Download On Integral Zeros of Krawtchouk Polynomials

On Integral Zeros of Krawtchouk
Polynomials
Ilia Krasikov
Tel-Aviv University
School of Mathematical Sciences,
Ramat-Aviv 69978 Tel-Aviv, Israel and
Beit-Berl College, Kfar-Sava, Israel
Simon Litsyn*
Tel-Aviv University
Department of Electrical Engineering { Systems
Ramat-Aviv 69978
Tel-Aviv, Israel
Abstract
We derive new conditions for nonexistence of integral zeros of binary Krawtchouk polynomials. Upper bounds for the number of integral roots of Krawtchouk polynomials are presented.
Krawtchouk polynomials, integral roots, perfect codes, switching reconstruction, Radon transform.
Keywords:
3 Research
supported by the Guastallo Fellowship and a grant from the Israeli Ministry of Science and
Technology.
1.
Introduction
The binary Krawtchouk polynomial Pkn(x) (of degree k ) is dened by the following generating function:
1
X
Pkn (x)zk = (1 0 z )x (1 + z )n0x:
(1)
k=0
Usually n is xed, and when it does not lead to confusion it is omitted.
The question of existence of integral zeros of Krawtchouk polynomials (or, that is essentially the same, existence of zero coecients in the expansion of (1 0 z )x (1 + z )n0x ) arises
in many combinatorial and coding theory problems. Let us state some of them.
1.
2.
n
Radon transform on Z2
f is
[14]. Let f : Zn ! R; then the Radon transform FT of
2
FT (x) =
X
f (y );
2T x
where T + x means the set ft + x : t 2 T g. The question is whether it is invertible.
Switching reconstruction problem [45]. Given a graph G = G(V; E ); jV j = n;
for U V the switching GU of G at U is the graph obtained from G by replacing
all edges between U and V n U by the nonedges. The multiset of unlabeled graphs
Ds (G) = fGU : jU j = sg is called the s-switching deck of G. The question is
y
+
whether G is uniquely dened up to isomorphism by Ds(G).
. Given a digraph 0 = 0(V; E ); jE j = e,
that is an orientation of edges of an ordinary graph, for any subset A of E denote
by 0A the graph obtained from 0 by the reorientation of all arcs in A. Dene the
s-reorientation deck Ds(0) = f0A : jAj = sg. The question again is whether 0 is
uniquely dened up to isomorphism by Ds(0).
3.
Reorientation reconstruction problem
4.
Sign reconstruction problem
. Let G = G(V; E ); jE j = e; be a sign graph that
is the graph with the edges marked by + or 0. Similarly we dene s-sign deck of
G as the multiset of signed graphs obtained from G by switching signs in all the
s-subsets of E . The question is the same as above. Note that if it is permitted only
+ signs to be switched in the last problem it turns out to be a generalization of
well-known edge-reconstruction problem.
The number of such examples can be easily increased. The essential feature of
all of them is that the operations on graphs (or on another structures) should be
involutions, and this is just the reason of arising Krawtchouk polynomials .
1
Let Fn be the binary Hamming space of dimension n, and S (x; r) (resp. B (x; r ))
be the Hamming sphere (resp. ball) of radius r centered at x 2 Fn.
(see, e.g. [29, 33]) Perfect code is a set C Fn with the
property that the union of the balls centered at the points of C covers the space Fn
without intersections. The question is whether such a code does exist for given n
and r .
5.
Perfect binary codes
6.
Multiple perfect coverings
[10, 50, 17]. Multiple perfect s-covering of radius r
is a (multi)set C Fn with the property that the union of balls centered at the
points of C covers every point of the space Fn precisely s times. The question is
whether such a covering does exist for given s, n and r.
Connections of the listed problems with integral roots of Krawtchouk polynomials is
reected by the next theorem.
Theorem 1
1. If Pns(x) has no integer roots then
a) the Radon transform is invertible provided S is S (x; s) 2 Fn or B (x; s) 2 Fn
+1
[14];
b) in Problems 3 and 4 the corresponding graphs (digraphs) are reconstructible.
2. If Psn(x) has no even integer roots then
in Problem 2 the graph is reconstructible [45].
3. If Psn(x) has at least one noninteger root then
there is no perfect code for radius s in Fn .
+1
4. If Psn(x) has less than N integer roots then
there is no perfect -fold covering of radius s in Fn , where N is the minimum integer
such that
N n!
s n!
X
X
j :
i
j
i
+1
=0
=0
Regarding the reconstruction Probems 2{4 note that the, so called, balance equations
(see [23]) for all the three problems are the same. The graphs are reconstructible if (but
not only if) those equations have the unique solution. The last is true whenever the
corresponding Krawtchouk polynomial (the characteristic polynomial of the matrix of
the balance equations) has no integral roots . In Problem 2 only the even roots are
relevant since the switching of the subset of vertices coincides with the switching of the
2
complement subset. For perfect multiple coverings the presented condition is a particular
case of a more general theorem in [8].
At present our knowledge about integral roots of Krawtchouk polynomials is quite poor.
For example, in the binary case we even can not assure in general that there is at least
one nonzero root. In coding theory it was overcome by using some extra conditions, such
as sphere-packing condition. Actually [48], we know all the possibilities for parameters of
binary perfect codes (see also [29, 47, 49, 51] for the corresponding results for nonbinary
case, when the base equals a power of prime). For nonbinary (a prime power) Krawtchouk
polynomials existence of at least one noninteger roots for polynomials of degree greater
than 2 was proved nally by Y.Hong basing on previous works (see, e.g. [2, 3, 5]). In
1985 P. Diaconis and R. L. Graham wrote in [14]: "We do not know of any systematic
study of integer zeros of Krawtchouk polynomials ". A detailed study of integral roots
of binary Krawtchouk polynomials was undertaken in [9, 16]. For general properties of
roots of Krawtchouk polynomials see [27, 35, 46].
We would like to mention several questions which appear to be out of the scope of the
paper but very much similar to its problematics. Namely, they are problems of existence
of perfect L-codes [11, 21], and perfect weighted coverings [8, 12]. In these cases we are
interested in integral roots of linear combinations of Krawtchouk polynomials of dierent
degrees.
Note that since Pkn (x) and Pnn0k (x) have common set of integral roots (see (14) below) we
assume, if the opposite is not stated explicitly, that k n=2. Computer search supports
the conjecture that the "typical" Krawtchouk polynomial does not have integral zeros
at all, and in general can possess only a few (we conjecture 4 to be the right number)
such roots. In [16] a list of innite in n families of integral roots is presented. Namely,
such families have been found only for k = 1; 2; 3 and k = (n 0 i)=2; i = 0; . . . ; 8; i 6= 7.
For other values of n only some sporadic zeros are known. It is tempting to conjecture
that the known list is complete but, maybe, a small number of sporadic roots. A partial
explanation of this phenomenon will be given.
In the paper we make an attempt of systematic study of existence of integral roots of
Krawtchouk polynomials . We start with some relevant properties of Krawtchouk polynomials in Section 2, assemble known facts about the integral roots of Krawtchouk
polynomials in Section 3, derive new upper bounds for the number of integral roots in
Section 4, obtain conditions for existence of at least one nonintegral root in Section 5,
present conditions for nonexistence of integral roots in Section 6. In Section 7 we discuss
behavior of zeros of Krawtchouk polynomials and distances between them.
3
2.
Properties of
Krawtchouk polynomials
Here we assemble some properties of Krawtchouk polynomials . Many of them can be
found in [5, 30, 28, 33, 35, 49], we present them without proofs. Recall that we deal only
with the binary case.
There are several explicit expressions for Krawtchouk polynomials :
!
!
x n0x
Pk (x) = (0
j k0j
j
!
!
k
X
n
0
j
x
= (02)j
j k0j
j
!
!
k
X
n0x n0k+j
j
k
0
j
:
= (01) 2
j
k0j
j
k
X
1)j
(2)
=0
(3)
=0
(4)
=0
A remarkable property of Krawtchouk polynomials is that in every variable they satisfy
a linear recurrence relation with linear coecients:
(k + 1)Pkn (x) = (n 0 2x)Pkn(x)
+1
0 (n 0 k + 1)Pkn0 (x);
(5)
1
(n 0 x)Pkn (x + 1) = (n 0 2k)Pkn (x) 0 xPkn (x 0 1)
(6)
(in [33] the last relation is given only for integer values of x. Using Lagrange interpolation
one can see that it holds as well in general, see [5]).
The following relation we did not nd in literature, so we supply it with a proof.
(n 0 k + 1)Pkn (x) = (3n 0 2k 0 2x + 1)Pkn (x)
+1
Proof.
0 2(n 0 x)Pkn0 (x):
1
(7)
We start from the following easy to check identity:
@
(n + 1)(1 0 z )x (1 + z )n 0x 0 z ((1 0 z )x (1 + z )n 0x ) =
@z
@
(3n 0 2x + 1)(1 0 z )x (1 + z )n0x 0 2z ((1 0 z )x (1 + z )n0x) 0 2(n 0 x)(1 0 z )x (1 + z )n0 0x:
@z
Using (1) and comparing the coecients at the equal powers of z we get (7).
+1
+1
1
2
Herein are listed several values of Krawtchouk polynomials :
4
(n 0 2x)
2
P (x) = 1; P (x) = n 0 2x; P (x) =
0
1
P (x) =
3
2
2
(n 0 2x)((n 0 2x)
6
2
0 n;
0 3n + 2) ;
!
!
n
n
; Pk (1) = (1 0 2k=n)
Pk (0) =
k
k
Pk (n=2) = 0; for k odd ; Pk (n=2) = (0
If Pkn(x) =
Pk
i=0 ci x
i
(8)
1)k=2
!
n=2
; for k even :
k=2
(9)
(10)
then
ck = (02)k =k !; ck0 = (02)k0 n=(k 0 1)!:
1
1
(11)
Note also that k!Pnk (x=2) is a polynomial with integral coecients.
The following relations reect some symmetry properties of Krawtchouk polynomials
with respect to their parameters:
!
!
n n
n n
P (k); (for nonnegative integer x);
Pk (x) =
k x
x
(12)
Pkn(x) = (01)k Pkn(n 0 x);
(13)
Pkn(x) = (01)x Pnn0k (x); (for integer x; 0 x n):
(14)
We would like to emphasize that Pkn (x) = 0 for k < 0, and also for k > n if x is an
integer, 0 x n. This follows from (1) and (5).
For the Krawtchouk polynomials the following orthogonality relations hold:
n
X
i=0
!
!
n n
n n
2;
Pk (i)Pln (i) = kl
k
i
(15)
n
X
P n(i)P n(k ) = (16)
i=0
l
n
lk 2 :
i
5
For any polynomial S (x) of degree d there is the unique Krawtchouk expansion:
S (x) =
where the coecients are
d
X
k=0
k = 20n
k Pkn (x) ;
n
X
i=0
(17)
S (i)Pin(k ):
Particularly, the derivative of Krawtchouk polynomial is
k0
X= Pk0 i0 (x)
0
Pk (x) = 02
:
[(
1) 2]
2
(18)
1
2i + 1
i=0
As orthogonal polynomials Krawtchouk polynomials satisfy the Christoel-Darboux formula:
! k n
Pi (x)Pin(y )
2 n X
Pkn (x)Pkn (y) 0 Pkn (x) Pkn (y )
=
:
n
y0x
k+1 k i
i
+1
+1
(19)
=0
It can be derived from the above formula (see e.g. [27]) that the ratios
Pkn0 (x) Pkn0 (x 0 1)
;
;
Pkn (x) Pkn (x)
1
1
+1
+1
increase in x for k = 0; . . . ; n 0 1; and
Pkn (x) 6= 0.
Pkn (x)
Pkn+1 (x)
increases in x for k = 0; . . . ; n; provided
+1
Krawtchouk polynomials satisfy the multiplication theorem, which will be presented here
for less restrictive conditions than in [30]:
Pkn(x)Pin (x) =
where
Xk;i
min(
)
j =max(0;k+i0n)
k + i 0 2j
a(k; i; j ) =
k0j
!
a(k; i; j )Pk i0 j (x);
+
2
(20)
!
n 0 k 0 i + 2j
:
j
Throughout the proof n and x are supposed to be xed, so we omit them in the
notation of Krawtchouk polynomials .
Proof.
First we prove that
6
Pkn(x)Pin (x) =
Xk;i
min(
)
a(k; i; j )Pkn i0 j (x):
+
j =0
(21)
2
The proof is by induction on (k + i). For small values of (k + i) (21) and (20) can be
veried directly. Using (5) and by the induction hypothesis we have:
(k + 1)Pk Pi = (n 0 2x)Pk Pi 0 (n + 1 0 k )Pk0 Pi =
+1
(n 0 2x)
Xk;i
min(
j =0
Xk;i
min(
j =0
Xk;i
min(
+
2
0 (n + 1 0 k)
a(k; i; j )((n 0 2x)Pk i0 j
+
2
(n + 1 0 k 0 i + 2j )a(k; i; j )Pk i0 j 0
+
Xk;i
j =0
a(k; i; j )Pk i0 j
)
j =0
min(
)
)
1
)
2
a(k; i; j )(k + i 0 2j +1)Pk i0 j +
+
2 +1
(n + 1 0 k )
1+min(
1
j =1
0 (n + 1 0 k)
Xk;i
j =1
1
)
j =0
1
)
a(k 0 1; i; j )Pk i0 j 0 =
+
0 (n + 1 0 k 0 i + 2j )Pk
1+min(
Xk0 ;i
k0 ;i
X
min(
)
k0 ;i
X
min(
1
j=0
)
+
2
1
i02j01 )+
a(k 0 1; i; j )Pk i0 j 0 =
+
2
a(k; i; j 0 1)(n 0 1 0 k 0 i +2j )Pk i0 j
+
1
2 +1
0
a(k 0 1; i; j 0 1)Pk i0 j :
+
2 +1
In the last expression the terms which do not appear in all three sums, sum up to zero.
Also the direct calculation shows:
a(k; i; j )a(k + i 0 2j + 1) + a(k; i; j 0 1)(n 0 1 0 k 0 i + 2j ) 0 a(k 0 1; i; j 0 1)(n + 1 0 k ) =
(k + 1)a(k + 1; i; j ):
Hence we get
7
(k + 1)Pk Pi = (k + 1)
k ;i
X
min( +1
+1
)
a(k + 1; i; j )Pk i0 j :
+
j =0
2 +1
That proves (21).
To prove (20) observe that a(k; i; j ) = 0 whenever j < k + i 0 n. The proof is complete.
2
Now we present some known facts about zeros of Krawtchouk polynomials .
Pkn (x) has k dierent roots 0 < r ;n(k) < r ;n(k) < . . . rk;n (k) < n;
The roots are symmetric with respect to n=2, that is
1
2
ri;n(k ) + rk 0i;n (k ) = n; i = 1; . . . ; k:
+1
More information on location of the roots can be easily derived from the following
elegant result due to V.Levenshtein [27]:
kX
02
r ;n (k ) = n=2 0 max(
1
i=0
xi xi
q
(i + 1)(n 0 i) );
+1
where maximum is taken over all xi subjected to
Particularly, we get for k n=2
Pk 0
1
i=0
xi = 1.
q
2
r ;n (k ) n=2 0 (k 0 1)(n 0 k + 2) max(
1
n=2 0
q
v
uk 0
0
u X kX
x
(k 0 1)(n 0 k + 2)t x
2
1
2
i=0
i
i=1
(22)
q
kX
02
i=0
xi xi ) +1
i n=2 0 (k 0 1)(n 0 k + 2):
2
(23)
Evidently, (22) enables getting other upper and lower bounds for the rst root. For
instance, in [28] the following estimate is given:
q
r ;n (k) n=2 0 k (n 0 k ) + k
1
=
1 6
p
n 0 k for k [n=2];
(24)
The roots of Krawtchouk polynomials for small k can be approximated by the
corresponding roots of Hermite polynomials [2]:
8
If (n 0 k ) ! 1 then the zeros of Pkn (x) approach
p
n0k01
hi (k);
2
where h (k) < . . . < hk (k ) are the roots of the Hermite polynomial Hk (z ).
n=2 +
(25)
1
The roots of Krawtchouk polynomials satisfy some interlacing properties (see e.g.
[46, 9]), namely,
ri;n (k ) < ri;n(k 0 1) < ri
+1
ri;n(k ) < ri;n (k ) < ri
+1
+1
;n (k )
+1
;n+1 (k );
< ri
+1
i = 1; . . . ; k 0 1;
;n+1 (k ); i
= 1; . . . ; k 0 1;
(26)
(27)
Moreover, for i = 1; . . . ; n;
ri;n(k ) 0 ri;n (k ) < 1 :
+1
For xed n and k < n=2 (see [9]):
ri
3
(28)
+1
;n (k )
0 ri;n(k) > 2:
(29)
What is known about integral roots ?
Very few is known at present about integral roots of binary Krawtchouk polynomials
(without extra conditions being imposed in coding theory). To the best of our knowledge,
the list of papers dealing with the problem is not very long [9, 14, 16, 24, 25]. As we have
have mentioned this problem can be restated via (1) and (12) as follows:
How many zero coecients does the expansion (1 0 z )i (1 + z )j have?
We denote by N (n; k ) = N (k) the number of integral roots of Pkn(x) . The polynomial
Pkn (x) with an integer root r denes the triple (n; k; r). By virtue of the relations (12){
(14) the set of these triples is closed under action of the group of order eight generated
by two involutions: (n; k; r) $ (n; r; k) and (n; k; r) $ (n; k; n 0 r). So, it is enough to
point out representatives of the orbits.
Several innite families of integral roots of Krawtchouk polynomials are known. Evidently, for n even and k odd we always have integer n=2 to be a root. We call such a
root trivial. The known values of k for which there exists an nontrivial integer root for
innitely many n are
9
n03 n04 n05 n06 n08
;
;
;
;
;
2
2
2
2
2
k = 2; 3;
see [9, 16]. For k = 2 and 3 it can be found from (8). For k close to n=2 the following
lemma is useful:
Lemma 1
Let t = n 0 2k. Then
P t=
1. Pk (2i) = 0 i
01)j
j =0 (
[
2. Pk (2i + 1) = 0 i
2]
P t0 =
[(
j =0
k
i0 j
1) 2]
t
j
2
= 0;
(01)j i0k j
j
t
2 +1
= 0:
Using (12) and (1) one can see that to nd the even and odd zeros of Pk (x) one
should nd zero coecients with even and odd indices respectively of (1 0 z )k (1 + z )n0k =
(1 0 z )k (1 + z )t . Now the result follows from calculating the coecient at z i and z i
respectively.
Proof.
2
2
2 +1
2
The lemma yields that the nontrivial even roots can be found from the following equations
(for t > 3 reducing to Pell equations) :
1. t = 3 : r = n0 ;
1
4
0 8nr + n 0 2n = 0;
16r 0 12nr + 4r + n 0 4n + 3 = 0;
16r 0 16nr + n 0 6n + 8 = 0.
2. t = 4 : 8r
3. t = 5 :
4. t = 6 :
2
2
2
2
2
2
For the nontrivial odd roots:
1. t = 3 : r =
n+1 ;
4
0 20nr 0 4r + 5n + 3 = 0;
16r 0 16nr + 3n 0 2n + 8 = 0;
8r 0 8nr + n 0 2n + 16 = 0.
2. t = 5 : 16r
2
3. t = 6 :
2
4. t = 8 :
3
2
2
2
2
10
For other values of t we get either only the trivial root or a diophantine equation of degree
greater than two.
Here is the list of sporadic roots for n < 8400 (we hope it is complete in the range):
n
k
r
36
5
14
66
4
30
67
5
22
67
5
28
67
6
31
67
23
31
98
14
47
n
k
r
132
19
62
177
61
86
214
31
103
289
5
133
345
6
155
465
44
230
514
34
254
n
k
r
576
84
286
774
113
383
932 1029 1219 1219 1252
62
7 116 421 183
463 496 607 607 622
n 1521 3193 3362 4516 7172 7302 8361
k
4 1103 492 661 480 1069 798
r 715 1594 1679 2254 3583 3647 4178
Observe that in all the cases either k < 8 or t < 9, as well it is valid for the innite families
presented earlier. A partial explanation of this phenomenon is given in the following three
theorems.
[24] For each xed k 4; Pkn(x) can have nontrivial integer roots only for
nitely many n.
Theorem 2
[25] Let t > 6 be either an odd prime, a power of 2 or of the form 2pq , where
p is an odd prime, q is odd, and p does not divide q . Then for k = (n 0 t)=2; Pkn(x) can
possess nontrivial even roots only for nitely many n.
Theorem 3
Dene the polynomials
Vt (x; y ) =
t=
X
[
! j0
Y
t
(x 0 i)
(01)j
2j i
2]
1
j =0
=0
t=2]Y
0j 01
[
i=0
(y 0 i);
Vt3(x; y) = Vt (x; y) for t 6= 2 (mod 4); and Vt (x; y )=(x 0 y ) otherwise ;
Ut (x; y ) =
[(
t0
1)=2]
X
j =0
(0
1)j
! j0
t0 Y
= 0j 0
Y
t
(y 0 i):
(x 0 i)
2j + 1 i
i
1
=0
11
[(
1) 2]
=0
1
Ut3 (x; y) = Ut (x; y ) for t 6= 0 (mod 4); and Ut(x; y)=(x 0 y) otherwise ;
[25] Let k = (n 0 t)=2; t xed. There are at most nitely many n such that
has an even nontrivial root provided t 7 whenever Vt3 (x; y ) is irreducible over
n
3
Z[x; y ]; and Pk (x) has an odd nontrivial root provided t = 7 or t 9 whenever Ut (x; y )
is irreducible over Z[x; y].
Theorem 4
Pkn (x)
Multiplying the expressions from Lemma 1 by i!y!=k !, where y = (k +[t=2] 0 i) for
the even case and y = (k + [(t 0 1)=2] 0 i) for the odd case, and putting x = i, we get that
Pkn (x) has an even (resp. odd) root i Vt (x; y ) = 0 (resp. Ut (x; y) = 0). Observe also that
by the symmetry of Vt (x; y ) in x and y for t 2 (mod 4); x 0 y is a factor of Vt (x; y )
and gives the trivial root. Similarly, for t 0 (mod 4); x 0 y is a factor of Ut (x; y )
and gives the trivial root. Now the claim follows from the Runge theorem [34, 37] (the
extra condition - the polynomial is not a constant multiple of a power of an irreducible
polynomial - can be easily checked).
Proof.
2
Let us make some remarks on eectivity of the three theorems above. Theorems 2 and 3
are based on a result due to A. Shinzel [39] which does not provide an eective upper bound
on n. However, for the cases k = 4; 5; one can use an eective version of Siegel theorem
[43] due to A. Baker [1] (see also [40]). As usual the bounds occur to be enormously large.
For these cases the equations get the form:
k = 4; 3n
k = 5; 15n
2
2
0 6n((n 0 2r)
0 10n((n 0 2r)
2
2
+ 1) + (n 0 2r) + 8(n 0 2r) = 0;
4
2
+ 5) + (n 0 2r) + 20(n 0 2r ) + 24 = 0:
4
2
For the case k = 4 the following solutions (n; r ) , n > 8 and r < n=2 satisfy the equation
: (17; 7); (66; 30); (1521; 715); (15043; 7476) [14]. This case was studied in [16]. For k =
5; n > 10; the list of solutions is (17; 3);(36; 14);(67; 22);(67; 28);(289; 133);(10882; 5292).
Both lists are conjectured to be complete. Note also that, as it was pointed out in [14],
for k = 4 (it is valid as well for k = 5) the above diophantine equations possess innite
number of rational solutions.
The third theorem is based on the Runge theorem (for our purposes it is enough to use
a partial case presented in [34]). Eective bounds for this case were derived in [19, 20].
12
Irreducibility of Vt3 (x; y ) and Ut3 (x; y ) for small t could be checked directly. We conjecture
that actually they are always irreducible.
Of course, for k and t depending on n this method is not applicable.
We nish the section with some conjectures.
[9] Let k > 3 or t > 8 or t = 7. Then the number of nontrivial zeros of
for n < N is o(N ).
Conjecture 1
Pkn (x)
Apparently, the conjecture states that the described families of integral roots contain
almost all of them.
Conjecture 2
root.
For 3 < k < n=2; any Krawtchouk polynomial possesses an noninteger
For k < n=2 there exists an absolute constant c such that N (k) c. That
is the number of zero coecients in the expansion (1 0 z )i (1 + z )j ; i < j; does not exceed
c.
Conjecture 3
Actually, according to a numerical evidence we guess c to be 4 for n even, and 3 for n
odd. In this context it is worth mentioning [26, 6] where it was proven that the number
of zeros arising from binary nondegenerated recurrences with integer constant coecients
does not exceed 4 (see also [7, 36]). Unfortunately, in our case the recursion turns out to
be with linear coecients.
The following particular case of the above conjecture is of some importance for switching
reconstruction.
[25] The only integer zeros of Pkn (x); k =
and, m =2 for m 2 (mod 4).
Conjecture 4
m
2
; n = m ; are 2; m
2
2
0 2;
2
4
How many integral roots can occur ?
In this section we derive some bounds for the number of integral roots of Krawtchouk
polynomials . We start with the following simple
13
Theorem 5
N (k) min(k; n 0 2k ):
Notice that deg Pkn (x) = k , and the degrees of Ut (x; y ) and Vt (x; y ) in x are at
most [ t0 ]; [ t ]; respectively, t = n 0 2k . The result follows now from Lemma 1.
Proof.
1
2
2
2
For convenience we use change of variable y = n 0 2x, dening Qk (y) = Pk ( n0y ). Let
yi ; i = 1; . . . ; k; be its roots. Notice that they are symmetric with respect to zero. Hence
from (11) we have:
2
(
Qk (y ) =
Qk=2 2
1
yi2)
k! Qi=1 (y
(
k
0
1)
=
2
y
(y2 yi2)
k! i=1
0
0
k even
k odd
(30)
Note also, that if Pk (xi ) = 0 and xi is integral then
yi n (mod 2):
(31)
Here are a few values of Qk (y ) for small y 's:
0
(01)k kY
(n 0 2i);
Q k (0) =
(2k )!! i
(32)
Q k (0) = 0;
(33)
0
(01)k kY
(n 0 2i 0 1);
(2k )!! i
(34)
1
2
=0
2 +1
Q k (1) =
2
1
=0
Q k (1) =
2 +1
0
(01)k kY
(n 0 2i 0 1);
(2k )!! i
1
(35)
=0
kY
0
(01)k
(n 0 4k) (n 0 2i);
Q k (2) =
(2k )!!
i
1
2
(36)
=1
Qk
2 +1
k
(01)k Y
(n 0 2i);
(2) = 2
(2k )!! i
=1
14
(37)
Q k (3) =
2
kY
0
(01)k
(n 0 8k 0 1) (n 0 2i 0 1);
(2k )!!
i
(38)
kY
0
(01)k
(3n 0 8k 0 3) (n 0 2i 0 1);
(2k )!!
i
(39)
k
Y
(01)k
(n 0 4k 0 2) (n 0 2i):
(2k )!!
i
(40)
1
=1
Q k (3) =
2 +1
1
=1
Q k (4) = 4
2 +1
=2
In general we have
Lemma 2
For integer y the following relations hold:
Q k (2y ) = (01)k
2
y0
F y (k; n) k0Y
(n 0 2y 0 2i);
(2k )!! i
1
2
(41)
=0
Q k (2y + 1) = (01)k
2
y0
F y (k; n) k0Y
(n 0 2y 0 1 0 2i);
(2k )!!
i
1
2 +1
(42)
=0
Q k (2y) = (01)k
2 +1
0y
G y (k; n) kY
(n 0 2y 0 2i);
(2k)!! i
2
(43)
=0
Q k (2y + 1) = (0
2 +1
y 01
G2y +1 (k; n) k0Y
k
(n
1)
(2k)!!
i=0
0 2y 0 1 0 2i):
(44)
where Fj (k; n) and Gj (k; n) for xed j are polynomials in k and n, with integer coecients.
For k = o(n);
Fj (k; n) = n j= (1 + o(1)); G j (k; n) = nj 0 (1 + o(1));
[
2]
1
2
(45)
G j (k; n) = (2j + 1)nj (1 + o(1)):
2 +1
Using (1) one easily gets that the above relations hold for y = 0; 1; 2; 3. The
proof is accomplished by induction on y . We will demonstrate it for the rst case, i.e. for
Q k (2y).
Proof.
2
15
Rewriting (6) in terms of Qk (y ) and y we get:
(n 0 y )Qk (y + 2) = 2(n 0 2k )Qk (y) 0 (n + y )Qk (y 0 2):
By induction (omitting the common factor (01)k =(2k )!!) we get
(n 0 2y )F y (k; n)
2 +2
(n 0 2k)F y (k; n)
k=2Y
0y 01
2
i=0
k=2Y
0y 02
i=0
(n 0 2y 0 2 0 2i) =
(n 0 2y 0 2i) 0 (n + 2y )F y0 (k; n)
2
2
k=Y
20y
i=0
(n 0 2y + 2 0 2i):
After cancellation we obtain
F y (k; n) = (n 0 2k )F y (k; n) 0 (n + 2y )(n 0 2y + 2)F y 0 (k; n);
2 +2
2
2
2
that proves the claim.
For the other cases the proof is similar.
2
Dene for a integer the function E (a) as the maximum power of 2 dividing a.
Theorem 6
N (k ) is at most
1. for k and n even
1
(n=2 0 1)!
(5k + 4E (n=2 0 k ) + 6E (n=2) + 10E (
));
16
((n 0 k )=2)!
2. for k odd and n even
1
(n=2 0 2)!
(5k + 11 + 4E (n=2 0 1) + 6E (n=2 0 k) + 10E (
));
16
((n 0 k 0 1)=2)!
3. for k even and n odd
((n 0 1)=2))!
1
(k + 2E ((
));
3
((n 0 k 0 1)=2)!
16
4. for k and n odd
((n 0 1)=2)!
1
(k 0 1 + 2E ((
)):
3
((n 0 k )=2)!
We give a complete proof only for the rst case. In the other cases the arguments
are analogous. We put n = 2m and k = 2l . Furthermore, let Pkn(x) have 2s integral roots
62i; i = 1; . . . ; s (recall that i is integer by (31)). Then from (30)
Proof.
(2l )!Q l (y ) =
2
s
Y
(y
i=1
2
0 4i )R(y);
2
where R(y) is a polynomial with integer coecients. By (41) we have
(2l)!Q l (0) = (01)sR(0)
2
(2l)!Q l (2) = R(2)
2
s
Y
s
Y
i=1
0
(2l)! lY
(2m 0 2i);
(2l)!! i
(46)
lY
0
(2l )!
(2m 0 4l) (2m 0 2i):
(2l)!!
i
(47)
4i = (01)l
2
1
=0
(4 0 4i ) = (01)l
2
i=1
1
=1
Multiply the cube of (46) by the square of (47), and estimate the maximum power of 2
dividing the LHS and the RHS (in what follows, denoted by E (LHS ) and E (RHS )).
E (RHS ) = 3(l + E ((m)!) 0 E ((m 0 l)!)) + 2(l + E (m 0 2l ) + E ((m 0 1)!) 0 E ((m 0 l)!)) =
5l + 3E (m) + 2E (m 0 2l) + 5E ((m 0 1)!) 0 5E ((m 0 l)!):
For the LHS we have
E (LHS ) E ((2 s )
2
3
10s + E (
Note that a (1 0 a )
6
2
2
0
s
Y
i=1
s
Y
i=1
i ((2 s )
6
2
2
s
Y
(1 0 i ) ) =
i=1
i (1 0 i ) );
6
(mod 2 ) hence
6
17
2
2
2
2
E (LHS ) 16s;
and the sought result follows.
In other cases we consider correspondingly (Qk (2)) (Qk (4)) ; Qk (1)Qk (3) and Qk (1)Qk (3).
Using (41)-(40) and the arguments analogous to those above we get the claim.
2
3
2
Corollary 1
N (k ) is at most
1. for k and n even
5k 0 13
1
+ log n + log k;
8
4
(48)
1
5k 0 10
+ log n + log k;
8
4
(49)
2
4k 0 8 6
+ log n + log k ;
7
7
7
(50)
4k + 6 6
2
+ log n + log k:
7
7
7
(51)
2
2
2. for k odd and n even
2
2
3. for k even and n odd
2
2
4. for k and n odd
2
Proof.
2
We use inequalities
E (a) [log a];
(52)
a 0 [log a] 0 1 E (a!) a 0 1:
(53)
2
2
We will prove here only the rst case. The three others are similar. First we estimate
E 3 = 2E (
n
n
0
k) + 3E ( ):
2
2
18
Observe, that GCD( n 0 k; n ) k . That yields
2
E(
2
n
n
n
n
0
k ) + E ( ) log k + max(E ( 0 k ); E ( )):
2
2
2
2
2
If E ( n 0 k ) E ( n ) then
2
2
E 3 = 2(E (
n
n
n
n
n
0
k ) + E ( )) + E ( ) 3E ( ) + 2 log k 3[log ] + 2[log k];
2
2
2
2
2
2
2
2
otherwise, E ( n ) E (k) and
2
E 3 2E (
n
n
0
k ) + 3 log k 2[log ( 0 k )] + 3[log k ]:
2
2
2
2
2
Comparing the last two inequalities notice that the rst estimate is always greater. So,
E 3 3[log
2
n
] + 2[log k]:
2
2
Routine calculations lead now to (48).
2
The theorem states that for suciently large degrees k of Krawtchouk polynomials
( k n ! 1) the number of integer roots does not exceed 5/8 of the total amount. This
coecient may be improved in expense of the coecient at n using products of more than
two values of Qi (k). Actually, we did not try to get the best possible bound achievable
by this method since it does not seem possible to get to the coecient at k less than 0.5 .
log
Theorem 7
For k = o(n); N (k) is less than
1. for k and n even, and any = 2i k
2
01
s
q
3
k
2 + 2
k + log n + o( log n) + k log n + o( k log n);
4 + 1
4
2
8
2
2
2. for k odd and n even, and any = 2i k0
(54)
1
2
q
k 1q
k + log n + o( log n) + k log n + o( k log n);
2 0 1
4
2 2
2
2
19
(55)
3. for n odd, and any = 2i 0 1 [ k0 ]
2
2
q
k 1q
+1
k + log n + o( log n) + k log n + o( k log n):
2 + 1
4
2 2
2
2
(56)
We will prove the theorem for the case n = 2m + 1; k = 2l + 1. The other
cases are similar. Let 2i + 1; i = 1; . . . ; s; be integer roots of Qnk (y) corresponding to the
integral roots of Pkn(x) . Then from (45)
Proof.
(2l + 1)!Q
l
2 +1
(2z + 1) = (2z + 1)A(z )R(2z + 1) = (01)l
(2l + 1)!
(2z + 1)B (z );
(2l)!!
(57)
where
A(z ) =
s
Y
((2z + 1)
2
i=1
0 (2i + 1) );
2
B (z ) = (G z (l; n))z (1 + o(1))
2 +1
Pick some integer of the form 2i 0 1
Q
B 3( ) = z B (z ). Observe
lY
01
i= z
(2m 0 2i):
l 0 1, and consider A3 () = Qz A(z) and
=0
=0
A3( ) = 2 s 2 ( +1)
Evidently,
E(
t
Y
s Y
Y
i=1 z=0
(i 0 z )(i + z + 1):
(i 0 z )(i + z + 1)) E ((2 + 2)!) = 2 + 1;
z =0
thus
E (A3( )) 2s( + 1) + sE ((2 + 2)!) s(4 + 2):
Further,
E (B 3 ()) E ((1 + o(1)) (G z (l; n))
+1
0+1+...+
2 +1
20
lY
01
( (2m 0 2i)) ) =
i=0
+1
(by (45)
E ((1 + o(1))
(1 + o(1))
+1
Y
z
G z (l; n))2l (
( +1)
2 +1
m! )
(m 0 l)!
+1
( + 1)( + 2)
log n + l( + 1) + (l + log m)( + 1):
2
2
2
Hence, from E (A3 ()) E (B 3 ()) follows
s
+1
l + log n + o( log n):
2 + 1 8
2
Recalling that the number of integer roots equals 2s + 1 for this case, conclude that it
does not exceed
+1
k + log n + o( log n):
2 + 1
4
2
q
Choosing to be about k= log n we get the last inequality in (56).
2
2
p
For k growing faster than n we will derive bounds basing on another ideas.
Theorem 8
N (k ) 3 q3
2(k 0 1)(n 0 k + 2):
2
We start by showing that the number of pairs of integer roots at distance i apart
can not be too large.
Proof.
Let r < n=2 be an integer root of Pk (x) . Then, by (29), r + 1 is not a root. Dene
S (x) =
(n 0 r 0 1)!Pk (r + x)
:
(n 0 r 0 x)!Pk (r + 1)
Using (6) we conclude that S (x) satises the following recurrence:
S (x + 1) = (n 0 2k)S (x) 0 (x + r)(n 0 r 0 x + 1)S (x 0 1);
and S (0) = 0; S (1) = 1.
21
Considering S (x) as a polynomial in r it is easy to check that
d(x) = degr S (x) (
x 0 2 if x even;
x 0 1 if x odd .
Thus the number of integer roots r, such that there is another integer root at distance x
from r , does not exceed d(x). Hence, we may remove from the set of integer roots not
more than (d(x) + 1)=2 elements so that the distance
q x does not appear in the resulting
set. Let L denote rk (k ) 0 r (k ). From (23) L < 2 (k 0 1)(n 0 k + 2). Now remove
1
hX
01
i=3
!
0
1 h
d(i) + 1 1 hX
i<
<
2
2i
2 2
1
01
=3
roots so that in the resulting set the minimum distance becomes
at least h. The minimum
length of an interval containing this set is at least (N (k ) 0 h )h < L. Choosing h to be
(2L) = we obtain the claim.
1
2
2
1 3
2
Corollary 2
Proof.
N (k ) ( n) =
9
2 3
4
Follows from Theorem 5.
2
5
When there exists at least one noninteger root?
The question in the section's title is crucial for proofs of nonexistence of perfect codes.
The problem is far from being trivial. A signicant eort has been made to achieve
the goal. For nonbinary case (q 6= 2) nonsymmetry of Krawtchouk polynomials with
respect to (q 0 1)n=q was essentially exploited [2, 5, 18], see also [29, 47, 51]. In binary
case Pkn(x) is symmetric with respect to n=2 so the mentioned approach fails to work.
A.Tietavainen [48] overcame it using the sphere-packing condition. Thus for the binary
case the question is still open.
In the section we will derive several conditions for existence of noninteger root. Unfortunately, we were unable to give a complete answer.
First observe that if n and k are odd then n=2 is such a root. Furthermore, an immediate
consequence of Theorem 6 and Corollary 1 is
22
Theorem 9
root.
There exists a constant c such that for k > c log n; Pkn (x) has an noninteger
For example for n > 22 one may choose c = 3.
Now we will use a particular case of a result due to T. N. Shorey and R. Tijdeman [41]:
[41] Let > 0 and m > k . Then there exists an eectively computable
number c depending only on such that for k > c one can nd a prime p > (1 0 )k log log k
such that for some nonnegative integer l ; p l j mm0k and p l 6 j mm0k .
2
Theorem 10
!
2 +1
(
!
2 +2
)!
(
)!
This theorem is used to obtain the following
For n odd and k even, or n even and k even and suciently large (eectively computable and independent on n) , Pkn(x) possesses an noninteger root.
Theorem 11
Consider Qk (0) dened in (46). Assuming by the contrary that all the roots
y ; . . . ; yk ; of Qk (y ) are integer we have for n odd:
Proof.
1
k !Qk (0) =
k=
Y
k! k=Y0
(n 0 2i) = yi :
k!! i
i
2
2
1
2
=1
=0
The LHS is odd while, in contradiction, the RHS is even.
For n even we have:
!
k=
Y
n=2
k ! = yi :
k!Qk (0) =
k=2
i
2
2
=1
This yields that the number
k!
(n=2)!
(k=2)! ((n 0 k )=2)!
is a perfect square. If n is odd then LHS is not an integer. Otherwise, for k < 2n
Theorem 10 shows that this is impossible. For k 2n the result follows from Corollary
1.
2
2
2
Since for n and k odd n=2 is a noninteger root we get:
23
Corollary 3
For n odd Pkn(x) always have a noninteger root.
The case n even and k odd turns out the most dicult (at least for us). In this case
Qk (0) = 0, but one may consider its derivative (see (18)):
!
k0
X=
n=2 k !
i
0
(01)
k!Qk (0) =
;
(
1) 2
i
i=0
k 0 2i
that is obligatory a perfect square if all the roots are integer. Although we conjecture it
is never the case we failed to prove it. For this situation (as well as for the cases having
been considered in Theorem 11, but now for all k's) we give some partial results using
another method.
The following expressions for sums of powers of roots of Qk (y ) prove to be useful:
k
X
i=1
k
X
i=1
k
X
i=1
yi =
1
k (k 01)(30kn
15
yi =
1
k (k 0 1)(105n k
21
4
6
2
2
4
2
2
2
3
0 357n k + 315n 0 252n k
3
0 1680nk
3
2
3
2
+ 1449n k
2
2
0 6888nk + 3528n 0 60k
2496k + 5064k 0 5072k + 1984):
3
3
Theorem 12
(58)
2
045n 045k n+165kn0150n+18k 0102k +188k 0112);
3
1890n + 210nk
1
yi = k (k 0 1)(3n 0 2k + 4);
3
+ 5082nk
2
2
For
n 0 (mod 4) and k 3 (mod 4), or
n k 2 (mod 4); or
n 6 (mod 8) and k 4 (mod 8), or
n 0 (mod 8) and k 5 (mod 8),
24
0 2835n k+
2
5
+ 612k
4
0
(59)
(60)
Pkn (x) possesses an noninteger root.
Recall that if all the roots are integral then for n even all yi 's are also even. Now
P
consider ki yi . Then yi 0 (mod 4). Since the roots are symmetric with respect to
zero the sum is divisible by 8. Checking divisibility by 8 of the RHS of (58) we get the
rst two items. The other two (as well as the rst two) can be derived in the same fashion
from (60).
Proof.
2
2
=1
2
For the reasons not completely clear to us, (59) does not yield any additional congruences
in comparison with (58). Evidently, the process could be continued but the expressions
become unavoidably cumbersome. We wonder if one can tackle it in general.
Another conditions could be derived from considering divisors of Q k (0) and Q k (2).
2
q
2 +1
q
1. If for n and k both even, k < n=2; there exists a prime p n=2;
dividing one of the integers in the interval [ n0k + 1; n ] then Pkn(x) possesses an
noninteger root;
Theorem 13
2
2
q
2. If for n even and k odd there exists a prime p (k 0 1)(n 0 k + 2) + 2 dividing
one of the integers in the interval [ n0k ; n 0 1] then Pkn (x) possesses an noninteger
root.
+1
2
Proof.
2
In the rst case consider
k !Qk (0) = (02)k=
k
Y
k ! k=Y0
(n=2 0 i) = yi ;
k !! i
i
2
2
1
2
=0
=1
here k is even and assume all yi are integers. Suppose that the announced prime p does
exist. Then p must divide the RHS. Since at most one of the integers in the interval is
divisible by p this one must be divisible by p as well. Hence, p n=2, a contradiction.
2
2
2
In the second case we consider Qk (2). As above we see that p must
q divide either (yi =2 0 1)
or (yi =2 + 1) for some i. But from (23) we get that jyi =2 6 1j < (k 0 1)(n 0 k + 2) + 1.
2
Now we apply Theorem 6 to establish existence of noninteger root for the case when the
interval [ n0k ; n ] does not contain numbers divisible by too large powers of 2.
2
2
25
We extend the denition of function E (a) equal to the maximum degree of 2 dividing a
to arbitrary set A of integers by setting
E (A) = max E (a):
a
If A is an interval then E (
Lemma 3
Q
a
2A
2 A a) can be easily estimated through E (A).
Let A = [a; b], then
E(
b!
(a 0 1)!
) b 0 a 0 1 + E (A):
The number of numbers in A divisible by 2i does not exceed b0a
up by i from 1 to E (A) and taking the integer part we get the claim.
Proof.
i
+1+2
2
i
0 . Summing
1
2
Theorem 14
Pkn (x) has an noninteger root if
1. n and k > 8 even, and
n
n0k+2 n
3
18
0
kg [ [
; 0 1]) k 0 log k + ;
2
2
2
5
5
2. n even and k > 18 odd, and
E (f
E (f
Proof.
u=
E( n
2
2
n
n0k+1 n
3
0
kg [ [
; 0 1]) k 0 log k + 4:
2
2
2
5
2
We give a proof for the second case, the rst one can be proved similarly. Put
0 k [ [ n0k ; n 0 1]). We consider three subcases:
+1
2
2
1. 2u ja; a 2 A = [ n0k ; n 0 2].
Since 8b 2 A GCD( n 0 1; b) k0 and GCD( n 0 k; b) k 0 2; then E ( n 0 1) log (k 0 3)=2 and E ( n 0 k) log (k 0 2). Thus, the bound for N (k) given in
Theorem 6 can be estimated by Lemma 3 as follows:
+1
2
2
3
2
2
2
2
2
2
5
N (k) < (k + log k 0 4 + u);
8
which, by conditions of the theorem, does not exceed k .
2
26
2
2. 2u j( n 0 1).
Since GCD( n 0 1; n 0 k ) k 0 1 then E ( n 0 k) log (k 0 1). Let
for some l and u; then
2
2
2
2
2
n
2
0 1 = (2l +1)2u
k03
; (2l + 1)2u 0 1]:
2
So, if b = (2l + 1)2u 0 i 2 A; then E (b) = E (i). Therefore,
A = [(2l + 1)2u 0
E(
(n=2 0 2)!
k03
k05
) = E ((
)!) <
:
((n 0 k 0 1)=2)!
2
2
As in the previous subcase we get
1
N (k ) < (5k + 3 log k 0 7 + 2u);
8
which, by conditions of the theorem, does not exceed k .
2
3. 2u j( n 0 k ). This case is treated similarly, and we get
2
1
N (k ) < (5k + 7 log k 0 7 + 3u);
8
and again we are through.
2
2
Note that the restrictions on k in the statement of the theorem are of no importance since
for small k existence of noninteger roots can be checked directly.
We would like to mention other possible approaches to a proof of Conjecture 2. They are
based on an observation that if all the roots of Pk (x) are integral, then by (14)
fk (x) =
Pn0k (x)
Pk (x)
might be a polynomial of degree n 0 2k . Considering the expansion
fk (x) =
nX
02k
i=0
and, thus,
27
i Pi (x)
Pn0k (x) = Pk (x)
nX
02k
i=0
i Pi (x):
Employing (21) one gets a system of linear equations in i 's which seems to be incompatible, but we were unable to prove it. On the other hand, notice that fk (x), by assumption,
a polynomial of degree n 0 2k , takes alternatively on 61 in all n + 1 0 k integer points of
the interval [0; n] which are not the roots of Pk (x) . This also seems to be quite restrictive.
6
When there are no integral roots at all?
In many applications nonexistence of integral roots is of essential importance [14, 45].
In view of Theorems 2 { 4 the most interesting case of the problem arises when k (or t)
grows with n. Then the diophantine equations become nonpolynomial, and a few is known
about their solutions. On the other hand the generating function (1) of Krawtchouk
polynomials resembles that of binomial coecient. This suggests that some consideration
in the style of the famous Lucas theorem [31] can be useful. As far as we know L. Chihara
and D. Stanton were the rst to notice it [9]. In this section we try to push it further
along this line.
P
Let us recall the Lucas theorem. Digits of p-adic expansion of a number a = i ap (i)pi
will be denoted by ap (i) . If ap (i) bp (i) for all i then we will write it as a pb.
Theorem 15
[31, 15] For p prime
a
b
!
Y
i
ap (i)
bp (i)
!
(mod p):
(61)
(mod 2) i a b:
(62)
In particular,
a
b
Theorem 16
Proof.
!
6 0
2
N (k ) = 0 if n k.
2
(1 0 z )k (1 + z )n0k (1 + z )n (mod 2) and the result follows from (62).
2
28
Now we pass to modules being powers of 2. The following theorem is a slightly more
general version of Theorem 4.3 from [9].
Theorem 17
Let n k + 2m . Then
Pk (x)
Pk (x + 2m)
(mod 2m ):
(1 0 z)
(mod 2m );
+1
Using an easy identity
Proof.
(1 + z )
2
m
m
2
+1
we have
(1 0 z )k (1 + z )n0k 0 (1 0 z )k
(1 0 z )k (1 + z )n0k0 ((1 + z )
m
2
2
m
+2
m
(1 + z )n0k0
0 (1 0 z )
2
m
m
2
=
) 0 (mod 2m ):
+1
2
Let n k + 2m and let Pk (i) 6 0 (mod 2m ) for i = 0; . . . ; 2m 0 1; then
Pk (x) 6 0 (mod 2m ), and, therefore, N (k ) = 0.
+1
Corollary 4
+1
For instance, if modulo equals 4 one has to check values of Pk (0) = nk and Pk (1) =
n0 k n0 modulo 4. For powers of primes there exists a generalization of the Lucas
k k0
theorem [13]. So one can check in general the rst condition. What is about the second
one?
2
1
1
Already for modulo 3 the situation is much more complicated, and we do not know
necessary and sucient conditions for Pkn(x) 6 0 (mod 3). We give only partial results.
We will use the following identities for Krawtchouk polynomials modulo a prime p.
Lemma 4
For any integer s 0 and q = ps; p is a prime
Pkn(x)
Pkn0q (x) + Pkn00qq (x)
29
(mod p);
(63)
Pkn (x)
Pkn0q (x 0 q) 0 Pkn00qq (x 0 q)
(mod p);
(64)
Pkn (x)
0Pkn (x 0 q) + 2Pkn0q (x 0 q)
(mod p);
(65)
For p = 3:
Pkn (x)
Pkn0q (x 0 2q) + Pkn00 qq (x 0 2q)
2
2
(mod 3):
(66)
We prove (63). The other identities are derived using the same arguments. The
following series of identities is valid
Proof.
1
X
k=0
(Pkn(x)
0 Pkn0q (x))z k = (1 0 z)x(1 + z)n0x 0 (1 0 z)x(1 + z)n0x0q =
(1 0 z )x (1 + z )n0x0q ((1 + z )q 0 1) (1 0 z )x (1 + z )n0x0q zq =
1
X
q
z
Pkn0q (x)z k
k=0
=
Hence the generating function for Pkn(x)
p.
1
X
P n0q (x)zk
k=0
k 0q
(mod p):
0 Pkn0q (x) 0 Pkn00qq (x) is identically zero modulo
2
Now we are in a position to prove
Let k = ks3s + . . . + k ; and n = nl 3l + . . . + n ; be the ternary expansions
for k and n. Let, moreover, n k; and ki ; ni 2 f0; 2g; i = 0; . . . ; s: Then Pkn (x) 6 0
(mod 3) for all integer x, 0 x n.
Theorem 18
0
0
3
The proof is by induction on n. Notice that for small n 8 the claim is easily
checked. We will consider several cases.
Proof.
1. nl = 2. Assume by (13) that x < n=2. If necessary, replacing k by n 0 k (see (14))
we can suppose that s = l and kl = 2. Choose q = 3l . Then by (63):
30
Pkn0q (x) + Pkn00qq (x)
Pkn (x)
(mod 3):
Since n 0 q n=2, x n 0 q , the rst summand at the RHS is zero (follows from
(1)). The second term, after replacing k 0 q by (n 0 k + q ), satises the conditions
of the theorem, and the claim follows from the induction hypothesis.
2. nl = 1; kl = kl0 = 0. Choosing q = 3l0 , and assuming x n=2, analogously to
the previous case we get
1
1
Pkn0q (x)
Pkn (x)
(mod 3);
which proves the claim.
3. nl = 1; nl0 = 2; kl = 0; kl0 = 2.
1
1
(a) xl = 1. Then we use (64), q = 3l , and the proof is as above;
(b) xl = 0; xl0 = 1. Then we use (63), q = 3l , and the proof is as above;
1
(c) xl = 0; xl0 = 2. Choose q = 3l0 . By (66) we get:
1
1
Pkn(x)
Pkn0q (x 0 2q) + Pkn00 qq (x 0 2q)
2
2
(mod 3):
We claim that the rst summand vanishes. To demonstrate it apply recursively
(65) to it till x becomes 0. Observe that nal sum is the form
X
i Pkn0q 0 (0);
i
where i and i are all integers, and x . However, n 0 q 0 i does not
majorize k. Hence, by Theorem 15, it equals 0, and the sum vanishes, and
induction holds.
3
Since all the possibilities are exhausted the proof is complete.
2
For arbitrary p prime we have got the following
Let n = nl pl + . . . + n , k = ks ps + . . . + k , x = xl pl + . . . + x ; and
ni xi for i = 0; . . . ; s. Then
Theorem 19
0
Pkn(x)
0
s
Y
P ni (x )
i=0
31
ki
i
(mod p):
0
(67)
Proof.
Put mi = ni 0 xi . The following series of congruences hold due to (1):
n
X
P n(x)z s
s=0
s
= (1 0 z )x (1 + z )n0x =
(1 0 z )x0 (1 + z )m0 . . . (1 0 z )x p (1 + z )m p
l
l
l
l
(1 0 z )x0 (1 + z )m0 (1 0 z p )x1 (1 + z p )m1 . . . (1 0 z p )x (1 + z p )m =
l
1 n
l X
Y
j
j
ip
Pi j (xj )zip )
Pi (xj )z )(
(
j =s+1 i=0
j =0 i=0
nj
s X
Y
nj
l
l
l
(mod p);
Comparing the coecients at z k , and taking into account that P n (x) = 1, we obtain
the result.
i
0
2
This theorem easily permit tackling the case n 01 (mod ps ), since the majorization in the claim evidently holds for all 0 x n. For such n; k must consist
from those digits for which Pkp0 (x) 6 0 (mod p). For small p we present the
explicit list.
+1
1
i
Corollary 5
Let n 01 (mod ps ); k < ps. Then N (k ) = 0 if
(a) p = 3 and ki 2 f0; 2g ;
(b) p = 5 and ki 2 f0; 4g ;
(c) p = 7 and ki 2 f0; 2; 4; 6g ;
(d) p = 11 and ki 2 f0; 2; 4; 6; 8; 10g ;
(e) p = 13 and ki 2 f0; 4; 8; 12g ;
(f) p = 17 and ki 2 f0; 4; 12; 16g ;
(g) p = 19 and ki 2 f0; 2; 4; 6; 8; 10; 12; 14; 16; 18g;
(h) p an arbitrary prime and ki 2 f0; p 0 1g.
Note that in the list of possible values for ki we do not meet odd digits since for
odd ki ; Pkn (x) has zero at x = (p 0 1)=2.
Further investigations of the values of Krawtchouk polynomials modulo primes
seem to be promising. Let us state some conjectures:
i
i
32
The conditions of Theorem 18 are actually necessary and sucient
6 0 (mod 3) for all integral x.
Conjecture 5
for
Pkn(x)
Dene Hk (y ) = k !Pk ( n0y ).
2
Conjecture 6
Hka r (y ) (Ha (y))k Hr (y ) (mod a):
+
In connection with the last conjecture, notice that for p being a prime we have
Hp (y ) = p!Qp (y) 0 (mod p)
for all y n (mod 2). This follows from the fact that Qp(y ) takes on integer values
for such y 's. Since the degree of Hp (y ) is exactly p then Hp (y ) yp 0 y (mod p).
It is also an open problem whether the conditions of Theorem 19 could be weakened.
7
Distribution of zeros
In this section we will study the graph of Krawtchouk polynomials between consecutive zeros (integer or noninteger). Our approach is similar to that of [5] (see
also [32, 22, 35, 38, 42, 44, 49]). We will use essentially the dierence equation (6)
in our estimates. In the sequel a! for noninteger a should be understood as the
gamma-function.
We need the following auxiliary
Lemma 5
Let L(x) be the solution of equation
L(x + 2) 0 L(x + 1) + L(x) = 0;
with initial conditions L(0) = 0; L(1) = 1, where is a constant.
Then the solution of
F (x + 2) 0 F (x + 1) + F (x) = G(x)
is given by
F (x) = L(x) +
xX
01
i=0
33
L(x 0 1 0 i)G(i):
(68)
Proof.
Routine.
2
Note that for 1=4,
L(x) = x0
(
=
1) 2
sin x
;
sin (69)
where
= () = arccos
p1 :
(70)
2 Let t stand, as above, for n 0 2k , and r be a root of Pkn(x) . Dene
f (x) =
(n 0 r 0 1)!t 0x Pk (r + x)
;
(n 0 r 0 x)! Pk (r + 1)
1
g (x) =
(r 0 1)!t 0x Pk (r 0 x)
:
(r 0 x)! Pk (r 0 1)
1
Recall that Pk (r 6 1) 6= 0.
Put
(r + 1)(n 0 r)
r(n 0 r + 1)
; (r ) = =
:
t
t
From (6) by substitution we get
(r) = =
2
2
f (x + 2) = f (x + 1) 0
(r + x + 1)(n 0 r 0 x)
f (x) =
t
2
f (x + 1) 0 f (x) 0
(71)
x(n 0 2r 0 1 0 x)
f (x);
t
2
f (0) = 0; f (1) = 1;
from (23) one can get > 1=4 for k n=2;
g (x + 2) = g (x + 1) 0
(r 0 x)(n 0 r + x + 1)
g (x) =
t
2
34
(72)
x(n 0 2r + 1 0 x)
g (x);
t
g (x + 1) 0 g (x) +
2
g(0) = 0; g(1) = 1:
Similarly, using (23) and (29) one gets > 1=4 for k n=2 providing r is not the
rst root.
Let ri ; ri be two consecutive roots of Pk (x) . We use (68) to establish
+1
Theorem 20
Let ri < n=2. Then
q
()x
jPk (ri + x)j < ( (ri + 1)(n 0 ri) )x0 ((nn 00 rri 00 x1)!)! sin
jP (r + 1)j;
sin () k i
i
1
provided 0 x ri
+1
0 ri;
q
( )x
jP (r 0 1)j;
jPk (ri 0 x)j > ( ri (n 0 ri 0 1) )x0 ((rri 00 x1)!)! sin
sin ( ) k i
i
+1
1
+1
+1
(73)
+1
+1
(74)
+1
provided 0 x sin
;r
( ) i+1
< n=2.
Moreover,
sin ( )
ri 0 ri sin () :
(75)
+1
W.l.o.g. assume that Pk (x) is nonnegative on the interval [ri ; ri ]. Applying (68) to (71) with
Proof.
+1
G(x) = 0
x(n 0 2ri 0 1 0 x)
;
t
2
we conclude that
L(x) = x0
(
=
1) 2
sin ()x
sin ()
f (x);
whenever f (x) 0; n 0 2ri 0 x 0; sin ()x 0; simultaneously. The rst two
inequalities evidently hold in the range [ri ; ri ]. The third one holds since assuming
the contrary we would get that sin ()x vanishes somewhere within (ri ; ri ), say in
+1
+1
35
ri + x3 , and, thus, for some x < x3 ; f (x) > L(x), a contradiction. Now (73) follows
from the denition of f . The proof of (74) is similar, note only that ri 0 > ri .
The estimate (75) follows from (73) and (74).
+1
( )
2
Note that the estimate (74) is valid only for a part of [ri ; ri ]. A lower
estimate valid for all the interval is easily obtained by replacing
Remark 1
+1
x(n 0 2ri 0 1 0 x)
t
in (71) by its maximum on the interval, and applying (68).
2
Corollary 6
ward n=2.
Proof.
For k < n=2 the distance between the consecutive roots decreases to-
Let ri0 < ri < ri . From (75) we have
1
+1
ri
+1
0 ri < ((r )) ;
i
:
((ri ))
ri 0 ri0 >
1
Using the denitions of ; and observe that it is enough to check > for
ri < n=2. The last trivially holds.
2
The minimal distance d between consecutive roots can be easily estimated using
the previous corollary and (75). We need the following simple inequality (the proof
is routine):
min
1
1
1 x
0
<
< ;
2x 8 arccos(1 0 2x ) 2x
(76)
2
provided jxj < p .
1
2
Corollary 7
For k 9;
rn
>d
2 k
min
>
s
arccos pnn0n k
2
(
36
+2)
p
n+1 2
>
0 16 :
4k + 2
(77)
We consider separately the cases of k being odd and even. For k odd we
have using (76):
Proof.
d
min
>
p
s
arccos pnn0n k
2
(
n+1 2
0 16 :
>
>
n
0
k
4k + 2
arccos n
2
+2)
+1
For deriving the upper bound observe that the minimal distance can not exceed the
average one, that is by (23)
q
d
min
s
2 (k 0 1)(n 0 k + 2)
n
< dav <
<2
:
k01
k01
Then by (75) we get
d
min
<
arccos p n0 d n0 k n
2
(
Replacing d
min
2
min +2)(
+2
dmin )
:
by dav one can check for k > 8 the RHS does not exceed
;
arccos n0n k
2
and the result follows from (76).
For k even the upper bound is derived similarly to the odd case. For the lower
bound we replace in (71), by virtue of (68), the last term by its maximum, getting:
(d 0 1)
= f (x + 1) + f (x);
f (x + 2) > f (x + 1) 0 f (x) 0
2
2
where
=
(
n+1)2
2 .
4t
Now, acting similarly to the proof of Theorem 20 we get
d
min
>
arccos
p
1
2
and the result follows from (76).
Corollary 8
toward n=2.
0
=
;
arccos nn0 k
2
+1
2
For k < n=2 the absolute values of local extrema of Pk (x) decrease
37
Proof.
have
Let r < n=2; be a root (not the rst one) of Pk (x) . From (71) and (72) we
jPk (r + x)j < ((nn 00 rr 00 x1)!)! ((r + 1)(n 0 r)) x0
(
jPk (r 0 x)j > ((rr 00 x1)!)! (r(n 0 r + 1)) x0
(
=
1) 2
=
1) 2
sin ()x
jP (r + 1)j;
sin () k
(78)
sin ( )x
jP (r 0 1)j:
sin ( ) k
(79)
We need the following simple inequality (for l > 1):
(a(a 0 l)) l
=
( +1) 2
< a(a 0 1) . . . (a 0 l) < (a 0 l=2)l :
+1
Then we have
(n 0 r 0 x)!
((r + 1)(n 0 r)) x0
(n 0 r 0 1)!
(
=
<(
1) 2
(r + 1)(n 0 r )
) x0
(n 0 r 0 1)(n 0 r 0 x + 1)
(
One can see that the RHS is less than 1. Thus,
k (r + 1)j
jPk (r + x)j < jPsin
:
()
Similarly,
(r 0 x)!
(r(n 0 r + 1)) x0
(r 0 1)!
(
=
1) 2
>(
r(n 0 r + 1) x0
) > 1:
(r 0 x=2)
1
2
Hence, on the interval between r and the previous root
max jPk (r 0 x)j >
jPk (r 0 1)j :
sin ( )
From (6)
jPk (r 0 1)j = n 0r r jPk (r + 1)j > jPk (r + 1)j:
Moreover,
sin () > sin ( );
and the result follows.
38
=
1) 2
:
2
Dene the following linear transformation
(x ; . . . ; xm ) = (x
1
1
0 x ; x 0 x ; . . . ; xm0 0 xm):
2
2
3
1
We end the section with the following
Let y < y < . . . < y k = be the rst half of the roots of Qk (y) =
. Then all the components of the vectors
Conjecture 7
Pk
( n0 y )
2
1
2
[( +1) 2]
i (y ; . . . ; y
1
k
=
[( +1) 2]
); i = 1; . . . ; [
k01
];
2
are negative.
8
Acknowledgement
We are grateful to Laurent Habsieger for helpful comments. We have also received
from him a proof of Conjecture 6.
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