On the divisibility of binomial coefficients
Abstract.
In Pacific J. Math. 292 (2018), 223–238, Shareshian and Woodroofe asked if for every positive integer there exist primes and such that, for all integers with , the binomial coefficient is divisible by at least one of or . We give conditions under which a number has this property and discuss a variant of this problem involving more than two primes. We prove that every positive integer has infinitely many multiples with this property.
Key words and phrases:
Binomial coefficients, divisibility, primorials2010 Mathematics Subject Classification:
Primary 11B65; Secondary 05A101. Introduction
Binomial coefficients display interesting divisibility properties. Conditions under which a prime power divides a binomial coefficient are given by Kummer’s Theorem [10] and also by a generalized form of Lucas’ Theorem [5, 12].
Still, there are problems involving divisibility of binomial coefficients that remain unsolved. In this article we investigate the following question, which was asked by Shareshian and Woodroofe in [15].
Question 1.1.
Is it true that for every positive integer there exist primes and such that, for all integers with , the binomial coefficient is divisible by or ?
As in [15], we say that satisfies Condition if such primes and exist for . In this article we discuss sufficient conditions under which an integer satisfies Condition 1. In Sections 2 and 3 we prove a variation of the Sieve Lemma from [15] and use it to show that satisfies Condition 1 if certain inequalities hold. In Section 5 we infer that every positive integer has infinitely many multiples for which Condition 1 is satisfied.
The collection of numbers for which Condition 1 is not known to hold has asymptotic density assuming the truth of Cramér’s conjecture (as first shown in [15]) and includes most primorials , where are the first primes, namely those primorials such that is not a prime.
In addition, we introduce the following variant of Condition 1:
Definition 1.2.
A positive integer satisfies the variation of Condition 1 if there exist different primes such that if then is divisible by at least one of .
For example, it follows from Kummer’s Theorem or from Lucas’ Theorem that a positive integer satisfies the variation of Condition if and only if is a prime power, and every integer satisfies the variation of Condition 1 if where are distinct primes. In Section 4 we discuss upper bounds on so that a given satisfies the variation of Condition 1.
2. An extended sieve lemma
Our results in this section will be based on Lucas’ Theorem:
Theorem 2.1 (Lucas [12]).
Let be a prime and let
be base expansions of two positive integers, where and for all , and . Then
By convention, a binomial coefficient is zero if . Hence, if any of the digits of the base expansion of is whereas the corresponding digit in the base expansion of is nonzero, then is divisible by . As a particular case, if a prime power with divides and does not divide , then is divisible by .
Observe that, if satisfies Condition 1 with two primes and , then at least one of these primes has to be a divisor of , because otherwise would not be divisible by any of them. The next two results are elementary consequences of Lucas’ Theorem.
Proposition 2.2.
If with a prime and , then satisfies Condition with and any prime dividing .
Proof.
If is a prime power then the two summands in the lefthand term of the equality
are divisible by by Lucas’ Theorem if , and hence is also divisible by . When or , we have that , so any prime factor of divides . ∎
Proposition 2.3.
If a positive integer is equal to the product of two prime powers and with , , and , then satisfies Condition with and .
Proof.
The base expansion of ends with zeroes and the base expansion of ends with zeroes. Because a positive integer smaller than cannot be divisible by both and , it is not possible that ends with zeroes in base and zeroes in base . Consequently, we can apply Lucas’ Theorem modulo if does not divide or modulo if does not divide . ∎
Proposition 2.3 generalizes as follows.
Proposition 2.4.
If are distinct primes and with for all , then satisfies the variation of Condition with .
Proof.
If , then the base expansion of ends with less zeroes than the base expansion of for at least one prime factor of . ∎
The following result extends [15, Lemma 4.3]. It is the starting point of our discussion of Question 1.1 in the next sections.
Theorem 2.5.
Let be a positive integer and suppose that divides where is a prime and . Suppose that there is a prime with , where . Then is divisible by or except possibly when is a multiple of belonging to one of the intervals with and .
Proof.
By symmetry, we only need to consider those values of with . Moreover, we may restrict our study further to those values of that are multiples of , since otherwise is divisible by .
Since , the number is positive. If then is in the interval , which is the case in the statement of the theorem.
The assumption that is equivalent to assuming the inequality , which implies that the last digit in the base expansion of is equal to . Hence, if then we may infer from Lucas’ Theorem that is divisible by .
The remaining range of values of to be considered is . In this case we look at the last digit of the base expansion of . If this last digit is bigger than , then is again divisible by . Thus the undecided cases are those in which the residue of mod is smaller than or equal to . This happens when for some positive integer , and if then . ∎
By the Bertrand–Chebyshev Theorem [2], for every integer there exists a prime such that . This yields the following particular instance of Theorem 2.5, which is also a special case of [15, Lemma 4.3].
Corollary 2.6.
For a positive integer , suppose that divides where is a prime and . If is a prime such that and , then satisfies Condition with and .
Proof.
Pick in Theorem 2.5. ∎
Note that, under the assumptions of Corollary 2.6, the equality cannot hold, since divides and because does not divide . Hence there remains to study the case when and is the largest prime smaller than while is the largest prime power dividing . In other words, Condition 1 holds for whenever there is a prime between and .
The sequence of integers for which there is no prime between and can be found in The OnLine Encyclopedia of Integer Sequences (OEIS [3]) with the reference A290203. Its first terms are the following:
(2.1) 
Banderier’s conjecture [1] claims that if denotes the th primorial, that is,
where are the first primes, and is the largest prime below , then either or is a prime.
Proposition 2.7.
If Banderier’s conjecture is true, then the sequence (2.1) contains all primorials such that is not a prime.
Proof.
If is not a prime, then is a prime according to Banderier’s conjecture. Since does not divide , we infer that is bigger than , which is the largest prime power dividing . ∎
The first primorials such that is not a prime are
, , , . 
Inspecting this list could be a strategy to seek for a counterexample for Question 1.1. The complementary list of primorials can be found in OEIS with reference A057704.
For any fixed value of , the number in Theorem 2.5 is smallest when is as close as possible to . For this reason, we focus our attention on the largest prime below for various values of . This motivates the next definition.
Definition 2.8.
For positive integers and , let be the largest prime smaller than and let . For each integer with , we call a dangerous interval.
By Theorem 2.5, if we attempt to prove that Condition 1 holds with and assuming that —that is, assuming that the dangerous intervals are disjoint— we only need to care about values of that lie in a dangerous interval and are multiples of the largest power of dividing .
In the case , the only dangerous interval below is . When , we have that and are dangerous intervals. Since , the second interval may be replaced by to carry our study further, as we do in the next section.
Example 2.9.
The largest prime below is and the largest prime dividing is . Here and therefore is divisible by or for all except for .
On the other hand, the largest prime below is . Thus and therefore and are dangerous intervals. The second interval contains a multiple of , namely . However, since
we infer from Lucas’ Theorem that is divisible by . Consequently, is divisible by or for all .
3. Using the nearest prime below
Nagura showed in [13] that, if , then there is a prime between and . Therefore, there is a prime such that when . This implies that, if and the largest primepower divisor of satisfies , then there is a prime between and and hence Condition 1 holds for with and .
The following result is sharper.
Proposition 3.1.
If and the largest primepower divisor of satisfies , then satisfies Condition with and the nearest prime below .
Proof.
The following are consequences of Nagura’s and Schoenfeld’s bounds.
Lemma 3.2.
Let be the largest prime below for positive integers and .

If and , then .

If and , then .
Proof.
In order to apply Theorem 2.5 with for a given , we need that there is a prime such that . If denotes the nearest prime below , then the inequality holds if by Lemma 3.2. Since by (2.1) we have that if , we may assume that without any loss of generality.
Note that the inequality is equivalent to , so the intervals and are disjoint.
Theorem 3.3.
For an odd positive integer and a prime power dividing , suppose that there is a prime with and . Then satisfies Condition with and .
Proof.
By Theorem 2.5, in order to infer that is divisible by or , the only cases that we need to discuss are those values of that are multiples of with or . By assumption, there are no multiples of in . Since , we may focus on the interval . Since is odd, is not an integer; hence we are only left to prove that there is no multiple of with . We will prove this by contradiction.
Thus suppose that for some integer . The assumption that implies that and hence
Consequently, . If we now write , we obtain that , which is impossible for an integer . ∎
The rest of this section is devoted to the case when is even.
Lemma 3.4.
Suppose that is even and there is a prime with and , where is the largest power of dividing . If there is a multiple of in the interval , then is odd and .
Proof.
Suppose first that is odd. Then the integer is a multiple of , so we may write for some integer . If there is another multiple of in the interval , then , and this implies that
Hence , which is incompatible with our assumption that .
In the case (so that is the largest power of dividing ), we have that is divisible by , and we may write with odd. If there is a multiple of in the interval , then , so and because is odd. Therefore
Hence, as above, , which contradicts that . ∎
Theorem 3.5.
For an even positive integer , suppose that there is a prime with and , where is the largest power of dividing .

If , then satisfies Condition with and .

If , then satisfies Condition with and if and only if is divisible by .
Proof.
Our last remarks in this section correspond to the case when is even, and they are only relevant if , by Theorem 3.5. Sufficient conditions are given to infer that a prime divides . The greatest integer less than or equal to a real number is denoted by , and we write if is the maximum power of such that divides .
Proposition 3.6.
Suppose that is even. A prime divides if and only if at least one of the numbers with is odd.
Proof.
By comparing and we see that, for each ,
if is even. If is even for all , we conclude that , and hence does not divide . However, if is odd, then
and consequently is greater than . ∎
Corollary 3.7.
If is even and is odd, then divides .
Corollary 3.8.
Suppose that is even.

If any of the digits in the base expansion of is larger than , then divides .

If one of the digits in the base expansion of is odd, then divides .
Proof.
If a digit of in base is larger than , then when we add to itself in base to obtain there is at least one carry. Similarly, if has an odd digit in base , then there is a carry when adding and in base . Hence, by Kummer’s Theorem [10] with , if there is at least one carry when adding to itself in base , then divides . ∎
Corollary 3.9.
Let be an even positive integer. Suppose that there is a prime such that and , where denotes the largest power of dividing . If , then divides and therefore satisfies Condition with and .
Proof.
In those cases when the inequalities and both fail for the largest prime power dividing , a possible strategy is to analyze the inequality for bigger values of , where is the largest prime below .
Up to there are integers that do not satisfy , where is the largest prime power dividing . The OnLine Encyclopedia of Integer Sequences has published these numbers [4] with the reference A290290. Among these, there are that do not satisfy the inequality ; there are that do not satisfy the inequality either; there are for which the inequality also fails, and there is only one integer for which the inequality still fails (namely, ). However, the value of need not decrease as grows, and the number of dangerous intervals that one needs to inspect when increases linearly with . Therefore this strategy is not conclusive, although it often works in practice.
Example 3.10.
The largest prime power dividing is . In this case, and . Thus, Condition 1 fails for and and it also fails for and . Nevertheless, works, as the dangerous interval contains one multiple of , namely , and is divisible by . Therefore Condition 1 holds for and .
Example 3.11.
For , the inequality fails while is true. However, is not divisible by . Hence we look for greater values of and find that with . Now and , yet and are both divisible by . Hence Condition 1 is satisfied with and .
Example 3.12.
For , the inequality is satisfied with but not with any smaller value of . There are dangerous intervals of length . Each of these intervals (except the first) contains one multiple of , and in each case the corresponding binomial coefficient happens to be divisible by . Therefore Condition 1 is satisfied with and .
4. On the variation of Condition 1
Recall from Definition 1.2 that satisfies the variation of Condition 1 if there are primes such that if then is divisible by at least one of .
Theorem 4.1.
If an even positive integer satisfies for a prime with , where is the largest power of dividing and , then satisfies the variation of Condition with , and any prime that divides .
Proof.
According to part (b) of Theorem 3.5, the only binomial coefficient with that might fail to be divisible by or is . Hence it suffices to add an extra prime with this purpose. ∎
Proposition 4.2.
For a positive integer , let be the largest prime smaller than , let be the largest primepower divisor of and let be the second largest primepower divisor of . If , then satisfies the variation of Condition with , and .
Proof.
By Lucas’ Theorem, for any such that , the binomial coefficient is divisible by , and for any such that the binomial coefficient is divisible by . Thus we need to add a prime that divides at least the binomial coefficients with in which is a multiple of . For this, we pick and therefore we only need to consider those values of that are, in addition, multiples of . The least that is a multiple of both prime powers is . Therefore, if , then all values of lying in the interval are such that is divisible by or . ∎
In the statement of Proposition 4.2, the condition that holds by Nagura’s bound [13] if we impose instead that .
For each , we are interested in the minimum number of primes such that satisfies the variation of Condition 1. We next discuss upper bounds for .
Proposition 4.3.
For positive integers and , suppose that there is a prime such that and a primepower divisor of such that . Then satisfies the variation of Condition with .
Proof.
By Theorem 2.5, the binomial coefficients are divisible by except possibly if lies in a dangerous interval. In the dangerous intervals we only need to consider those integers that are multiples of , since otherwise is divisible by . Since we are assuming that , we know that in each dangerous interval there is at most one multiple of . This means that the worst case is the one in which there is a multiple of in every dangerous interval with . Hence we pick one extra prime for each such interval. ∎
Corollary 4.4.
If and where divides and is the largest prime below , and , then satisfies Condition with and .
Proof.
By Lemma 3.2, we may assume that . If , then equals or . If , then the assumption that implies that no multiple of falls into any dangerous interval until . If , then we need to check that in order to ensure that does not fall into the third dangerous interval. The minimum value of such that our assumption holds is . The next multiple of is , which is greater than and therefore does not fall into the third dangerous interval. ∎
In order to refine the conclusion of Proposition 4.3, we consider the Diophantine equation
(4.1) 
for , where is a primepower divisor of a given number and is the largest prime below with . We keep assuming, as above, that . We will also assume that , which guarantees that (4.1) has infinitely many solutions for each value of . Specifically, if is a particular solution for some value of , then the general solution for this is
where is any integer. In the next theorem we denote by the number of solutions of (4.1) with and for each value of with . Thus precisely when (4.1) has no solution subject to these conditions.
Theorem 4.5.
For positive integers and , suppose that the largest prime below satisfies , and let . Let be a prime power dividing with . Then satisfies the variation of Condition with
where is the number of solutions of with and for each value of with .
Proof.
The number counts how many times a multiple of falls into a dangerous interval at a distance from the origin of that interval. Thus we pick an extra prime for each such case, and add two to the sum in order to account for and . ∎
Example 4.6.
The largest primepower divisor of is . For we find that and . Since mod , the only solution of the Diophantine equation with and is for . Thus, and for . In other words, the only occurrence of a multiple of in a dangerous interval for is . This example shows that the bound given in Proposition 4.3 can be lowered.
The number given by Theorem 4.5 is not a sharp bound. For those multiples of falling into a dangerous interval , it often happens that the corresponding binomial coefficient is divisible by , as in Example 4.6 or in other examples given in the previous sections. It could also be divisible by if . When , we have that satisfies Condition 1 with and if and only if the binomial coefficient is divisible by for every solution of (4.1) with and , since and with and , so is not divisible by by Lucas’ Theorem. Note also, for practical purposes, that mod .
5. Every number has multiples for which Condition 1 holds
We next prove that every positive integer has infinitely many multiples for which Condition 1 holds. We are indebted to R. Woodroofe for simplifying and improving our earlier statement of this result, which was based on prime gap conjectures.
It follows from the Prime Number Theorem [7] that, given any real number , there is a prime between and for sufficiently large . This fact can be used to prove the following:
Theorem 5.1.
For every positive integer and every prime , the number satisfies Condition with and another prime, for all sufficiently large values of .
Proof.
For any prime and any , let . Then
Therefore, by the Prime Number Theorem, there is a prime between and for all sufficiently large values of . Choose the largest prime with this property. Thus,
so , from which it follows, according to Corollary 2.6, that satisfies Condition 1 with and . ∎
Theorem 5.2.
For every positive integer there is a number such that if is any prime with then satisfies Condition with and another prime.
Proof.
Given , let . Choose such that there is a prime between and for all , and let . If is any prime such that , then for we have
Therefore, by our choice of , there is a prime between and . If is the largest prime with this property, then , and consequently satisfies Condition 1 with and . ∎
Prime gap conjectures provide information relevant to our problem. For example, if denotes the th prime, then Cramér’s conjecture [6] claims that there exist constants and such that if then
Proposition 5.3.
Let be the number of distinct prime factors of . If Cramér’s conjecture is true and grows sufficiently large keeping fixed, then satisfies Condition .
Proof.
If has distinct prime factors, then , where is the largest primepower divisor of . Let and be the constants given by Cramér’s conjecture. Pick such that if then . For every such that and (where denotes the th prime), we have
from which it follows that satisfies Condition 1 with and . ∎
We note that the argument used in the proof of Proposition 5.3 yields an alternative proof of the fact that Condition 1 holds for a set of integers of asymptotic density if Cramér’s conjecture holds, a result first found in [15, § 5]:
Theorem 5.4 ([15]).
If Cramér’s conjecture is true, then the set of numbers in the sequence (2.1) has asymptotic density zero.
Proof.
Suppose that Cramér’s conjecture holds with constants and , and denote by the number of distinct prime divisors of . Thus , where is the largest primepower divisor of . According to [8, § 3.2], for every the inequality
(5.1) 
holds for all except those of a set of asymptotic density zero. Since
for all , there is an such that if . Now, if is bigger than and satisfies , and moreover is not in the set of integers for which (5.1) fails, then
Therefore, satisfies Condition 1 with and . ∎
6. Multinomials
We also consider a generalization of Condition 1 to multinomials. We say that a positive integer satisfies Condition for multinomials of order if there are primes and such that the multinomial coefficient
is divisible by either or whenever with for all .
Proposition 6.1.
If satisfies Condition with two primes and , then satisfies Condition for multinomials of any order with and .
Proof.
This follows from the equality
and the fact that is divisible by or by assumption. ∎
Therefore, if Condition 1 is proven for binomial coefficients, then it automatically holds for multinomial coefficients.
Acknowledgements
I am indebted to my mentor Oscar Mickelin for his guidance throughout this research and to Prof. Russ Woodroofe for correspondence and kind suggestions. This work was carried out during the Research Science Program at MIT in the summer of 2017 and was supported by the Center for Excellence in Education, the MIT Mathematics Department, and the Youth and Science Program of Fundació Catalunya La Pedrera (Barcelona).
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