Binomial predictors

TL;DR

This paper characterizes binomial predictors in base p by relating the set of binomial coefficients divisible by p^k to the base p expansion of n+1, providing a complete classification of such predictors.

Contribution

It offers a full description and characterization of binomial predictors in base p, connecting divisibility properties of binomial coefficients to base p expansions.

Findings

Identifies the structure of binomial predictors in base p.

Provides explicit formulas linking binomial coefficient divisibility to base p digits.

Completes the classification of binomial predictors in the given setting.

Abstract

For a prime p and nonnegative integers n,k, consider the set A_{n,k}^{(p)}={x is in [0,1,...,n]: p^k||binom {n} {x}}. Let the expansion of n+1 in base p be: n+1=alpha_{0} p^{\nu}+alpha_{1}p^{nu-1}+...+alpha_{nu}, where 0<=alpha_{i}<= p-1,i=0,...,nu. Then the number n is called a binomial predictor in base p,if |A_{n,k}^{(p)}|=alpha_{k}p^{nu-k},k=0,1,...,nu. We give a full description of the binomial predictors in base p.

Equations

$$A_{n,k}^{(p)}=\{x\in [0,1,...,n]:p^k||(\genfrac{}{}{0ex}{}{n}{x})\}.$$

$$|A_{n,0}^{(p)}|,|A_{n,1}^{(p)}|,...$$

$$1,11,55,165,330,462,462,330,165,55,11,1.$$

$$p^{a_p(n)}||n.$$

$$n=n_0{p}^m+n_1{p}^{m-1}+...+n_m,0\le t_i\le p-1.$$

$$s_p(n)=n_0+n_1+...+n_m,$$

$$a_p(n!)=(n-s_p(n))/(p-1).$$

$$a_p((\genfrac{}{}{0ex}{}{n}{x}))=a_p(n!)-a_p(x!)-a_p((n-x)!)=$$

$$((n-s_p(n))-(x-s_p(x))-(n-x-s_p(n-x))/(p-1)=$$

$$(s_p(x)+s_p(n-x)-s_p(n))/(p-1).$$

$$a_2((\genfrac{}{}{0ex}{}{2n}{n}))=s_2(n);a_2((\genfrac{}{}{0ex}{}{2n+1}{n}))=s_2(n+1)-1.$$

$$s_p((p-1)n)=s_q((q-1)m),$$

$$(p-1)a_p((\genfrac{}{}{0ex}{}{pn}{p}))=(q-1)a_q((\genfrac{}{}{0ex}{}{qm}{q})).$$

$$s_p(x+y)\le s_p(x)+s_p(y).$$

$$s_p(x)+s_p(n-x)-s_p(n)=k(p-1),x\in [0,1,...,n].$$

$$|A_{n,k}^{(p)}|={\lambda }_p^{(k)}(n),k=0,1,...$$

$$t=t_0{p}^m+t_1{p}^{m-1}+...+t_m,0\le t_i\le p-1.$$

$$\left(\genfrac{}{}{0ex}{}{n}{t}\right)\equiv \prod _{i=0}^m\left(\genfrac{}{}{0ex}{}{n_i}{t_i}\right)(\mathrm{mod}p).$$

$${\lambda }_p^{(0)}(n)=(n_0+1)(n_1+1)...(n_m+1).$$

$${\lambda }_2^{(0)}(n)=2^{s_2(n)}.$$

$$|A_{n,0}^{(p)}|=(p-1)p^m.$$

$$n=\underset{t}{\underbrace{(p-1)\vee ...\vee (p-1)}}\vee (p-2)\vee \underset{m-t}{\underbrace{(p-1)\vee ...\vee (p-1)}}.$$

$$n+1=\underset{t+1}{\underbrace{(p-1)\vee ...\vee (p-1)}}\vee \underset{m-t}{\underbrace{0\vee ...\vee 0}}=$$

$$(p-1)(p^m+p^{m-1}+...+p^{m-t}).$$

$$x_0\vee ...\vee x_{t-1}\vee x_t\vee x_{t+1}\vee ...\vee x_m.$$

$$|A_{n,0}^{(p)}|=p^t(p-1)p^{m-t}=(p-1)p^m.$$

$$x_t=x_{t-1}=x_{t-2}=...=x_r=p-1,$$

$$1\le r\le t.$$

$$x\in A_{n,t-r+1}^{(p)}.$$

$$t-r+1=k,$$

Full text

1. Introduction

Let $p$ be a prime. For nonnegative integers $n,k,$ consider the set

[TABLE]

Quite recently (2008), W.B. Everett [1] solved the following important problem.

Question 1**.**

How, knowing $n,$ to find the finite sequence

[TABLE]

(of course, without the direct calculations)?

Due to generality, the Everett’s formula is sufficiently complicated. In this paper we indicate an infinite set of $n^{\prime}s,$ for which Question 1 has especially simple solution which immediately follows from the expansion of $n+1$ in base $p.$ Conversely, in the limits of this set, knowing the sequence $\{|A_{n,\enskip k}^{(p)}|\},$ we can ”predict” the expansion of $n+1$ in base $p.$ In the connection with this, we introduce the following notion.

Definition 1**.**

Let the expansion of $n+1$ in base $p$ be: $n+1=\alpha_{0}p^{\nu}+\alpha_{1}p^{\nu-1}+...+\alpha_{\nu},$ where $0\leq\alpha_{i}\leq p-1,\enskip i=0,...,\nu.$ Then the number $n$ is called a binomial predictor in base $p$, if $|A_{n,\enskip k}^{(p)}|=\alpha_{k}p^{\nu-k},\enskip k=0,1,...,\nu.$

Example 1**.**

It is easy to see that $n=0$ is a binomial predictor in every base $p.$

Indeed, $|A_{0,\enskip 0}^{(p)}|=1,$ that is the binary expansion of 1 in every base.

Example 2**.**

Let $p=2,\enskip n=11.$

Then $n+1=8+4.$ The row of the binomial coefficients $\{\binom{n}{x}\enskip,x=0,1,...,11\}$ is:

[TABLE]

Here $|A_{11,0}^{(2)}|=8,\enskip|A_{11,1}^{(2)}|=4,\enskip|A_{11,k}^{(2)}|=0,\enskip k\geq 2.$ Thus, by the definition, 11 is a binomial predictor in base 2.

Example 3**.**

Let $p=3,\enskip n=23.$ Then $n+1=2\cdot 3^{2}+2\cdot 3.$

Here $|A_{23,0}^{(3)}|=18,\enskip|A_{23,1}^{(3)}|=6,\enskip|A_{23,k}^{(3)}|=0,\enskip k\geq 2.$ Thus 23 is a binomial predictor in base 3.

Our aim is to give a full description of binomial predictors in base $p.$

Definition 2**.**

A nonnegative integer $n$ is called a Zumkeller’s number in base $p$ (in the case of $p=2$ see sequence $A089633\enskip in\enskip[12]$), if either it is 0 or its expansion in base $p$ has all digits $p-1,$ except, maybe, one; if the exceptional digit is the first ( it can occur only in case of $p\geq 3$), then it could take an arbitrary value from $1,...,p-2;$ otherwise, it is only $p-2.$

Our result is the following.

Theorem 1**.**

$n\geq 0$* is a binomial predictor in base $p$ if and only if it is a Zumkeller’s number in the same base.*

2. Some classical results on binomial coefficients

The binomial coefficients play a very important role in numerous questions of number theory. For example, it is very known proof of the beautiful Chebyshev’s theorem, using the binomial coefficients (see, for example, a Finsler’s version of the proof in [13]). A connection between some questions of divisibility of the binomial coefficients and the old conjecture of the infinity of tween primes is appeared in the author’s article [10].

The first important contributions into theory of binomial coefficients belong to Legendre (1830), Kummer (1852) and Lucas (1878). Let $p$ a prime and $a_{p}(n)$ be such exponent that

[TABLE]

Let, furthermore,

[TABLE]

be the expansion of $n$ in base $p.$ Denote

[TABLE]

A.-M. Legendre [7, p.12] empirically noticed that (in our notations)

[TABLE]

A proof see, e.g., in [9]. From (5) we immediately obtain:

[TABLE]

[TABLE]

[TABLE]

Example 4**.**

Since $s_{2}(2n)=s_{2}(n),$ then for the central binomial coefficients we find

[TABLE]

Notice that in [11] was posed a question which remains open up to now.

Question 2**.**

Does the diophantine equation $s_{p}(n)=s_{q}(n),$ where $p\neq q$ are fixed primes, have infinitely many solutions?

It easy to see that, by (6), the following equalities are equivalent:

[TABLE]

[TABLE]

Furthermore, note that (6) implies the following simple corollary.

Corollary 1**.**

For every lattice pair $(x,y)\geq(0,0),$ we have the triangle inequality:

[TABLE]

The equality attains if and only if $\binom{x+y}{x}$ is not multiple of $p.$

Now we can treat of Question 1 in a different foreshortening. Consider the equation

[TABLE]

For $k\geq 0,$ denote $\lambda_{p}^{(k)}(n)$ the number of solutions of (8). Thus we see that

[TABLE]

In 1852, Kummer [6] made an important observation (a proof one can find, e.g., in [3]):

$a_{p}(\binom{n}{x})$ is the number of ”carries” which appear in adding $x$ and $n-x$ in base $p.$

This statement plays a large role in the Everett’s matrix method which was used for receiving his general formula.

Another important result was obtained by Lucas [8]. He proved that if together with (3)

[TABLE]

then

[TABLE]

From (11) immediately follows the next corollary ([2]).

Corollary 2**.**

[TABLE]

**Proof. **Indeed, (12) gives the number of all nonzero products in (11), when $0\leq t_{i}\leq n_{i},\enskip i=0,...,m.$ Since $t_{i},n_{i}\in[0,...,p-1],$ then none of the considered products is divisible of $p.\blacksquare$

Note that in the binary case in (12) sufficiently consider only factors with $n_{i}=1.$ Therefore, we have

[TABLE]

The latter is known result of J.Glaisher (1899; in [4] A.Granville gives a new elegant proof; generalizations in other direct see in [4]-[5]).

3. Proof of necessity

A proof of necessity we easily derive from  only the first condition of Definition 1, i.e. from the equality $|A_{n,0}^{(p)}|=\alpha_{0}p^{\nu}.$ In connection with (3), note that $\nu\neq m$ only when $n_{0}=n_{1}=...=n_{m}=p-1.$ This a Zumkeller’s number in base $p.$ In other cases we consider the equality $|A_{n,0}^{(p)}|=\alpha_{0}p^{m}.$ In its turn,$\alpha_{0}\neq n_{0}\leq p-2$ only when $n_{1}=...=n_{m}=p-1.$ In this case we also have a Zumkeller’s number in base $p.$ Let now $n$ be a binomial predictor in base $p,$ when $n_{0}=p-1,$ but not all $n_{i}$ equal $p-1.$ Thus, we have

[TABLE]

From this equality and Corollary 2 we conclude that exactly $m$ from $m+1$ brackets in product (12) equal $p,$ while some one bracket equals $p-1.$ This means that exactly $m-1$ digits from $n_{1},...,n_{m}$ equal to $p-1,$ while some one digit equals $p-2.$ Thus in this case we again have a Zumkeller’s number in base $p.\blacksquare$

4. Proof of sufficiency

Here we use the Kummer’s theorem in the following equivalent form:

$a_{p}(\binom{n}{x})$ is the number of ”carries” which appear in subtracting $x$ from $n$ in base $p.$

Let $n$ be an Zumkeller’s number. Evidently, in the trivial case when $n=p^{m+1}-1=\underbrace{(p-1)\vee(p-1)\vee...\vee(p-1)}_{m+1}$ we have a binomial predictor in base $p$ ( we use $\vee$ as operator of concatenation). Indeed, here $n+1=p^{m+1}$ and, by (9) and (12), $|A_{n,\enskip 0}^{(p)}|=p^{m+1}.$ On the another hand, for every $x\leq n,$ in the subtracting $x$ from $n$ in base $p$ we have no any ”carries”, i.e. $|A_{n,\enskip k}^{(p)}|=0,\enskip k\geq 1.$ Analogously, we have a binomial predictor in case $n=n_{0}\vee\underbrace{(p-1)\vee(p-1)\vee...\vee(p-1)}_{m},\enskip n_{0}\leq p-2.$ Indeed, here $n+1=(n_{0}+1)p^{m}$ and, by (9) and (12), $|A_{n,\enskip 0}^{(p)}|=(n_{0}+1)p^{m}.$ and again we have no any ”carries” during the subtracting $x$ from $n$ in base $p.$ Let now $n$ have a unique digit $p-1$ in its expansion in base $p.$ Consider such $n$ of the general form:

[TABLE]

Then

[TABLE]

[TABLE]

Let the $x$ in base $p$ has the form:

[TABLE]

If $x_{t}\leq p-2,$ then in subtracting $x$ from $n$ in base $p$ we have not ”carries”. Evidently, that the number of such $x^{\prime}s$ is

[TABLE]

If $x_{t}=p-1,$ such that also

[TABLE]

then, in view of $x\leq n,$ the length of this chain is not more than $t,$ i.e.

[TABLE]

In this case the number of ”carries” is equal to the length of the chain, i.e. $t-r+1,$ and, by Kummer’s theorem, we have

[TABLE]

Putting

[TABLE]

we easily calculate $|A_{n,k}^{(p)}|:$

[TABLE]

where the factor $\frac{p-1}{p}$ corresponds to the digit $x_{r-1},$ i.e. the place after the end of chain (19). Comparison of (18) and (23) with (16) shows that $n$ is a binomial predictor. This completes the proof of Theorem 1. $\blacksquare$

Acknowledgment. The author is grateful to J.-P. Allouche for sending a scanned copy of pp. 10-12 of Legendre book [7], and to D. Berend for important advises.