q -QUASIADDITIVE
FUNCTIONS
SARA KROPF AND STEPHAN WAGNER
q -quasiadditivity of arithmetic functions,
q -quasimultiplicativity, which generalises complete q -additivity
Abstract. In this paper, we introduce the notion of
as well as the related concept of
and -multiplicativity, respectively.
We show that there are many natural examples for these
f (q k+r a + b) = f (a) + f (b)
f (q k+r a+b) = f (a)f (b) for all b < q k and a xed parameter r. In addition to some elementary
properties of q -quasiadditive and q -quasimultiplicative functions, we prove characterisations of
q -quasiadditivity and q -quasimultiplicativity for the special class of q -regular functions. The
concepts, which are characterised by functional equations of the form
or
nal main result provides a general central limit theorem that includes both classical and new
examples as corollaries.
1. Introduction
Arithmetic functions based on the digital expansion in some base
q -additive function
called q -additive if
e.g., [26]) The notion of a
nonnegative integers) is
q
have a long history (see,
is due to [4]: an arithmetic function (dened on
f (q k a + b) = f (q k a) + f (b)
whenever
0 ≤ b < qk .
said to be completely
A stronger version of this concept is
q -additive
complete q-additivity:
a function
f
is
if we even have
f (q k a + b) = f (a) + f (b)
whenever
0 ≤ b < qk .
q -multiplicative functions is dened in an analogous
q -additivity of a function means that it can be evaluated
Typical examples of completely q -additive functions are the
The class of (completely)
fashion. Loosely speaking, (complete)
by breaking up the base-q expansion.
q -ary
sum of digits and the number of occurrences of a specied digit.
There are, however, many simple and natural functions based on the
not
q -additive.
A very basic example of this kind are
a certain block of digits in the
q -ary
block counts :
q -ary
expansion that are
the number of occurrences of
expansion. This and other examples provide the motivation
for the present paper, in which we dene and study a larger class of functions with comparable
properties.
Denition. An arithmetic function (a function dened on the set of nonnegative integers) is called
q -quasiadditive if there exists some nonnegative integer r such that
f (q k+r a + b) = f (a) + f (b)
(1)
whenever
0 ≤ b < qk .
Likewise,
f
is said to be
f (q
(2)
for some xed nonnegative integer
Key words and phrases.
q -additive
r
k+r
if it satises the identity
a + b) = f (a)f (b)
whenever
function,
q -quasimultiplicative
0 ≤ b < qk .
q -quasiadditive
function,
q -regular
function, central limit theorem.
The rst author is supported by the Austrian Science Fund (FWF): P 24644-N26. The second author is supported
by the National Research Foundation of South Africa under grant number 96236. The authors were also supported
by the Karl Popper Kolleg ModelingSimulationOptimization funded by the Alpen-Adria-Universität Klagenfurt
and by the Carinthian Economic Promotion Fund (KWF). Part of this paper was written while the second author
was a Karl Popper Fellow at the Mathematics Institute in Klagenfurt. He would like to thank the institute for the
hospitality received.
1
2
SARA KROPF AND STEPHAN WAGNER
We remark that the special case
term completely
q -quasiadditive
r=0
q -additivity,
is exactly complete
so strictly speaking the
function might be more appropriate. However, since we are not
considering a weaker version (for which natural examples seem to be much harder to nd), we do
not make a distinction.
In the following section, we present a variety of examples of
q -quasiadditive and q -quasimultipli-
cative functions. In Section 3, we give some general properties of such functions. Since most of
our examples also belong to the related class of
q -regular
functions, we discuss the connection in
Section 4. Finally, we prove a general central limit theorem for
q -quasiadditive and -multiplicative
functions that contains both old and new examples as special cases.
2. Examples of
q -quasiadditive
and
q -quasimultiplicative
q -quasiadditivity
Let us now back up the abstract concept of
functions
by some concrete examples.
Block counts. As mentioned in the introduction, the number of occurrences of a xed digit is
a typical example of a
B = 1 2 · · · `
q -additive
function. However, the number of occurrences of a given block
n, which we denote by cB (n),
q -ary expansion of q k a + b is
obtained by joining the expansions of a and b, so occurrences of B in a and occurrences of B in b
are counted by cB (a) + cB (b), but occurrences that involve digits of both a and b are not.
However, if B is a block dierent from 00 · · · 0, then cB is q -quasiadditive: note that the
k+`
representation of q
a + b is of the form
of digits in the expansion of a nonnegative integer
does not represent a
q -additive
function. The reason is simple: the
a1 a2 · · · aµ 00 · · · 0 b1 b2 · · · bν
| {z } | {z } | {z }
expansion of
a `
zeros expansion of
b
0 ≤ b < q k , so occurrences of the block B have to belong to either a or b only, implying
cB (q a + b) = cB (a) + cB (b), with one small caveat: if the block starts and/or ends with
whenever
that
k+`
a sequence of zeros, then the count needs to be adjusted by assuming the digital expansion of a
nonnegative integer to be padded with zeros on the left and on the right.
For example, let B be the block 0101 in base 2. The binary representations of 469 and 22 are
111010101 and 10110 respectively, so we have cB (469) = 2 and cB (22) = 1 (note the occurrence
of 0101 at the beginning of 10110 if we assume the expansion to be padded with zeros), as well as
cB (240150) = cB (29 · 469 + 22) = cB (469) + cB (22) = 3.
Indeed, the block
B
occurs three times in the expansion of
240150,
which is
111010101000010110.
The number of runs and the Gray code. The number of ones in the Gray code of a nonnegative integer
n,
which we denote by
hGRAY (n),
is also equal to the number of runs (maximal
n (counting the number
hGRAY (n) is A005811 in Sloane's
sequences of consecutive identical digits) in the binary representations of
of runs in the representation of
0
as
0);
the sequence dened by
On-Line Encyclopedia of Integer Sequences [16]. An analysis of its expected value is performed
hGRAY is 2-quasiadditive
f (n) = hGRAY (n) + 1 if n
in [8]. The function
if
n
is even and
up to some minor modication: set
is odd.
the total number of occurrences of the two blocks
01
The new function
and
10
f
f (n) = hGRAY (n)
can be interpreted as
in the binary expansion (considering
binary expansions to be padded with zeros at both ends), so the argument of the previous example
applies again and shows that
f
is
2-quasiadditive.
The nonadjacent form and its Hamming weight. The nonadjacent form (NAF) of a nonnegative integer is the unique base-2 representation with digits
as
1
0, 1, −1 (−1
is usually represented
in this context) and the additional requirement that there may not be two adjacent nonzero
digits, see [17]. For example, the NAF of
27
is
100101.
It is well known that the NAF always
has minimum Hamming weight (i.e., the number of nonzero digits) among all possible binary
representations with this particular digit set, although it may not be unique with this property
(compare, e.g., [17] with [14]).
q -QUASIADDITIVE
The Hamming weight
hNAF
FUNCTIONS
3
of the nonadjacent form has been analysed in some detail [11, 19],
and it is also an example of a
2-quasiadditive
function.
It is not dicult to see that
hNAF
is
characterised by the recursions
hNAF (2n) = hNAF (n),
hNAF (4n + 1) = hNAF (n) + 1,
together with the initial value
hNAF (0) = 0.
hNAF (4n − 1) = hNAF (n) + 1
The identity
hNAF (2k+2 a + b) = hNAF (a) + hNAF (b)
can be proved by induction. In Section 4, this example will be generalised and put into a greater
context.
The number of optimal
{0, 1, −1}-representations.
As mentioned above, the NAF may not
be the only representation with minimum Hamming weight among all possible binary representations with digits
n
0, 1, −1.
The number of optimal representations of a given nonnegative integer
is therefore a quantity of interest in its own right. Its average over intervals of the form
was studied by Grabner and Heuberger [10], who also proved that the number
representations of
n
rOPT (n)
[0, N )
of optimal
can be obtained in the following way:
Lemma 1 (GrabnerHeuberger [10]).
Let sequences ui (i = 1, 2, . . . , 5) be given recursively by
u1 (0) = u2 (0) = · · · = u5 (0) = 1,
u1 (1) = u2 (1) = 1, u3 (1) = u4 (1) = u5 (1) = 0,
and
u1 (2n) = u1 (n),
u1 (2n + 1) = u2 (n) + u4 (n + 1),
u2 (2n) = u1 (n),
u2 (2n + 1) = u3 (n),
u3 (2n) = u2 (n),
u3 (2n + 1) = 0,
u4 (2n) = u1 (n),
u4 (2n + 1) = u5 (n + 1),
u5 (2n) = u4 (n),
u5 (2n + 1) = 0.
The number rOPT (n) of optimal representations of n is equal to u1 (n).
A straightforward calculation shows that
u1 (8n) = u2 (8n) = · · · = u5 (8n) = u1 (8n + 1) = u2 (8n + 1) = u1 (n),
(3)
u3 (8n + 1) = u4 (8n + 1) = u5 (8n + 1) = 0.
This gives us the following result:
The number of optimal {0, 1, −1}-representations of a nonnegative integer is a 2quasimultiplicative function. Specically, for any three nonnegative integers a, b, k with b < 2k , we
have
Lemma 2.
rOPT (2k+3 a + b) = rOPT (a)rOPT (b).
Proof.
We will prove a somewhat stronger statement by induction on
t:
write
u(n) = (u1 (n), u2 (n), u3 (n), u4 (n), u5 (n))t .
We show that
u(2k+3 a + b) = rOPT (a)u(b)
and
u(2k+3 a + b + 1) = rOPT (a)u(b + 1)
for all
a, b, k
satisfying the conditions of the lemma, from which the desired result follows by
considering the rst entry of the vector
u(2k+3 a + b).
Note rst that both identities are clearly
4
SARA KROPF AND STEPHAN WAGNER
true for
k=0
in view of (3). For the induction step, we distinguish two cases: if
b
is even, we
have

1
1

u(2k+3 a + b) = 
0
1
0

1
1

=
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0
0
0
0
0
1

0
0

k+2
0
a + b/2)
 · u(2

0
0

0
0

0
 · rOPT (a)u(b/2)
0
0
= rOPT (a)u(b)
by the induction hypothesis, as well as

0
0

u(2k+3 a + b + 1) = 
0
0
0

0
0

=
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
0
0


0
0
0
0


k+2
0
a + b/2) + 
0
 · u(2
0
0
0
0


0
0
0
0



0
 · rOPT (a)u(b/2) + 0
0

0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0

0
0

k+2
0
a + b/2 + 1)
 · u(2

1
0

0
0

0
 · rOPT (a)u(b/2 + 1)
1
0
= rOPT (a)u(b + 1).
The case that
b
is odd is treated in an analogous fashion.
In Section 4, we will show that this is also an instance of a more general phenomenon.
The run length transform and cellular automata. The
run length transform
of a sequence
is dened in a recent paper of Sloane [18]: it is based on the binary representation, but could in
principle also be generalised to other bases. Given a sequence
s1 , s2 , . . .,
its run length transform
is obtained by the rule
t(n) =
Y
si ,
i∈L(n)
where
L(n)
is the multiset of run lengths of
n
(lengths of blocks of consecutive ones in the binary
1910
t(1910) = s2 s23 .
representation). For example, the binary expansion of
of run lengths would be
{3, 3, 2},
giving
is
11101110110,
so the multiset
L(n)
A typical example is obtained for the sequence of Jacobsthal numbers given by the formula
sn = 31 (2n+2 −(−1)n ).
The associated run length transform tn (sequence A071053 in the OEIS [16])
counts the number of odd coecients in the expansion of
as the number of active cells at the
n-th
(1+x+x2 )n , and it can also be interpreted
generation of a certain cellular automaton.
Further
examples stemming from cellular automata can be found in Sloane's paper [18].
The argument that proved
q -quasiadditivity
of block counts also applies here, and indeed it is
easy to see that the identity
t(2k+1 a + b) = t(a)t(b),
0 ≤ b < 2k , holds for the run length transform of any sequence, meaning that any such transform is 2-quasimultiplicative. In fact, it is not dicult to show that every 2-quasimultiplicative
function with parameter r = 1 is the run length transform of some sequence.
where
q -QUASIADDITIVE
FUNCTIONS
5
3. Elementary properties
Now that we have gathered some motivating examples for the concepts of
q -quasimultiplicativity,
q -quasiadditivity and
let us present some simple results about functions with these properties.
First of all, let us state an obvious relation between
q -quasiadditive
and
q -quasimultiplicative
functions:
If a function f is q-quasiadditive, then the function dened by g(n) = cf (n)
for some positive constant c is q-quasimultiplicative. Conversely, if f is a q-quasimultiplicative
function that only takes positive values, then the function dened by g(n) = logc f (n) for some
positive constant c 6= 1 is q-quasiadditive.
Proposition 3.
The next proposition deals with the parameter
Proposition 4.
r
in the denition of a
q -quasiadditive
function:
If the arithmetic function f satises
f (q k+r a + b) = f (a) + f (b)
for some xed nonnegative integer r whenever 0 ≤ b < qk , then it also satises
f (q k+s a + b) = f (a) + f (b)
for all nonnegative integers s ≥ r whenever 0 ≤ b < qk .
Proof. If a, b are nonnegative integers with 0 ≤ b < qk , then clearly also 0 ≤ b < qk+s−r
if
s ≥ r,
and thus
f (q k+s a + b) = f (q (k+s−r)+r a + b) = f (a) + f (b).
If two arithmetic functions f and g are q-quasiadditive functions, then so is any
linear combination αf + βg of the two.
Proof. In view of the previous proposition, we may assume the parameter r in (1) to be the same
Corollary 5.
for both functions. The statement follows immediately.
Finally, we observe that
by breaking the
q -ary
q -quasiadditive
and
q -quasimultiplicative
functions can be computed
expansion into pieces.
If f is a q-quasiadditive (q-quasimultiplicative) function, then
• f (0) = 0 (f (0) = 1, respectively, unless f is identically 0),
• f (qa) = f (a) for all nonnegative integers a.
Proof. Assume rst that f is q-quasiadditive. Setting a = b = 0 in the
Lemma 6.
dening functional
equation (1), we obtain
f (0) = f (0) + f (0),
b = 0 while a is arbitrary,
and the rst statement follows. Setting
f (q
for all
k ≥ 0.
k+r
we now nd that
a) = f (a)
In particular, this also means that
f (a) = f (q r+1 a) = f (q r · qa) = f (qa),
which proves the second statement.
For
q -quasimultiplicative
functions, the proof is analogous
(and one can also use Proposition 3 for positive functions).
Suppose that the function f is q-quasiadditive with parameter r, i.e. f (qk+r a +
b) = f (a) + f (b) whenever 0 ≤ b < q k . Going from left to right, split the q -ary expansion of n
into blocks by inserting breaks after each run of r or more zeros. If these blocks are the q-ary
representations of n1 , n2 , . . . , n` , then we have
Proposition 7.
f (n) = f (n1 ) + f (n2 ) + · · · + f (n` ).
Moreover, if m1 , m2 , . . . , m` are obtained from n1 , n2 , . . . , n` by dividing o the highest possible
powers of q, then
f (n) = f (m1 ) + f (m2 ) + · · · + f (m` ).
6
SARA KROPF AND STEPHAN WAGNER
Analogous statements hold for q-quasimultiplicative functions, with sums replaced by products.
Proof. This is obtained by a straightforward induction on ` together with the fact that f (qh a) =
f (a),
which follows from the previous lemma.
Example 1.
Recall that the Hamming weight of the NAF (which is the minimum Hamming
{0, 1, −1}-representation) is 2-quasiadditive with parameter r = 2. To determine
hNAF (314 159 265), we split the binary representation, which is 10010101110011011000010100001,,
weight of a
into blocks by inserting breaks after each run of at least two zeros:
100|101011100|110110000|1010000|1.
n1 , n2 , . . . , n` in the statement of the proposition are now 4, 348, 432, 80, 1 respecm1 , m2 , . . . , m` are therefore 1, 87, 27, 5, 1. Now we use the values hNAF (1) =
1, hNAF (5) = 2, hNAF (27) = 3 and hNAF (87) = 4 to obtain
The numbers
tively, and the numbers
hNAF (314 159 265) = 2hNAF (1) + hNAF (5) + hNAF (27) + hNAF (87) = 11.
Example 2.
In the same way, we consider the number of optimal representations
2-quasimultiplicative with parameter r = 3. Consider for instance the binary
204 280 974, namely 1100001011010001010010001110.. We split into blocks:
rOPT ,
which is
representation of
110000|101101000|101001000|1110.
The four blocks correspond to the numbers
rOPT (3) = 2, rOPT (45) = 5, rOPT (41) = 1
4.
48 = 16·3, 360 = 8·45, 328 = 8·41 and 14 = 2·7. Since
rOPT (7) = 1, we obtain rOPT (204 280 974) = 10.
and
q -Regular
functions
q -regular functions and examine the connection to our concepts.
q -regular sequences.
t
if f = u f for a vector u and a vector-valued function f with matrices
In this section, we introduce
See [1] for more background on
A function
Mi , 0 ≤ i < q
f
is
q -regular
satisfying
f (qn + i) = Mi f (n)
(4)
0 ≤ i < q , qn + i > 0.
for
v = f (0).
q -regular if and
We set
Equivalently, a function
f
is
only if
f (n) = ut
(5)
L
Y
f
can be written as
M ni v
i=0
nL · · · n0
q -ary expansion of n.
q -regular functions is a generalization of q -additive and q -multiplicative functions.
However, we emphasise that q -quasiadditive and q -quasimultiplicative functions are not necessarily
q -regular: a q -regular sequence can always be bounded by O(nc ) for a constant c, see [1, Thm.
16.3.1]. In our setting however, the values of f (n) can be chosen arbitrarily for those n whose
q -ary expansion does not contain 0r . Therefore a q -quasiadditive or -multiplicative function can
where
is the
The notion of
grow arbitrarily fast.
(u, (Mi )0≤i<q , v) a representation of the q -regular function f . Such a representation is
if M0 v = v , meaning that in (5), leading zeros in the q -ary expansion of n
do not change anything. We call a representation minimal if the dimension of the matrices Mi is
minimal among all representations of f .
Following [7], every q -regular function has a zero-insensitive minimal representation.
We call
called
zero-insensitive
4.1. When is a
of
q -regular
q -regular
function
functions that are
q -quasimultiplicative?
q -quasimultiplicative.
We now give a characterisation
Let f be a q-regular sequence with zero-insensitive minimal representation (5). Then
the following two assertions are equivalent:
• The sequence f is q -quasimultiplicative with parameter r.
• M0r = vut .
Theorem 8.
q -QUASIADDITIVE
Proof.
1}, I
Let
d
be the dimension of the vectors.
FUNCTIONS
We prove that
nite} is a generating system of the whole
7
{ut
d-dimensional
Q
i∈I
Mni | ni ∈ {0, . . . , q −
vector space by contradiction:
d0 < d unit vectors form
M
|
n
∈
{0,
.
. . , q − 1}, I nite}. This
ni
i
i∈I
coordinate transform denes a dierent representation of f with matrices M̂i and vectors û and
Q
v̂ . However, only the rst d0 coordinates of any vector ut i∈I Mni are nonzero. Thus we can
reduce the dimension of the matrices and vectors from d to d0 to obtain a new representation of
f . This contradicts
Q the minimality of the original representation.
Analogously, { j∈J Mnj v | nj ∈ {0, . . . , q − 1}, J nite} is also a generating system for the
assume that there is a coordinate transformation such that the rst
a basis of the transformed space spanned by
{ut
Q
whole vector space.
The
q -quasimultiplicativity
f (n) with parameter r is equivalent
Y
Y
ut
Mni (M0r − vut )
M nj v = 0
of
i∈I
to the identity
j∈J
{ut
Q
Q
i∈I Mni } and { j∈J
t
r
t
ating systems of the entire vector space, this is equivalent to x (M0 − vu )y =
r
t
and y , which in turn is equivalent to M0 = vu .
for all nite tuples
Example
(ni )i∈I
and
(nj )j∈J .
Since both
Mnj v} are gener0 for all vectors x
{0, 1, −1}-representations). The number of optimal {0, 1, −1}2-regular sequence by Lemma 1. A minimal zerot
vector (u1 (n), u2 (n), u3 (n), u1 (n + 1), u4 (n + 1), u5 (n + 1)) is
3 (The number of optimal
representations as described in Section 2 is a
insensitive representation for the
given by

1
1

0
M0 = 
0

0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0

0
0

0
,
0

1
0

0
0

0
M1 = 
0

0
0
ut = (1, 0, 0, 0, 0, 0) and v = (1, 1, 1, 1, 0, 0)t .
3
t
As M0 = vu , this sequence is 2-quasimultiplicative
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
1
0
with parameter
1
0
0
0
0
1
3,

0
0

0
,
0

0
0
which is the same result
as in Lemma 2.
Remark.
The condition on the minimality of the representation in Theorem 8 is necessary as
illustrated by the following example:
f (n) = 2s2 (n) where s2 (n) is the binary sum of digits function. This
sequence is 2-regular and 2-(quasi-)multiplicative with parameter r = 0. A minimal representation
0
t
is given by M0 = 1, M1 = 2, v = 1 and u = 1. As stated in Theorem 8, we have M0 = vu = 1.
1
13
27
If we use the zero-insensitive non-minimal representation dened by M0 = 0 2 , M1 = 2
0 5 ,
v = (1, 0)t and ut = (1, 0) instead, we have rank M0r = 2 for all r ≥ 0. Thus M0r 6= vut .
Consider the sequence
4.2. When is a
q -regular function q -quasiadditive? The characterisation of q -regular funcq -quasiadditive is somewhat more complicated. Again, we consider a zeroinsensitive (but not necessarily minimal) representation. We let U be the smallest vector space
Q
t
t
t
t
such that all vectors of the form u
i∈I Mni lie in the ane subspace u + U (U is used as a
t
t
shorthand for {x : x ∈ U }). Such a vector space must exist, since u is a vector of this form
(corresponding to the empty product, where I = ∅). Likewise, let V be the smallest vector space
Q
such that all vectors of the form
j∈J Mnj v lie in the ane subspace v + V .
tions that are also
Let f be a q-regular sequence with zero-insensitive representation (5). The sequence
is q-quasiadditive with parameter r if and only if all of the following statements hold:
• ut v = 0,
• U t is orthogonal to (M0r − I)v , i.e. xt (M0r − I)v = xt M0r v − xt v = 0 for all x ∈ U ,
• V is orthogonal to ut (M0r − I), i.e. ut (M0r − I)y = ut M0r y − ut y = 0 for all y ∈ V ,
• U t M0r V = 0, i.e. xt M0r y = 0 for all x ∈ U and y ∈ V .
Theorem 9.
f
8
SARA KROPF AND STEPHAN WAGNER
Proof.
ut v = 0
The rst statement
necessary condition by Lemma 6.
f (0) = 0, which we already know to
ut M0r v = ut v = 0 by the assumption
is equivalent to
Note also that
be a
that
the representation is zero-insensitive. For the remaining statements, we write the quasiadditivity
condition in terms of our matrix representation as we did in the quasimultiplicative case:
ut
Y
Mni M0r
i∈I
J = ∅,
Specically, when
Y
M nj v = u t
j∈J
Y
M ni v + u t
i∈I
Y
Mnj v.
j∈J
we get
ut
Y
Mni M0r − I v = ut v = 0.
i∈I
I=∅
Setting also
gives us
u
t
(M0r
− I)v = 0, so together we obtain
Y
ut
Mni − ut M0r − I v = 0.
i∈I
t
i∈I Mni − u , the second statement follows.
The proof of the third statement is analogous. Finally, if we assume that the rst three statements
U
Since
t
ut
is spanned by all vectors of the form
Q
hold, then we nd that
ut
Y
Mni M0r
i∈I
Y
M nj v
j∈J
= ut
Y
Y
Y
Y
Mni − ut M0r
M nj v − v + u t
Mni − ut M0r v + ut M0r
M nj v − v
i∈I
j∈J
i∈I
j∈J
+ ut M0r v
Y
Y
Y
Y
= ut
Mni − ut M0r
M nj v − v + u t
Mni − ut v + ut
M nj v − v
i∈I
= u
t
Y
j∈J
t
M ni − u
M0r
i∈I
Thus
Y
being valid for all choices of
U
and
Y
j∈J
M ni v + u
i∈I
t
Y
Mnj v.
j∈J
is equivalent to
ut
Example 4.
M nj v − v + u
j∈J
q -quasiadditivity
denition of
i∈I
t
Y
Y
Mni − ut M0r
M nj v − v = 0
i∈I
j∈J
I, J , ni
and
nj .
The desired fourth condition is clearly equivalent by
V.
For the Hamming weight of the nonadjacent form, a zero-insensitive (and also minimal)
(hNAF (n), hNAF (n + 1), hNAF (2n + 1), 1)t is




1 0 0 0
0 0 1 0
0 0 1 0
 0 1 0 0



M0 = 
1 0 0 1 , M1 = 0 1 0 1 ,
0 0 0 1
0 0 0 1
representation for the vector
ut = (1, 0, 0, 0)
v = (0, 1, 1, 1)t .
w1 = ut M1 − ut , w2 = ut M12 − ut and w3 = ut M1 M0 M1 − ut are linearly
independent. If we let W be the vector space spanned by those three, it is easily veried that M0
t
t
and M1 map the ane subspace u + W to itself, so U = W is spanned by these vectors.
2
Similarly, the three vectors M1 v − v , M1 v − v and M1 M0 M1 v − v span V .
and
The three vectors
The rst condition of Theorem 9 is obviously true.
conditions with
2-regular
r = 2
for the base vectors of
sequence that is also
Finding the vector spaces
U
2-quasiadditive,
U
and
We only have to verify the other three
V,
which is done easily.
Thus
hNAF
is a
as it was also shown in Section 2.
V is not trivial. But in a special case of q -regular functions, we
q -additivity, which is easier to check. These q -regular functions
as dened in [12]: a transducer transforms the q -ary expansion
and
can give a sucient condition for
are output sums of transducers
of an integer
n
(read from the least signicant to the most signicant digit) deterministically
q -QUASIADDITIVE
into an output sequence and leads to a state
s.
FUNCTIONS
The output sum is then the sum of this output
sequence together with the nal output of the state
function evaluated at
n.
The function
hN AF
9
s.
This denes the value of the
q -regular
discussed in the example above, as well as many
other examples, can be represented in this way.
The output sum of a connected transducer is q-additive with parameter r if the
following conditions are satised:
• The transducer has the reset sequence 0r going to the initial state, i.e., reading r zeros
always leads to the initial state of the transducer.
• For every state, the output sum along the path of the reset sequence 0r equals the nal
output of this state.
• Additional zeros at the end of the input sequence do not change the output sum.
Proposition 10.
Proof.
The matrices of the representation for the output sum of a transducer have the structure

Nε
Mε =  0
0
δε
1
0

[ε = 0]I
0 
[ε = 0]I
Nε is a matrix with exactly one nonzero entry per row that is necessarily equal to 1,
δ ε are arbitrary vectors (see [12, Remark 3.10]). Furthermore, ut = (1, 0, . . . , 0) and v =
(b(0), 1, b(0) − N0 b(0) − δ 0 ) for a vector b(0).
The vector b(0) contains the nal outputs of the states. The vectors δ ε contain the outputs
of the transitions with input ε, and the matrices Nε are the adjacency matrices of the transitions
with input ε. The initial state corresponds to the rst coordinate.
where
and
Then by (4), the output sum of the transducer is the rst coordinate of
b(qn + ε) = Nε b(n) + δ ε
(6)
if
qn + ε > 0
and of
b(0)
otherwise.
The third condition ensures that that one leading zero does not change anything.
connectivity of the underlying graph implies that (6) also holds for
coordinates of
Let
J
v are
qn + ε = 0
Thus the
and the last
zero and we can reduce the dimension of the representation.
be nite and
nj ∈ {0, . . . , q − 1}
j ∈ J . The rst condition


1 0 ··· 0
Y


Nnj N0r =  ... ... . . . ... 
j∈J
1 0 ··· 0
for
implies that
and the second condition implies that
Y
Nnj b(0) =
j∈J
Y
Nnj (I + · · · + N0r−1 )δ 0 .
j∈J
Using (6) recursively together with these two conditions gives
b(q k+r m + n) =
k−1
Y
Nnj N0r b(m) + b(n) −
j=0

1
 ..
= .
1
of
Nnj b(0) +
j=0
···
0
.
.
.
..
.
.  b(m)
.
0
···
.
k−1
Y
Nnj (I + · · · + N0r−1 )δ 0
j=0

0
+ b(n)
0
n with q -ary digit expansion (nk−1 · · · n0 )
b(n) is q -quasiadditive.
for all
k−1
Y
and all
m.
This implies that the rst coordinate
10
SARA KROPF AND STEPHAN WAGNER
5. A central limit theorem for
q -quasiadditive
In this section, we prove a central limit theorem for
positive values.
and -multiplicative functions
q -quasimultiplicative
functions taking only
By Proposition 3, this also implies a central limit theorem for
q -quasiadditive
functions.
To this end, we dene a generating function: let
Mk
positive values, let
integers whose
q -ary
f
be a
q -quasimultiplicative function with
q k (i.e., those positive
be the set of all nonnegative integers less than
k
X
expansion needs at most
F (x, t) =
digits), and set
xk
X
f (n)t .
n∈Mk
k≥0
The decomposition of Proposition 7 now translates directly to an alternative representation for
F (x, t):
let
B
the function
q whose q -ary representation does
q -ary representation of n, and dene
be the set of all positive integers not divisible by
not contain the block
B(x, t)
0r ,
let
`(n)
denote the length of the
by
B(x, t) =
X
x`(n) f (n)t .
n∈B
q = 2 and r = 1, this
X
B(x, t) =
xk f (2k − 1)t .
We remark that in the special case where
(7)
simplies greatly to
k≥1
Proposition 11.
F (x, t) =
Proof.
The generating function F (x, t) can be expressed as
1
·
1−x 1−
1 + (1 + x + · · · + xr−1 )B(x, t)
1
.
1+(1+x+· · ·+xr−1 )B(x, t) =
1 − x − xr B(x, t)
1−x B(x, t)
xr
The rst factor stands for the initial sequence of leading zeros, the second factor for a
(possibly empty) sequence of blocks consisting of an element of
B
r or more
B with up
and
last factor for the nal part, which may be empty or an element of
zeros, and the
to
r−1
zeros
(possibly none) added at the end.
Under suitable assumptions on the growth of a
q -quasiadditive or q -quasimultiplicative function,
we can exploit the expression of Proposition 11 to prove a central limit theorem in the following
steps.
Denition.
at most polynomial growth if f (n) = O(nc ) and f (n) =
Ω(n ) for a xed c ≥ 0. We say that f has at most logarithmic growth if f (n) = O(log n).
Lemma 12. Assume that the positive, q -quasimultiplicative function f has at most polynomial
growth.
There exist positive constants δ and such that
• B(x, t) has radius of convergence ρ(t) > 1q whenever |t| ≤ δ .
• For |t| ≤ δ , the equation x + xr B(x, t) = 1 has a complex solution α(t) with |α(t)| < ρ(t)
and no other solutions with modulus ≤ (1 + )|α(t)|.
• Thus the generating function F (x, t) has a simple pole at α(t) and no further singularities
of modulus ≤ (1 + )|α(t)|.
• Finally, α is an analytic function of t for |t| ≤ δ .
Proof. The polynomial growth of f implies that C −1 φ−`(n) ≤ f (n) ≤ Cφ`(n) for some positive
We say that a function
f
has
−c
φ. Moreover, B contains O(β ` ) elements whose q -ary expansion has length at
r
r−1
most `, where β < q is a root of the polynomial x − (q − 1)x
− · · · − (q − 1)x − (q − 1). This
−1 δ
implies that B(x, t) is indeed an analytic function of x for |x| < β
φ whenever |t| ≤ δ . For
1
−1 δ
suitably small δ , β
φ is greater than q , which proves the rst part of our statement. Next note
constants
C
and
that
(q − 1)x
,
1 − (q − 1)x − · · · − (q − 1)xr
r
only solution of the equation x + x B(x, 0) = 1.
B(x, 0) =
1
q is the
statements are therefore a simple consequence of the implicit function theorem.
and it follows that
α(0) =
All remaining
q -QUASIADDITIVE
FUNCTIONS
11
Assume that the positive, q-quasimultiplicative function f has at most polynomial
growth.
With δ and as in the previous lemma, we have, uniformly in t,
Lemma 13.
[xk ]F (x, t) = κ(t) · α(t)−k 1 + O((1 + )−k )
for some function κ. Both α and κ are analytic functions of t for |t| ≤ δ , and κ(t) 6= 0 in this
region.
Proof.
This follows from the previous lemma by means of singularity analysis, see [9].
Assume that the positive, q-quasimultiplicative function f has at most polynomial
growth.
Let Nk be a randomly chosen integer in {0, 1, . . . , qk − 1}. The random variable Lk = log f (Nk )
has mean µk + O(1) and variance σ2 k + O(1), where the two constants are given by
Theorem 14.
Bt (1/q, 0)
q 2r
µ=
and
(8)
σ 2 = −Bt (1/q, 0)2 q −4r+1 (q − 1)−1 + 2Bt (1/q, 0)2 q −3r+1 (q − 1)−1 − Bt (1/q, 0)2 q −4r (q − 1)−1
− 4rBt (1/q, 0)2 q −4r + Btt (1/q, 0)q −2r − 2Bt (1/q, 0)Btx (1/q, 0)q −4r−1 .
2
If f is not the
√ constant function f ≡ 1, then σ 6= 0 and the renormalised random variable
(Lk − µk)/(σ k) converges weakly to a standard Gaussian distribution.
Proof.
This follows from the fact that
[xk ]F (x, t)/q k
is the moment generating function of
the Quasi-power theorem, see [13]. The only part that we actually have to verify is that
unless
f
Lk and
σ2 =
6 0
is constant.
Assume that
σ 2 = 0.
We rst consider the case that
be the least integer greater than
expansion of
log α(t)
at
t = 0,
1
such that
ts
log α(t)
is not a linear function. Let
s
occurs with a nonzero coecient in the Taylor
i.e.,
log α(t) = log α(0) + at + bts + O(ts+1 ).
By assumption, we have
s ≥ 3.
Since
α(0) =
1
q and
κ(0) = 1,
it follows that
[xk ]F (x, t)
= exp log κ(t) − k log α(t) − k log q + O (1 + )−k
k
q
= exp − akt − bkts + O kts+1 + t + (1 + )−k .
E(exp(tLk )) =
We see that
Lk ,
a = −µ.
Considering the renormalised version
Rk =
Lk −µk
of the random variable
k1/s
we get
for xed
τ.
It
E exp τ Rk = exp − bτ s + O k −1/s + (1 + )−k
s
follows that limk→∞ E(exp τ Rk ) = exp(−bτ ) for every complex τ ,
which is a
continuous function. By Lévy's continuity theorem, this would imply convergence in distribution
M (τ ) = exp(−bτ s ). However,
τ = 0 are nite and the second derivative of
exp(−bτ s ) at τ = 0 is 0, thus the second moment is 0. A random variable whose second moment
is 0 is almost surely equal to 0 and thus would have moment generating function 1.
The only remaining possibility is that log α(t) is linear: log α(t) = log α(0) + at, thus α(t) =
α(0)eat = eat /q . If we plug this into the dening equation of α(t), we obtain
of
Rk
to a random variable with moment generating function
there is no such random variable: All derivatives at
1=
eat
eart X −`(n) a`(n)t
+ r
q
e
f (n)t
q
q
n∈B
12
SARA KROPF AND STEPHAN WAGNER
|t| ≤ δ . However, the right side of this identity has strictly positive second derivative
t unless a = 0 and f (n) = 1 for all n ∈ B (in which case f (n) = 1 for all n). Thus σ 2 6= 0
f ≡ 1.
identically for
for real
unless
Assume that the q-quasiadditive function f has at most logarithmic growth.
Let Nk be a randomly chosen integer in {0, 1, . . . , qk − 1}. The random variable Lk = f (Nk )
has mean µ̂k + O(1) and variance σ̂2 k + O(1), where the two constants µ and σ2 are given by the
same formulas as in Theorem 14, with B(x, t) replaced by
Corollary 15.
X
B̂(x, t) =
x`(n) ef (n)t .
n∈B
√
If f is not the constant function f ≡ 0, then the renormalised random variable (Lk − µ̂k)/(σ̂ k)
converges weakly to a standard Gaussian distribution.
Remark. By means of the Cramér-Wold device (and Corollary 5), we also obtain joint normal
distribution of tuples of
q -quasiadditive
functions.
We now revisit the examples discussed in Section 2 and state the corresponding central limit
theorems. Some of them are well known while others are new. We also provide numerical values
for the constants in mean and variance.
Example
a
5 (see also [6, 15])
2-quasiadditive
.
The number of blocks
0101
occurring in the binary expansion of
n
is
function of at most logarithmic growth. Thus by Corollary 15, the standardised
random variable is asymptotically normally distributed, the constants being
Example 6 (see also [11,19]).
µ̂ =
The Hamming weight of the nonadjacent form is
at most logarithmic growth (as the length of the NAF of
n
1
16 and
σ̂ 2 =
17
256 .
2-quasiadditive with
is logarithmic). Thus by Corollary 15,
the standardised random variable is asymptotically normally distributed. The associated constants
are
1
3 and
µ̂ =
σ̂ 2 =
2
27 .
Example 7 (see Section 2).
0, 1, −1-representations is 2-quasimultiplicative.
2-regular, it has at most polynomial growth. Thus
The number of optimal
As it is always greater or equal to
1
and
Theorem 14 implies that the standardised logarithm of this random variable is asymptotically
normally distributed with numerical constants given by
Example
.
µ ≈ 0.060829, σ 2 ≈ 0.038212.
s1 , s2 , . . . satises sn ≥ 1 and sn = O(cn )
for a constant c ≥ 1. The run length transformation t(n) of sn is 2-quasimultiplicative. As sn ≥ 1
for all n, we have t(n) ≥ 1 for all n as well. Furthermore, there exists a constant A such that
sn ≤ Acn for all n, and the sum of all run lengths is bounded by the length of the binary expansion,
thus
Y
Y
t(n) =
si ≤
(Aci ) ≤ (Ac)1+log2 n .
8 (see Section 2)
Suppose that the sequence
i∈L(n)
Consequently,
t(n)
i∈L(n)
is positive and has at most polynomial growth. By Theorem 14, we obtain an
asymptotic normal distribution for the standardised random variable
and
σ2
log t(Nk ).
The constants
µ
in mean and variance are given by
µ=
X
(log si )2−i−2
i≥1
and
σ2 =
X
X
(log si )2 2−i−2 − (2i − 1)2−2i−4 −
(log si )(log sj )(i + j − 1)2−i−j−3 .
i≥1
j>i≥1
These formulas can be derived from those given in Theorem 14 by means of the representation (7),
P
log t(n) = i≥1 Xi (n) log si , where Xi (n) is the
number of runs of length i in the binary representation of n. The coecients in the two formulas
stem from mean, variance and covariances of the Xi (n).
1 n+2
In the special case that sn is the Jacobsthal sequence (sn = (2
− (−1)n ), see Section 2),
3
2
we have the numerical values µ ≈ 0.429947, σ ≈ 0.121137.
and the terms can also be interpreted easily: write
q -QUASIADDITIVE
FUNCTIONS
13
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Institut für Mathematik, Alpen-Adria-Universität Klagenfurt, Austria, and Institute of Statistical Science, Academia Sinica, Taipei, Taiwan
E-mail address :
sara.kropf@aau.at
and
sarakropf@stat.sinica.edu.tw
Department of Mathematical Sciences, Stellenbosch University, South Africa
E-mail address :
swagner@sun.ac.za