A Generalization of the Digital Binomial Theorem

0.19 MB·English

by  Hieu D. Nguyen

A GENERALIZATION OF THE DIGITAL BINOMIAL THEOREM
HIEUD.NGUYEN
Abstract. We prove a generalization of the digital binomial theorem by constructing a one-parameter
subgroup of generalized Sierpinski matrices. In addition, we derive new formulas for the coefficients of
5
Prouhet-Thue-Morse polynomials and describe group relations satisfied by generating matrices defined in
1
termsoftheseSierpinskimatrices.
0
2

n
a
1. Introduction
J

4 The classical binomial theorem describes the expansion of (x +y)N in terms of binomial coefficients,
2
namely for any non-negative integer N, we have

] N N
T (x+y)N = xkyN−k, (1)
N k
k=0(cid:18) (cid:19)
X
h. where N are defined in terms of factorials:
k
t
a (cid:0) (cid:1) N N!
m = .
k k!(N −k)!
(cid:18) (cid:19)
[
In [9], the author introduced a digital version of this theorem where the exponents appearing in (1) are

1 viewed as sums of digits. To illustrate this, consider the binomial theorem for N =2:
v
8 (x+y)2 =x2y0+x1y1+x1y1+x0y2. (2)
9
It is easy to verify that (2) is equivalent to
9
5 (x+y)s(3) =xs(3)ys(0)+xs(2)ys(1)+xs(1)ys(2)+xs(0)ys(3), (3)
0
. where s(k) denotes the sum of digits of k expressed in binary. For example, s(3) = s(1·21+1·20) = 2.
1
0 More generally, we have
5
Theorem 1 (Digital Binomial Theorem [9]). Let n be a non-negative integer. Then
1
v: (x+y)s(n) = xs(m)ys(n−m). (4)
Xi 0≤Xm≤n
(m,n−m) carry-free
r
a Here, a pair of non-negative integers (j,k) is said to be carry-free if their addition involves no carries when
performedinbinary. Forexample, (8,2)is carry-freesince 8+2=(1·23+0·22+0·21+0·20)+1·21=10
involves no carries. It is clear that (j,k) is carry-free if and only if s(j)+s(k) = s(j+k) (see [2, 9]. Also,
observe that if n=2N −1, then (4) reduces to (1).
In this paper we generalize Theorem 1 to any base b≥2. To state this result, we shall need to introduce
a digital dominance partial order on N as defined by Ball, Edgar, and Juda in [2]. Let
n=n b0+n b1+...+n bN−1
0 1 N−1
represent the base-b expansion of n and denote d :=d (n)=n to be the i-th digit of n in base b. We shall
i i i
say that m is digitally less than or equal to n if m ≤ n for all i = 0,1,...,N −1. In that case, we shall
i i
write m(cid:22)n. We are now ready to state our result.
Date:1-23-2015.
2010 Mathematics Subject Classification. Primary11.
Keywords and phrases. binomialtheorem;Sierpinskitriangle;Prouhet-Thue-Morsesequence.
1
Theorem 2. Let n be a non-negative integer. Then
N−1 N−1 N−1
x+y+d (n)−1 x+d (m)−1 y+d (n−m)−1
i i i
= . (5)
d (n) d (m) d (n−m)
i=0 (cid:18) i (cid:19) 0≤m(cid:22)n"i=0 (cid:18) i (cid:19) i=0 (cid:18) i (cid:19)#
Y X Y Y
Let µ (n):=µ(b)(n) denote the multiplicity of the digit j >0 in the base-b expansion of n, i.e.,
j j
µ (n)=|{i:d (n)=j}|.
j i
As a corollary,we obtain
Corollary 3. Let n be a non-negative integer. Then
b−1 x+y+j−1 µj(n) b−1 x+j−1 µj(m)b−1 y+j−1 µj(n−m)
= .
j  j j 
j=0(cid:18) (cid:19) 0≤m(cid:22)n j=1(cid:18) (cid:19) j=1(cid:18) (cid:19)
Y X Y Y
 
Observe that if b=2, then Corollary 3 reduces to Theorem 1.
The sourcebehind Theorem1 is a one-parametersubgroupof Sierpinski matrices,investigatedby Callan
in [3], which encodes the digital binomial theorem. To illustrate this, define a sequence of lower-triangular
matrix functions S (x) of dimension 2N ×2N recursively by
N
1 0
S (x)= , S (x)=S (x)⊗S (x), (6)
1 x 1 N+1 1 N
(cid:18) (cid:19)
where ⊗ denotes the Kroneckerproductoftwo matrices. For example, S (x) and S (x) canbe computed as
2 3
follows:
1 0 0 0
1·S (x) 0·S (x) x 1 0 0
S2(x)=S1(x)⊗S1(x)= x·S1(x) 1·S1(x) = x 0 1 0 ,
1 1
(cid:18) (cid:19)
 x2 x x 1 
 
 
1 0 0 0 0 0 0 0
x 1 0 0 0 0 0 0
 
x 0 1 0 0 0 0 0
S (x)=S (x)⊗S (x)= 1·S2(x) 0·S2(x) = x2 x x 1 0 0 0 0 .
3 1 2 x·S (x) 1·S (x)  x 0 0 0 1 0 0 0 
(cid:18) 2 2 (cid:19)  
 x2 x 0 0 x 1 0 0 
 
 x2 0 x 0 x 0 1 0 
 
 x3 x2 x2 x x2 x x 1 
 
 
Observe that when x=1, the infinite matrix S =lim S (1) is of course Sierpinski’s triangle.
N→∞ N
In [3], a formula for the entries of S (x) = (α (j,k,x)), 0 ≤ j,k ≤ 2N −1, is given in terms of the
N N
sum-of-digits function:
xs(j−k), if 0≤k ≤j ≤2N −1 and (k,j−k) are carry-free
α (j,k,x)= (7)
N 0, otherwise.
(cid:26)
Moreover,itwasproventhatS (x)formsaone-parametersubgroupofSL(2N,R),i.e.,thegroupof2N×2N
N
real matrices with determinant one. Namely, we have
S (x)S (y)=S (x+y). (8)
N N N
If we denote the entries of S (x)S (y) by t (j,k), then the equality
N N N
2N−1
t (j,k)= α (j,i,x)α (i,k,y)=α (j,k,x+y)
N N N N
i=0
X
2
corresponds precisely to Theorem 1 with j =0 and k =0. For example, if N =2, then (8) becomes
1 0 0 0 1 0 0 0 1 0 0 0
x 1 0 0 y 1 0 0 x+y 1 0 0
  = .
x 0 1 0 y 0 1 0 x+y 0 1 0
 x2 x x 1  y2 y y 1   (x+y)2 x+y x+y 1 
    
    
The rest of this paper is devoted to generalizing Callan’s construction of Sierpinski matrices to arbitrary
bases and considering two applications of them. In Section 2, we use these generalized Sierpinski matrices
to prove Theorem 2. In Section 3, we demonstrate how these matrices arise in the construction of Prouhet-
Thue-Morsepolynomialsdefinedin[10]. InSection4,wedescribeagrouppresentationintermsofgenerators
definedthroughthesematricesandshowthatthesegeneratorssatisfyarelationthatgeneralizesE.Ferrand’s
result in [4].
2. Sierpinski Triangles
To prove Theorem 2, we consider the following generalization of the Sierpinski matrix S (x) in terms of
N
binomial coefficients. Define lower-triangularmatrices S (x) of dimension bN ×bN recursively by
b,N
1 0 0 ... 0
x 1 0 ... 0
Sb,1(x)= x(cid:0)+21(cid:1)1 x1 1 ... 0 = x+jj−−kk−1 , if 0≤k ≤j ≤b−1
 (cid:0) ... (cid:1) (cid:0)...(cid:1) ... ... ...  (cid:26) (cid:0) 0, (cid:1) otherwise.
 
 x+b−2 x+b−3 x+b−4 ... 1 
 b−1 b−2 b−3 
 
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
and for N >1,
S (x)=S (x)⊗S (x).
b,N+1 b,1 b,N
Example 4. To illustrate our generalized Sierpinksi matrices, we calculate S (x) and S (x):
3,1 3,2
1 0 0
S (x)= x 1 0 ,
3,1  x+11 x 1 
(cid:0)2(cid:1) 1
S (x)=S (x)⊗S (x) 
3,2 3,1(cid:0) (cid:1) (cid:0)3,1(cid:1)
1 0 0 0 0 0 0 0 0
x 1 0 0 0 0 0 0 0
 x+11 x 1 0 0 0 0 0 0 
(cid:0)2(cid:1) 1
 x 0 0 1 0 0 0 0 0 
= (cid:0)x1x(cid:1) (cid:0)x(cid:1) 0 x 1 0 0 0 0 .
 x1(cid:0)x(cid:1)+11 x1x x x+11 x 1 0 0 0 
 1(cid:0) (cid:1)(cid:0)2(cid:1) 1(cid:0) (cid:1)1 1 (cid:0)2(cid:1) 1 
 x+1 0 0 x 0 0 1 0 0 
 (cid:0) (cid:1)(cid:0)2 (cid:1) (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) 1 (cid:1) (cid:0) (cid:1) 
 x+1 x x+1 0 x x x 0 x 1 0 
 (cid:0)2 (cid:1)1 2 1(cid:0) (cid:1)1 1 1 
 x+1 x+1 x+1 x x+1 x x+1 x x x x+1 x 1 
 (cid:0)2 (cid:1)(cid:0)2(cid:1) (cid:0)2 (cid:1)1 2 1(cid:0) (cid:1)(cid:0)2(cid:1) 1(cid:0) (cid:1)1 1 (cid:0)2(cid:1) 1 
 
We now generalizeC(cid:0)allan(cid:1)(cid:0)’s res(cid:1)ult(cid:0)for(cid:1)S(cid:0) (cid:1)(x)(cid:0)by p(cid:1)rese(cid:0)n(cid:1)ti(cid:0)ng a(cid:1)for(cid:0)m(cid:1)u(cid:0)la(cid:1)for(cid:0)th(cid:1)e e(cid:0)ntrie(cid:1)s o(cid:0)f S(cid:1) (x) (see [7] for
N b,N
a similar generalization of Callan’s result but along a different vein).
Theorem 5. Let α (j,k):=α (j,k,x) denote the (j,k)-entry of S (x). Then
N N b,N
x+d0−1 x+d1−1 ··· x+dN−1−1 , if 0≤k ≤j ≤bN −1 and k (cid:22)j
d0 d1 dN−1
α (j,k)= (9)
N  (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)
 0, otherwise.
where j−k =d b0+db1+...+d bL is the base-b expansion of j−k, assuming j ≥k.
0 1 L
3
Proof. We argue by induction on N. It is clear that (9) holds for S (x). Next, assume that (9) holds
b,1
for S (x) and let α (j,k) be an arbitrary entry of S (x), where pbN ≤ j ≤ (p+1)bN −1 and
b,N N+1 b,N+1
qbN ≤ k ≤ (q +1)bN −1 for some non-negative integers p,q ∈ {0,1,...,b−1}. Set j′ = j −pbN and
k′ =k−qbN. We consider two cases:
Case 1. p<q. Then j ≤k and α (j,k)=0·α (j′,k′)=0.
N+1 N
Case 2. p≥q. Then j ≥k and
x+p−q−1
α (j,k)= α (j′,k′). (10)
N+1 N
p−q
(cid:18) (cid:19)
Let j−k =d b0+d b1+...+d bN, where d =p−q. Then j′−k′ =d b0+d b1+...+d bN−1. By
0 1 N N 0 1 N−1
assumption,
x+d0−1 x+d1−1 ··· x+dN−1−1 if 0≤k′ ≤j′ ≤bN −1 and k′ (cid:22)j′,
α (j′,k′)= d0 d1 dN−1 (11)
N  (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)
 0 otherwise.
Since k (cid:22)j if and onlyif k′ (cid:22)j′, it follows from (10) and (11) that
x+d0−1 x+d1−1 ··· x+dN−1 if 0≤k ≤j ≤bN+1−1 and k (cid:22)j,
d0 d1 dN
α (j,k)= (12)
N+1  (cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)
 0 otherwise.
Thus, (9) holds for Sb,N+1. (cid:3)
Next, we show that S (x) forms a one-parameter subgroup of SL(bN,R). To prove this, we shall need
b,N
the following two lemmas; the first is due to Gould [6] and the second follows easily from the first through
an appropriate change of variables.
Lemma 6 (Gould [6]).
n
x+k y+n−k x+y+n+1
= . (13)
k n−k n
k=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
X
Proof. In [6], Gould derives (13) as a special case of a generalizationof Vandemonde’s convolution formula.
We shall proof (13) more directly by presenting two different proofs, one using a combinatorial argument
and the other using the beta function.
Combinatorial argument. Let A, B, and C = {0,1,...,n} denote three sets containing x, y, and n+1
elements(alldistinct),respectively,wherenisanon-negativeinteger. Foranynon-negativeintegerk,define
A =A∪{0,...,k−1}andB =B∪{k+1,...,n}. Thengivenanyn-elementsubsetS ofA∪B∪C,there
k k
exists a unique integer k in C−S, called the index of S with respect to A and B, such that |S∩A |=k
S kS S
and |S∩B |=n−k . To see this, define S =A∩S, S =B∩S, and T =C−S. We begin by deleting
kS S A B
|S | consecutive elements from T, in increasing order and beginning with its smallest element, to obtain a
A
subset T′. We then delete |S | consecutive elements from T′, in decreasing order and beginning with its
B
largest element, to obtain a subset T′′, which must now contain a single element denoted by k . It is now
S
clear that |S∩A |=k and |S∩B |=n−k .
kS S kS S
To prove (13), we count the n-element subsets S of A∪B∪C in two different ways. On the one hand,
since |A∪B∪C|=x+y+n+1, the number of such subsets is givenby x+y+n+1 . On the other hand, we
n
partition all such n-element subsets into equivalence classes according to each subset’s index value. Since
(cid:0) (cid:1)
|S∩A |=k and |S∩B |=n−k, it follows that the number of n-element subsets S having the same index
k k
k is given by x+k y+n−k and total number of n-element subsets is given by
k n−k
(cid:0) (cid:1)(cid:0) (cid:1) n
x+k y+n−k
.
k n−k
k=0(cid:18) (cid:19)(cid:18) (cid:19)
X
Lastly, we equate the two answers to obtain (13).
4
Analytic argument. Recall that the beta function is defined as B(x,y) = Γ(x)Γ(y)/Γ(x+y), where Γ(x) is
the gamma function. If x is a non-negative integer, then Γ(x+1)=x! and
(x−1)!(y−1)!
B(x,y)= .
(x+y−1)!
For convenience, we shall write x! to represent Γ(x+1) even when x is not non-negative. We now divide
(13) by n!x!y!/(x+y+n+1)! to obtain the following identity:
n
n
B(x+k+1,y+n−k+1)=B(x+1,y+1). (14)
n−k
k=0(cid:18) (cid:19)
X
But (14) is a generalizationof the following classical property of the beta function:
B(x,y−1)+B(x−1,y)=B(x−1,y−1). (15)
It is now straightforwardto prove (14) using an induction argument. (cid:3)
Lemma 7. Let p and q be positive integers with q ≤p. Then
p
x+p−v−1 y+v−q−1 x+y+p−q−1
= . (16)
p−v v−q p−q
v=q(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
X
Proof. Set w =v−q. Then (16) can be rewritten as
p−q
x+p−q−w−1 y+w−1 x+y+p−q−1
= ,
p−q−w w p−q
w=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
X
which follows from Lemma 6. (cid:3)
Theorem 8. For all N ∈N,
S (x)S (y)=S (x+y). (17)
b,N b,N b,N
Proof. We argue by induction on N. Lemma 6 proves that (17) holds for S (x). Next, assume that (17)
b,1
holds for S (x). Let t (j,k) denote the (j,k)-entry of S (x)S (y). Then
b,N N+1 b,N+1 b,N+1
bN+1−1
t (j,k)= α (j,m,x)α (m,k,y)
N+1 N+1 N+1
m=0
X
b−1bN−1
= α (j,vbN +r,x)α (vbN +r,k,y).
N+1 N+1
v=0 r=0
X X
As before, assume pbN ≤ j ≤ (p+1)bN −1 and qbN ≤ k ≤ (q+1)bN −1 for some non-negative integers
p,q ∈{0,1,...,b−1}. Set j′ =j−pbN and k′ =k−qbN. Again, we consider two cases:
Case 1: j <k. Then α (j,k,x+y)=0 by definition. On the other hand, we have
N+1
j bN+1−1
t (j,k)= α (j,m,x)α (m,k,y)+ α (j,m,x)α (m,k,y)
N+1 N+1 N+1 N+1 N+1
m=0 m=j+1
X X
j bN+1−1
= α (j,m,x)·0+ 0·α (m,k,y)
N+1 N+1
m=0 m=j+1
X X
=0
and thus (17) holds.
Case 2: j ≥k. Since S (x)=S (x)⊗S (x), we have
b,N+1 b,1 b,N
x+p−v−1 α (j′,r,x) if j ≥vbN +r,
α (j,vbN +r,x)= p−v N
N+1 0 if j <vbN +r.
(cid:26) (cid:0) (cid:1)
5
Similarly, we have
y+v−q−1 α (r,k′,y) if k ≤vbN +r,
α (vbN +r,k,y)= v−q N
N+1 0 if k >vbN +r.
(cid:26) (cid:0) (cid:1)
It follows that
p bN−1
x+p−v−1 y+v−q−1
t (j,k)= α (j′,r,x)α (r,k′,y)
N+1 N N
p−v v−q
v=q r=0 (cid:18) (cid:19)(cid:18) (cid:19)
X X
p bN−1
x+p−v−1 y+v−q−1
= α (j′,r,x)α (r,k′,y)
N N
p−v v−q
"v=q(cid:18) (cid:19)(cid:18) (cid:19)# r=0
X X
x+y+p−q−1
= α (j′,k′,x+y)
N
p−q
(cid:18) (cid:19)
=α (j,k,x+y),
N+1
where we have made use of the inductive assumption and Lemma 7. This proves that (17) holds. (cid:3)
As a corollary, we obtain Theorem 2, which we now prove.
Proof of Theorem 2. Let j =n and k =0. Then the identity
bN−1
α (j,m,x)α (m,k,y)=α (j,k,x+y),
N N N
m=0
X
which follows from (17), is equivalent to (5). (cid:3)
We end this section by describing the infinitesimal generator of S (x). Define X (x)=(χ ) to be a
b,N b,1 j,k
strictly lower-triangularmatrix whose entries χ are given by
j,k
x/(j−k), if j ≥k+1
χ = (18)
j,k 0, otherwise.
(cid:26)
For N >1, we define matrices
X (x)=X (x)⊕X (x)=X (x)⊗I +I ⊗X (x),
b,N+1 b,1 b,N b,1 bN b b,N
where⊕ denotes the KroneckersumandI denotes the bN×bN identity matrix. Observethat X (x) has
bN b,1
the following matrix form:
0 0 0 ... 0
x 0 0 ... 0
 
x/2 x 0 ... 0
Xb,1(x)= x/3 x/2 x ... 0  (19)
 
 ... ... ... ... ... 
 
 x/(b−1) x/(b−2) x/(b−3) ... 0 
 
 
The following lemmas will be needed. The first states a useful identity involving the unsigned Stirling
numbers of the first kind, c(n,k), defined by the generating function
n
x(x+1)...(x+n−1)= c(n,k)xk.
k=0
X
It is well known that c(n,k) counts the number of n-element permutations consisting of k cycles.
Lemma 9. Let l and n be positive integers with l≥n. Then
l−n+1
l
(i−1)! c(l−i,n−1)=nc(l,n). (20)
i
i=1 (cid:18) (cid:19)
X
6
Proof. We give a combinatorialargument. Let A={1,2,...,l}. We count in two different ways the number
of permutations π = σ σ ···σ of A consisting of n cycles where we distinguish one of the cycles σ of
1 2 n j
π. On the one hand, since there are c(l,n) such permutations π and n ways to distinguish a cycle of π, it
follows that the answer is given by nc(l,n). On the other hand, we can construct π by first choosing our
distinguished cycle σ consisting of i elements. The number of possibilities for σ is l (i−1)! since there
1 1 i
are l ways to choose i elements from A and (i−1)! ways to construct a cycle from these i elements. It
i (cid:0)(cid:1)
remains to constructthe remainingcycles σ ,···,σ , whichwe view as a permutationπ′ =σ ···σ onl−i
2 n 2 n
(cid:0)(cid:1)
elements consisting of n−1 cycles. Since there are c(l−i,n−1) such possibilities for π′, it follows that the
number of permutations π with a distinguished cycle is given by
l−n+1
l
(i−1)! c(l−i,n−1).
i
i=1 (cid:18) (cid:19)
X
Equating the two answers yields (20) as desired. (cid:3)
Lemma 10. Let n be a positive integer with 1≤n≤b−1. Then
Xn (x)=(χ (j,k)), (21)
b,1 n
where the entries χ (j,k) are given by
n
n! c(j−k,n)xn, if j ≥k+n
χ (j,k)= (j−k)! (22)
n
0, otherwise.
(cid:26)
Proof. We argue by induction on n. It is clear that (22) holds when n = 1. Suppose n > 1. If j < k+n,
then χ (j,k)=0 because Xm is a powerofstrictly lower-triangularmatrices. Therefore,assumej ≥k+n.
n b,1
Then
b−1 j−n+1
χ (j,k)= χ (j,i)χ (i,k)= χ (j,i)χ (i,k)
n n−1 1 n−1 1
i=0 i=k+1
X X
j−n+1
(n−1)! 1
=xn c(j−i,n−1)
(j−i)! i−k
i=k+1
X
(n−1)!xn l−n+1 l
= (i−1)! c(l−i,n−1),
l! i
i=1 (cid:18) (cid:19)
X
where l =j−k. It follows from Lemma 9 that
(n−1)!xn n!
χ (j,k)= ·nc(l,n)= c(j−k,n)xn
n
l! (j−k)!
as desired. (cid:3)
Lemma 11. We have
exp(X (x))=S (x). (23)
b,1 b,1
7
Proof. Denote the entries of exp(X (x)) by ξ(j,k). It is clear that ξ(j,k)=0 for j <k and ξ(j,k)=1 for
b,1
j =k since X is strictly lower triangular. Therefore, assume j ≥k+1. Since Xn =0 for n≥b, we have
b,1 b,1
b−1
χ (j,k)
n
ξ(j,k)=
n!
n=1
X
b−1
1
= c(j−k,n)xn
(j−k)!
n=1
X
x(x+1)···(x+j−k−1)
=
(j−k)!
x+j−k−1
=
j−k
(cid:18) (cid:19)
=α (j,k).
1
Thus, (23) holds. (cid:3)
Theorem 12. Let N be a positive integer. Then
exp(X (x))=S (x). (24)
b,N b,N
Proof. WearguebyinductiononN. Lemma(11)showsthat(24)istrueforN =1. Thensinceexp(A⊕B)=
exp(A)⊗exp(B) for any two matrices A and B, it follows that
exp(X (x))=exp(X (x)⊕X (x))=S (x)⊗S (x)=S (x),
b,N b,1 b,N−1 b,1 b,N−1 b,N
which proves (24). (cid:3)
3. Prouhet-Thue-Morse Polynomials
In this section we demonstrate how the generalized Sierpinski matrices S (1) arise in the study of
b,N
Prouhet-Thue-Morse polynomials, first investigated by the author in [10]. These polynomials were used in
the samepaper to give anew proofofthe well-knownProuhet-Tarry-Escottproblem, whichseeksb≥2 sets
of non-negative integers S , S , ..., S that have equal sums of like powers up to degree M ≥1, i.e.,
0 1 b−1
nm = nm =···= nm
nX∈S0 nX∈S1 n∈XSb−1
forallm=0,1,...,M. In1851,E.Prouhet[11]gaveasolution(butdidnotpublishaproof;seeLehmer[8])
bypartitioningthefirstbM+1non-negativeintegersintothesetsS ,S ,...,S accordingtotheassignment
0 1 b−1
n∈S .
ub(n)
Here, u (n) is the generalized Prouhet-Thue-Morse sequence, defined as the residue of the sum of digits of
b
n (base b):
d
u (n)= n mod b,
b j
j=0
X
where n=n bd+···+n b0 is the base-b expansionof n. When b=2, u(n):=u (n) generates the classical
d 0 2
Prouhet-Thue-Morse sequence: 0,1,1,0,1,0,0,1,....
Let A = (a ,a ,...,a ) be a zero-sum vector, i.e., an ordered collection of b arbitrary complex values
0 1 b−1
that sum to zero:
a +a +···+a =0.
0 1 b−1
We define F (x;A) to be the Prouhet-Thue-Morse (PTM) polynomial of degree bN −1 whose coefficients
N
belong to A and repeat according to u (n), i.e.,
b
bN−1
F (x;A)= a xn. (25)
N ub(n)
n=0
X
8
In the case where b=2, a =1, and a =−1, we obtain the classic product generating function formula
0 1
N 2N+1−1
(1−x2m)= (−1)u(n)xn. (26)
m=0 n=0
Y X
Lehmer generalizedthis formula to the case where A=(1,ω,ω2,...,ωb−1) consists of all b-th roots of unity
with ω =ei2π/b. The following theorem, proven in [10], extends this factorization to F (x;A) for arbitrary
N
zero-sum vectors.
Theorem 13 ([10]). Let N be a positive integer and A a zero-sum vector. There exists a polynomial P (x)
N
such that
N−1
F (x;A)=P (x) (1−xbm). (27)
N N
m=0
Y
Theorem13 is useful in that it allows us to establish that the polynomial F (x,A) has a zero of order N
N
atx=1,fromwhich Prouhet’ssolutionfollows easily by setting N =M+1 and differentiating F (x;A) m
N
times (see [10]).
We nowderiveformulasforthe coefficientsofP (x) intermsofgeneralizedSierpinskitriangles. Towards
N
this end, let
bN−1
P (x;C )= c xn
N N n
n=0
X
denote a polynomial whose coefficients are given by the column vector c =(c ,...,c )T. Also, let
N 0 bN−1
a =(a ,a ,...,a )T
n ub(0) ub(1) ub(bN−1)
be a column vector consisting of elements of A generated by the PTM sequence u (n). Next, define a
b
sequence of bN ×bN matrices M recursively by
N
1 0 0 ... 0 0
−1 1 0 ... 0 0
M1 = ... 
 
 0 0 0 ... −1 1 
 
 
and for N >1,
M 0 0 ... 0 0
N N N N N
0 −M M ... 0 0
N N N N N
MN+1 =M1⊗MN = .. , (28)
.
 
 0 0 0 ... −M M 
 N N N N N 
 
where M ⊗M denotes the Kronecker product. The following theorem establishes a matrix relationship
1 N
between the vectors a and c .
N N
Theorem 14. Let A=(a ,...,a ) be a zero-sum vector. The polynomial equation
0 b−1
N−1
F (x;A)=P (x;C ) (1−xbm) (29)
N N N
m=0
Y
is equivalent to the matrix equation
a =M c (30)
N N N
together with the condition c = 0 for any n that contains the digit b−1 in its base-b expansion, where
n
0≤n≤bN −1.
ToproveTheorem14,weshallneedthefollowinglemmas,whichwestatewithoutproofsincetheirresults
are easy to verify.
9
Lemma 15. Let A=(a ,a ,...,a ) and C ={c ,...,c }. Then the polynomial equation
0 1 b−1 0 b−1
b−1 b−1
a xn = c xn (1−x)
n n
!
n=0 n=0
X X
is equivalent to the system of equations
a =c
0 0
a =−c +c
1 0 1
···
a =−c +c
b−1 b−2 b−1
together with the condition c =0.
b−1
Lemma 16. The system of equations in Lemma 15 can be expressed in matrix form as
a =M c .
1 1 1
Proof of Theorem 14. WearguebyinductiononN. Itisclearthat(30)holdsforN =1sincethepolynomial
equationF (x;A)=P (x;C )(1−x)isequivalenttoa =M c becauseofLemmas15and16. Next,assume
1 1 1 1 1 1
that (30) holds for case N. We shall prove that (30) holds for case N +1. Define
bN−1
P (x;C (p))= c xn+pbN
N N n+pbN
n=0
X
to be a polynomialwithcoefficientsetC (p)={c ,...,c }. We then expandthe right-handside
N pbN (p+1)bN−1
of (27) for case N +1 as follows:
N N−1
P (x;C ) (1−xbm)=[P (x;C (0))+...+P (x;C (b−1))] (1−xbm) (1−xbN)
N+1 N+1 N N N N
" #
m=0 m=0
Y Y
=[Q (x;C (0))+...+Q (x;C (b−1))](1−xbN)
N N N N
=Q (x;C (0))+[Q (x;C (1))−xbNQ (x;C (0))]+...
N N N N N N
+[Q (x;C (b−1))−xbNQ (x;C (b−2))]−xbNQ (x;C (b−1)),
N N N N N N
where we define Q (x;C (p))=P (x;C (p))(1−xbN). Equating this result with
N N N N
bN+1−1
F (x;A)= a xn =F (x;A (0))+...+F (x;A (b−1)),
N+1 ub(n) N N N N
n=0
X
where
bN−1
F (x;A (p))= a xn+pbN,
N N u(n+pbN)
n=0
X
leads to the system of polynomial equations
F (x;A (0))=Q (x;C (0))
N N N N
F (x;A (1))=Q (x;C (1))−xbNQ (x;C (0))
N N N N N N
...
F (x;A (b−1))=Q (x;C (b−1))−xbNQ (x;C (b−2))
N N N N N N
10