1. Introduction
As it is known, Pascal’s triangle is constructed in the following way: Write the first row “”. Then each member of each subsequent row is given by taking the sum of the just above two members, regarding any blank as 0.
Example 1. Here is the Pascal’s triangle from the first row to the 7-th row: Remark 1. For any integers and , we putwhere we put . Then it is well-known that the n-th row in the Pascal’s triangle is equal to the sequenceconsisting of terms. In ([
1], Section 1.4), the construction above is generalized as follows:
Definition 1. Let be any integer. The k-Pascal’s triangle is constructed in the following way: Write the first row “”. Then each member of each subsequent row is given by taking the sum of the just above k members regarding the blank as 0.
Example 2. In the case where , the 4-Pascal’s triangle from the first row to the 5-th row is the following: Remark 2. In ([1], Section 1.4), for any integers and , it is mentioned that the n-th row in the k-Pascal’s triangle consists of integerssatisfying the equationof polynomials with indeterminate x and integral coefficients. A detailed proof of this fact is described in ([2], Lemma 1.1). (2) In ([1], Section 9.10), the following formula for is described:where is the greatest integer that is less than or equal to . In Example 1, we can see that the n-th row consists of odd integers when n is equal to the Mersenne number 1, 3 or 7. Actually, Lucas showed the following
Theorem 1 ([
3], Exemple I in Section 228).
Let be any integer. Then is odd for any if and only if n is a Mersenne number, i.e., n is of the form with some integer . In
Section 2 in this article, we generalize the Lucas’ result above as the following
Theorem 2. Let p be any prime number and e any positive integer. For any integer , the n-th row in the -Pascal’s triangle consists of integers which are congruent to 1 modulo p if and only if n is of the form with some integer .
Remark 3. (1) Theorem 2 is a generalization of ([2], Theorem 0.2) which is in the case where . (2) We can see that Example 2 gives a partial example of Theorem 2 in the case where , and .
As an application of Theorem 2, we can prove that ([
2], Conjecture 0.3) holds for
, i.e., there exists no row in the 4-Pascal’s triangle consisting of integers which are congruent to 1 modulo 4 except the first row as follows:
By Theorem 2, in the case where , we see that for any integer , the n-th row in the 4-Pascal’s triangle consists of odd integers if and only if n is of the form with some integer .
Moreover, we can see an essential property of the
-th row in the 4-Pascal’s triangle for any integer
as in the following theorem proved in
Section 3.2:
Theorem 3. For any integer , the -th row in the 4-Pascal’s triangle is congruent to the sequencemodulo 4, which consists of the repeated 1133’s and 3311’s whose numbers are the same . Therefore we can obtain the following
Corollary 1. ([2], Conjecture 0.3) holds for , i.e., there exists no row in the 4-Pascal’s triangle consisting of integers which are congruent to 1 modulo 4 except the first row. Remark 4. (1) By Example 2, in the case where , we can see that the 5-th row in the 4-Pascal’s triangle is congruent to the sequencemodulo 4, which matches the assertion of Theorem 3. (2) It seems that one could obtain the forms of the sequenece to which the -th row in the 4-Pascal’s triangle is congruent modulo 4 for some positive integers ℓ by means of Theorem 3. We would like to do these calculations in the future.
Lemma 1. For any prime number p and any positive integer e, we have the following coefficient-wise congruenceof binomial expansions with indetermiate x. 2. A Proof of Theorem 2
Although Theorem 2 can be proved by the same argument as the proof of ([
2], Theorem 0.2), we shall describe its detailed proof here to make this article self-contained.
Let n and e be any positive integers and p be any prime number.
Firstly, we assume that
n is of the form
with some integer
. In the algebra
of polynomials of one varible
x with coefficients in the finite field
of
p elements, we see that for any positive integer
ℓ,
Therefore we see that
in
. By Remark 2 (1), this implies that the
n-th row in the
-Pascal’s triangle consists of integers which are congruent to 1 modulo
p as desired.
Conversely, we now assume that
n is of the form
with some integers
and
. Moreover, we assume that we have
in
to obtain some contradiction. Since the left hand side is equal to
and the right hand side is equal to
, we then have the equality
in
. Since
, this implies that
Let
be the
p-adic valuation of any non-zero integer
a, i.e.,
and
. Since
, we see that
and then
Therefore we can put
with some positive integer
t which is prime to
p. Then we have
which implies that
since
is an integral domain. Since
, we see that
. Therefore substituting
leads a contradiction
in
as desired, and Theorem 2 is proved.
3. An Application to the 4-Pascal’s Triangle
By Theorem 2, in the case where and , we see that for any integer , the n-th row in the 4-Pascal triangle consists of odd integers if and only if n is of the form with some integer .
In this section, we shall prove Theorem 3 asserting that for any integer
, the
-th row in the 4-Pascal’s triangle is congruent to the sequence
modulo 4. Here we should note that
is the number of 1133’s and 3311’s, respectively.
Then Theorems 2 and 3 imply that ([
2], Conjecture 0.3) holds in the case where
, i.e., there exists no row in the 4-Pascal’s triangle consisting of integers which are congruent to 1 modulo 4 except the first row as we have seen in Corollary 1.
3.1. On a Congruence of Binomial Expansions
Before proving Theorem 3, we shall prove Lemma 1 on a congruence of binomial expansions in this subsection.
Let
p be any prime number and
e any positive integer. In order to prove the congruence
of binomial expansions with indeterminate
x, it suffices to see the following two congruences hold:
(1) For any integer
which is prime to
p,
(2) In the case where
, for any integers
and
i such that
and
,
Firstly, we shall prove the part (1). In the case where
, we see that
Moreover, in the case where
, we see that
Since
for any
and
ℓ is prime to
p, we then see that
Therefore , and part (1) is proved.
Secondly, we shall prove part (2). We see that
and that
Since , we then see that is divisible by as desired.
3.2. A Proof of Theorem 3
Now we shall prove Theorem 3 by means of Lemma 1 with
and
, i.e., the congruence of binomial expansions
By Remark 2 (1), proving Theorem 3 is equivalent to proving that for any integer
, the coefficient-wise congruence
holds with indeterminate
x by the induction on
m.
Before doing this, we see the following
Lemma 2. The polynomial in the right hand side of the congruence relation can be decomposed as Proof. By a direct calculation, we can see that there exists some positive integer
ℓ such that the polynomial in the right hand side of the congruence relation
can be decomposed as
Since the degree of the polynomial in the right hand side of the congruence relation
is equal to
, we then see that
which implies that
as desired. □
Let us start to prove Theorem 3 by the induction on
. Firstly, in the case where
, since
and
by the congruence relation
, we see that
Therefore the congruence relation holds for by Lemma 2.
Secondly, we assume that the congruence relation
holds for some
. By the congruence relation
, we see that
By Lemma 2, we then see that
i.e., the congruence relation
also holds for
as desired. This proves Theorem 3.