arXiv:1412.0392v1 [math.CO] 1 Dec 2014
ON THE EQUATION m = xyzw WITH x 6 y 6 z 6 w IN POSITIVE
INTEGERS
MADJID MIRZAVAZIRI AND DANIEL YAQUBI
Abstract. As a well-known enumerative problem, the number of solutions of the equation m = m1 + . . . + mk with m1 6 . . . 6 mk in positive integers is Π(m, k) =
Pk
i=0 Π(m − k, i) and Π is called the additive partition function. In this paper, we
give a recursive formula for the so-called multiplicative partition function µ1 (m, k) :=
the number of solutions of the equation m = m1 . . . mk with m1 6 . . . 6 mk in positive
integers. In particular, using an elementary proof, we give an explicit formula for the
cases k = 1, 2, 3, 4.
1. Introduction
Partitioning numbers and sets are two important enumerative problems. These problems can be stated in different ways. Given a positive integer m the total number of
ways of writing m as a sum of positive integers is clearly 2m−1 . To see this we can put
m balls in a row and we then realize that the ways of writing m as a sum of positive
integers correspond to putting some walls between the balls. As we can put walls in m − 1
places, we have the result. On the other hand, the number of solutions of the equation
m = m1 . . . + mk in positive integers for a fixed k, is equal to the number of ways of
putting k − 1 walls into m − 1 allowed places; i.e. m−1
.
k−1
We can also consider the equation m = m1 + . . . + mk in non-negative integers. For a
fixed k, the number of solutions is m+k−1
. This equals to the number of ways of putting
k−1
m balls and k − 1 walls in a row. The total number of ways of writing m as a sum of
non-negative integers cannot be a good question, since the answer is infinity.
Moreover, we can add some conditions to the problem. For instance, we can think
about the number of solutions of the equation m = m1 + . . . + mk with the conditions
ℓi 6 mi 6 ℓ′i for i = 1, . . . , k. A straightforward application of the Inclusion Exclusion
Principle solves this problem.
2010 Mathematics Subject Classification. 11P81, 05A17.
Key words and phrases. Multiplicative partition function.
1
2
MADJID MIRZAVAZIRI AND DANIEL YAQUBI
On the other hand, if we add the condition m1 6 . . . 6 mk to our problem then we have
a more difficult situation. The answer is denoted by Π(m, k) and a recursive argument
Pk
shows that Π(m, k) =
i=0 Π(m − k, i) with the initial values Π(m, m) = Π(m, 1) =
Π(0, 0) = 1. For a discussion about these problems see [1].
These considerations motivate us to think about a multiplicative version of the above
problems. Various types of this problem can be found in sequences A001055, A034836,
A122179, A122180 and A088432 of [7]. We deal with this problem in the present paper
and we give a solution for the problem whenever k = 1, 2, 3 or 4. A survey on this problem
can be found in [4]. The reader is also referred to [3] and [5].
There is a solution to the case k = 4 of multiplicative and additive forms of our problem
using the Polya Enumeration Theorem. For a complete solutions of these problems see
section 7.5 of [2] and for a background material on the Polya method and multiset cycle
indices see [6].
2. A Recursive Formula
Definition 2.1. Let m, k and ℓ be positive integers. We denote the number of solutions
of the equation
m = m1 . . . mk
with mi > ℓ by
νℓ (m, k);
m = m1 . . . mk
with ℓ 6 m1 6 . . . 6 mk by
µℓ (m, k);
m = m1 . . . mk
with ℓ 6 m1 , . . . , mk and k ∈ N by
νℓ (m);
m = m1 . . . mk
m = m1 . . . mk
with ℓ 6 m1 6 . . . 6 mk and k ∈ N by
with mi > ℓ and {m1 , . . . mk } = {m′1 , . . . , m′r }
µℓ (m);
and βj = |{i : mi = m′j }|, 1 6 j 6 r by
νℓ′ (m; β1 , . . . , βr );
and ℓ 6 m1 < . . . < mr by
µ′ℓ (m; β1 , . . . , βr ).
m = mβ1 1 . . . mβr r with βi ∈ N, β1 + . . . + βr = k
Lemma 2.2. Let m > 1, k be positive integers and pα1 1 . . . pαnn be the prime decomposition
of m. Then
and
n Y
αj + k − 1
ν1 (m, k) =
k−1
j=1
Y
n αj + k − i − 1
k
.
ν2 (m, k) =
(−1)
k
−
i
−
1
i
j=1
i=0
k−1
X
i
Proof. Suppose that for 1 6 j 6 n we have αj balls labelled pj and we want to put these
balls into k different cells. There is a one to one correspondence between these situations
ON THE EQUATION m = xyzw WITH x 6 y 6 z 6 w IN POSITIVE INTEGERS
3
and multiplicative partitioning m = m1 . . . mk . In fact we can consider mj as the product
+k−1
of the balls in cell j. There are αjk−1
ways to put balls labelled pj . Thus the first part
is obvious.
For the second part, let Er be the set of all situations in which the cell r is empty,
where 1 6 r 6 k. Then we have
n Y
αj + k − i − 1
,
|Er1 ∩ . . . ∩ Eri | =
k
−
i
−
1
j=1
1 6 i 6 k − 1.
Thus the Inclusion Exclusion Principle implies the result.
Example 2.3. We evaluate ν2 (m) for m = pα . Clearly
α+k−1
ν1 (m, k) =
k−1
and
ν2 (m) =
Y
n αj + k − i − 1
k
.
(−1)
k
−
i
−
1
i
j=1
i=0
α X
k−1
X
k=1
i
Suppose that there are k boys and α − 1 girls and we want to choose a team consisting
of k − 1 girls. This is obviously equal to α−1
.
k−1
Let Ar be the set of all situations in which the boy r is belonged to the chosen team.
Then
α+k−i−1
,
|Ar1 ∩ . . . ∩ Ari | =
k−i−1
1 6 i 6 k − 1.
By the Inclusion Exclusion Principle we therefore have
Y
n α X
k−1
X
αj + k − i − 1
i k
ν2 (m) =
(−1)
k−i−1
i j=1
k=1 i=0
=
α−1 X
k−1
k=1
α−1
= 2α−1 .
Proposition 2.4. Let m > 1 and k, ℓ be positive integers. Suppose that ℓs divides m but
ℓs+1 does not divide m. Then
min{k,s}
µℓ (m, k) =
X
i=max{k−s,1}
µℓ+1 (m, i).
4
MADJID MIRZAVAZIRI AND DANIEL YAQUBI
Proof. Let
E = {(m1 , . . . , mk ) : m = m1 . . . mk , ℓ 6 m1 6 . . . 6 mk }
and
Ei = {(m1 , . . . , mk ) ∈ E : m1 = . . . = mi = ℓ, mi+1 6= ℓ},
min{k−1,s}
Then E = ∪i=0
µℓ (m, k) =
i = 0, 1, . . . , min{k − 1, s}.
Ei and the union is disjoint. Thus
min{k−1,s}
min{k−1,s}
X
X
i=0
|Ei | =
i=0
min{k,s}
µℓ+1(m, k − i) =
X
µℓ+1 (m, i).
i=max{k−s,1}
Corollary 2.5. Let m > 1 and k, ℓ be positive integers. Then
µ1 (m, k) =
k
X
µ2 (m, i).
i=1
Proof. Using Proposition 2.4 we have the result, since 1s | m for each positive integer
s.
The following result is straightforward.
Lemma 2.6. Let m, k and ℓ be positive integers. Then
P
i. νℓ (m, k) = ℓ6d|m νℓ ( md , k − 1);
P
ii. µℓ (m, k) = ℓ6d|m µd ( md , k − 1);
P
iii. νℓ (m, k) = βi ∈N,β1 +...+βr =k νℓ′ (m; β1 , . . . , βr );
P
iv. µℓ (m, k) = βi ∈N,β1<...<βr ,β1 +...+βr =k µ′ℓ (m; β1 , . . . , βr );
v. µ′ℓ (m; β1 , . . . , βr ) =
k!
ν ′ (m; β1 , . . . , βr ),
β1 !...βr ! ℓ
βi ∈ N, β1 + . . . + βr = k.
3. An Explicit Formula For the Cases k = 1, 2, 3, 4 and ℓ = 1, 2
We now aim to give an explicit formula for the cases k = 1, 2, 3, 4 and ℓ = 1, 2. In the
following, we denote the number of natural divisors of m by τ (m). Moreover,
(
√
1 if i m ∈ N
εi (m) =
0 otherwise
Proposition 3.1. Let m > 1 be a positive integer. Then
µ1 (m, 1) = µ2 (m, 1) = 1,
τ (m)
τ (m)
⌉ and µ2 (m, 2) = ⌈
⌉ − 1.
µ1 (m, 2) = ⌈
2
2
ON THE EQUATION m = xyzw WITH x 6 y 6 z 6 w IN POSITIVE INTEGERS
5
Proof. The first part is obvious. For the second part, we note that ν1 (m, 2) is number of
ways to write m as xy, where x is a natural divisor of m. Thus ν1 (m, 2) = τ (m). Now if
m is not a perfect square then τ (m) is even and so µ1 (m, 2) =
a perfect square then µ1 (m, 2) =
µ2 (m, 2) = µ1 (m, 2) − 1.
τ (m)−1
2
τ (m)
2
= ⌈ τ (m)
⌉ and if m is
2
+ 1 = ⌈ τ (m)
⌉. Using Corollary 2.5, we now have
2
Theorem 3.2. Let m > 1 be a positive integer and pα1 1 . . . pαnn be its prime decomposition.
Then
n n
1 Y αj + 2
1 Y αj + 2
ε3 (m)
µ1 (m, 3) =
+
⌊
⌋+
,
2
6 j=1
2 j=1
2
3
µ2 (m, 3) = µ1 (m, 3) − ⌈
τ (m)
⌉.
2
Proof. We have
ν1 (m, 3) = ν1′ (m; 1, 1, 1) + ν1′ (m; 1, 2) + ν1′ (m; 3)
= 6µ′1 (m; 1, 1, 1) + 3µ′1 (m; 1, 2) + µ′1 (m; 3).
We know that µ′1 (m; 1, 2) is the number of ways to write m as xy 2 , where x 6= y. This is
equal to the number of y’s such that y 2 | m minus the number of ways such that ym2 = y,
Q
α +2
in which the latter is equal to ε3 (m). The number of y’s such that y 2 | m is nj=1 ⌊ j2 ⌋.
Moreover, µ′1 (m; 3) = ε3 (m). Thus Lemma 2.2 implies
µ′1 (m; 1, 1, 1)
n
Y
1
αj + 2
= ν1 (m, 3) − 3
⌊
⌋ − ε3 (m) − ε3 (m)
6
2
j=1
n
1
1 Y αj + 2
1
= ν1 (m, 3) −
⌊
⌋ + ε3 (m)
6
2 j=1
2
3
n n
1 Y αj + 2
ε3 (m)
1 Y αj + 2
−
⌊
⌋+
.
=
2
6 j=1
2 j=1
2
3
we have
µ1 (m, 3) = µ′1 (m; 1, 1, 1) + µ′1 (m; 1, 2) + µ′1 (m; 3)
n
n n
1 Y αj + 2
1 Y αj + 2
ε3 (m) Y αj + 2
=
−
⌋+
+ ⌊
⌋ − ε3 (m) + ε3 (m).
⌊
6 j=1
2 j=1
2
3
2
2
j=1
This proves the first part.
6
MADJID MIRZAVAZIRI AND DANIEL YAQUBI
For the second part, using Corollary 2.5, we have
µ2 (m, 3) = µ1 (m, 3) − µ2 (m, 2) − 1 = µ1 (m, 3) − ⌈
τ (m)
⌉ + 1 − 1.
2
Lemma 3.3. Let k, ℓ > 1 be a positive integer and pβ1 1 . . . pβnn be the prime decomposition
of k. Then
X
n Y
βj + 2
τ (d) =
2
j=1
X
n
Y
βj + ℓ
εℓ (d) =
⌊
⌋.
ℓ
j=1
d|k
and
d|k
Theorem 3.4. Let m > 1 be a positive integer and pα1 1 . . . pαnn be its prime decomposition.
then
n n
n
1 Y αi + 3
1 Y αi + 3
1 Y αi + 2
αi − 2
µ1 (m, 4) =
+
⌊
⌋+ ( ⌊
⌋(αi − ⌊
⌋))
3
24 i=1
3 i=1
3
4 i=1
2
2
√
n
ε2 (m) τ ( m)
3ε4 (m)
ε2 (m) Y αi + 2
⌊
⌋−
⌈
⌉+
.
+
4 i=1
2
4
2
8
Moreover,
µ2 (m, 4) = µ1 (m, 4) − µ1 (m, 3).
Proof. We have
ν1 (m, 4) = ν1′ (m; 1, 1, 1, 1) + ν1′ (m; 1, 1, 2) + ν1′ (m; 1, 3) + ν1′ (m; 2, 2) + ν1′ (m; 4)
= 24µ′1(m; 1, 1, 1, 1) + 12µ′1(m; 1, 1, 2) + 4µ′1 (m; 1, 3) + 6µ′1 (m; 2, 2) + µ′1 (m; 4).
We know that µ′1 (m; 1, 1, 2) is the number of ways to write m as xyz 2 , where x, y and z
are different positive integers. This is equal to the number of z’s such that z 2 | m minus
the number of ways to write m as xz 3 , x2 z 2 or z 4 , where x 6= z. The number of z’s such
P
τ( m )
that z 2 | m is z 2 |m ⌈ 2z2 ⌉. Thus
µ′1 (m; 1, 1, 2) =
X τ ( m2 )
⌈ z ⌉ − µ′1 (m; 1, 3) − µ′1 (m; 2, 2) − ε4 (m).
2
2
z |m
ON THE EQUATION m = xyzw WITH x 6 y 6 z 6 w IN POSITIVE INTEGERS
7
If m is not a perfect square then τ (m) is even. Let z = pβ1 1 . . . pβnn . There exists an index
i such that αi is odd and then
X 1
X τ ( m2 )
τ (pα1 1 −2β1 . . . pnαn −2βn )
⌈ z ⌉=
2
2
2
2
z |m
z |m
n
1Y X
=
(αi − 2βi + 1)
2 i=1
αi
0≤βi ≤
2
n
1 Y αi + 2
αi − 2
= ( ⌊
⌋(αi − ⌊
⌋)).
2 i=1
2
2
Moreover, µ′1 (m; 1, 3) is the number of ways to write m as xz 3 , where x 6= z. This is equal
to the number of z’s such that z 3 | m minus the number of ways to write m as z 4 . The
Q
α +3
number of z’s such that z 3 | m is nj=1⌊ j3 ⌋. Whence
µ′1 (m; 1, 3)
n
Y
αj + 3
⌋ − ε4 (m).
=
⌊
3
j=1
Since m is not a perfect square, µ′1 (m; 2, 2) = 0. Thus Lemma 2.2 implies
1
ν1 (m, 4) − 12µ′1 (m; 1, 1, 2) − 4µ′1 (m; 1, 3) − 6µ′1 (m; 2, 2) − ε4 (m)
24
n
n 1 Y αi + 3
αi − 2
1 Y αi + 2
=
⌋(αi − ⌊
⌋))
− ( ⌊
24 i=1
4 i=1
2
2
3
µ′1 (m; 1, 1, 1, 1) =
1
11
1
+ µ′1 (m; 1, 3) + µ′1 (m; 2, 2) + ε4 (m)
3
4
24
n
n
1 Y αi + 3
1 Y αi + 2
αi − 2
1
=
− ( ⌊
⌋(αi − ⌊
⌋)) + µ′1 (m; 1, 3).
3
24 i=1
4 i=1
2
2
3
Therefore we have
µ1 (m, 4) = µ′1 (m; 1, 1, 1, 1) + µ′1 (m; 1, 1, 2) + µ′1 (m; 1, 3) + µ′1 (m; 2, 2) + µ′1 (m; 4)
n n
n
1 Y αi + 3
1 Y αi + 2
αi − 2
1 Y αi + 3
+
⌊
⌋+ ( ⌊
⌋(αi − ⌊
⌋)).
=
3
24 i=1
3 i=1
3
4 i=1
2
2
8
MADJID MIRZAVAZIRI AND DANIEL YAQUBI
Now let m be a perfect square. Then τ (m) is odd and we have
X τ ( m2 ) + 1
X τ ( m2 )
( d
)
⌈ d ⌉=
2
2
2
2
z |m
z |m
=
X 1 α −2β
1X
1
τ (p1 1 1 . . . pnαn −2βn ) +
2
2 2
2
z |m
=
1
2
z |m
X
α
0≤βi ≤ 2i
n
Y
i=1
(αi − 2βi + 1) +
1
2
X
1
α
0≤βi ≤ 2i
n
n
1Y X
1 Y αi + 2
(⌊
⌋)
(αi − 2βi + 1) +
=
2 i=1
2 i=1
2
αi
0≤βi ≤
2
n
Y
n
1
αi + 2
αi − 2
1 Y αi + 2
(⌊
= ( ⌊
⌋(αi − ⌊
⌋)) +
⌋)
2 i=1
2
2
2 i=1
2
Thus
µ′1 (m; 1, 1, 1, 1)
n n
1 Y αi + 3
1 Y αi + 2
αi − 2
=
− ( ⌊
⌋(αi − ⌊
⌋))
3
24 i=1
4 i=1
2
2
n
1
1
1
1 Y αi + 2
(⌊
⌋) + µ′1 (m; 1, 3) − µ′1 (m; 2, 2) − ε4 (m).
+
2 i=1
2
3
4
24
Furthermore, µ′1 (m; 2, 2) is the number of ways to write m as x2 y 2 = (xy)2, where x 6= y.
√
If m is not a perfect square then this number is 0 and if m is a perfect square then m ∈ N
and thus
√
√
√ µ′1 (m; 2, 2) = ε2 (m)µ′2 ( m; 1, 1) = ε2 (m) µ2 ( m, 2) − ε2 ( m)
√
τ ( m)
⌉ − ε4 (m)
= ε2 (m) ⌈
2
√
τ ( m)
= ε2 (m)⌈
⌉ − ε4 (m).
2
Therefore
n n
n
1 Y αi + 3
1 Y αi + 3
1 Y αi + 2
αi − 2
+
⌊
µ1 (m, 4) =
⌋+ ( ⌊
⌋(αi − ⌊
⌋))
3
24 i=1
3 i=1
3
4 i=1
2
2
√
n
ε2 (m) τ ( m)
3ε4 (m)
ε2 (m) Y αi + 2
⌊
⌋−
⌈
⌉+
.
+
4 i=1
2
4
2
8
This proves the first part. The second part is obvious.
ON THE EQUATION m = xyzw WITH x 6 y 6 z 6 w IN POSITIVE INTEGERS
9
Note that if a positive integer m is a prime power, say m = pn , then µ2 (m, k) = Π(n, k),
where Π is the additive partition function.
Corollary 3.5. Let n be a positive integer. Then the number of all solutions of the
equation
x1 + x2 + x3 = n
in N, under the condition x1 ≤ x2 ≤ x3 , is
1 n+2
ε3 (pn )
1 n+2
+ ⌊
⌋+
.
2
6
2
2
3
Corollary 3.6. Let n be positive a integer. Then the number of all solutions of the
equation
x1 + x2 + x3 + x4 = n
in N, under the condition x1 ≤ x2 ≤ x3 ≤ x4 , is
1 n+3
1 n+3
1 n+2
n−2
+ ⌊
⌋ + (⌊
⌋(n − ⌊
⌋))
3
24
3
3
4
2
2
√
ε2 (pn ) n + 2
ε2 (pn ) τ ( pn )
3ε4 (pn )
+
⌊
⌋−
⌈
⌉+
.
4
2
4
2
8
Remark 3.7. We can compute µ1 (m, 4) using the following Mathlab code.
clear all
n=input(‘enter your number’)
for m=1:n
a=Power(m);
b=xyzw(m,a);
c(m,1)=b;
end
function a=Power(m1)
B=factor(m1);[n2 n1]=size(B);i=1;j=1;a(j)=1;
while i<n1
if B(i)==B(i+1)
a(j)=a(j)+1;
else j=j+1;a(j)=1;
end
i=i+1;
10
MADJID MIRZAVAZIRI AND DANIEL YAQUBI
end
end
function b=xyzw(m,a)
[n2 n1]=size(a);
b1=1;b2=1;b3=1;
e2=0;e4=0;
m1=floor(m^(1/2));
if m1==(m^(1/2))
e2=1;
end
m1=floor(m^(1/4));
if m1==m^(1/4)
e4=1;
end
for i=1:n1
b1=(1/6)*(a(i)+3)*(a(i)+2)*(a(i)+1)*b1;
end
for i=1:n1
b2=floor((a(i)+3)/3)*b2;
end
for i=1:n1
b3=floor((a(i)+2)/2)*(a(i)-floor((a(i)-2)/2))*b3;
end
b4=1;
for i=1:n1
b4=floor((a(i)+2)/2)*b4;
end
b5=1;
for i=1:n1
b5=b5*((a(i)/2)+1);
end
b=(1/24)*b1+(1/3)*b2+(1/4)*b3+(1/4)*e2*b4-(1/4)*e2*ceil(b5)+(3/8)*e4;
end
ON THE EQUATION m = xyzw WITH x 6 y 6 z 6 w IN POSITIVE INTEGERS
11
References
[1] G. E. Andrews, The Theory of Partitions, Cambridge, England: Cambridge University Press, 1998.
[2] J. Herman, R. Kucera and J. Simsa, Counting and Configurations, Problems in Combinatorics,
Arithmetic, and Geometry, Original Czech edition published by Masaryk University, Brno, 1997,
Translated by K. Dilcher, Springer, 2003.
[3] J. F. Hughes and J. Shallit, On the number of multiplicative partitions, American Mathematical
Monthly 90 (7) (1983), 468-471.
[4] A. Knopfmacher, M. E. Mays, A survey of factorization counting functions, International Journal of
Number Theory 1 (4) (2005), 563-581.
[5] A. Knopfmacher and M. E. Mays, Ordered and Unordered Factorizations of Integers, Mathematica
Journal 10 (2006), 72-89.
[6] M. Riedel, http://math.stackexchange.com/users/44883/marko-riedel.
[7] N. J. A. Sloane, The On-Line Encyclopaedia of Integer Sequences.
Department of Pure Mathematics, Ferdowsi University of Mashhad, P. O. Box 1159,
Mashhad 91775, Iran.
E-mail address: [email protected], danyal [email protected]