z2 ak+ 2 + S=a= #1 ak+1 + #2 ak+2 + SB~ (8)
(=2" #2) ak+2 = #1 ak+l + Sn; (orl ak+l + S= ), 3 Since ak+ 2 < 0 , 22 " #2 > 1, from (4) we have
(9)
LH.S. of (8) < ak+ 2 < - (h 2 - 2) a k + 1
(1) If =1 = h-2, i.e. a= < 1
R.H.S.of (8) > - ( ¢1 mk+l + 8= I ), ( ,ffh?oo Pl ink.1 + Sill = 0 ) - -[(h
-
2) ,,.,
+ S.]
- - [(h - 2) (h + 2) =, + S,,]
- -p,-
4) . , + s.]
- [(h = - 4) =, + =,],
- - (h = - 3) a, i.e.
(10)
R.H.S. of (8) > - (h 2 - 3) a k . From (9) and (10)
(=~
=,,
1)
Packing sums of pairs
173
R.H.S. of (8) > LH.S. of (8), (since a= :; 1) and we get contradiction. (2)
ff cr1 < h - 3 ,
i.e. a 3
<__h-1
R.H.S.of (S) > - [(h - 3)(h + 2) a~ + S,,] - - [(h' - h-
6) ,,+
S,,]
> - [ ( h = - h - 6 ) ak+ ( h - 1) ak], (sincees =; h - 1) -
-
(h = -
7)
a,
i.e°
(11)
R.H.S of (8) > - (h2- 7) a,. From (9) and (11) R.H.S. of (8) > L.H.S. of (8)
and we get also contradiction, and the proof of the theorem is completed.
References [1] H.Halberstam and K.F.Roth, sequences. Vol.l, Oxford Univ. Press, Oxford, 1966.