Divisibility properties of sum of powers of successive natural numbers and Bowen’s hypothesis

Abstract

The explicit formulae for remainder of sum of powers of successive natural numbers (SPSN) divided by any prime factor $p$ of the number $N$ of sum terms is derived. The remainder is zero, if $N$ contains $p^{j}$ , $j>1$ . It is also 0, if $j=1$ and the exponent is not divisible by $p-1$ . The expression ( $N-N/p)$ (mod $p$ ), named “eigen-remainder” introduced for $N$ is the remainder of SPSN in the rest of cases. All natural numbers, having all eigen-remainders equal to 1, are found not exceeding 42. It is demonstrated that in all other cases, there exists a prime divisor of $N$ , which dividing SPSN gives the remainder $\neq$ 1. This proves Bowen’s hypothesis.

Keywords

Stirling numbers Faulhaber formula Chinese remainder theorem Bowen’s hypothesis averaged natural power

1. Introduction

The sum of powers of successive natural numbers (SPSN), which appears in many chapters of mathematics, theoretical physics, statistics etc., attracted attention of known mathematicians J. Bernoulli, J. Faulhaber and others, but is casually mentioned in popular books on the number theory, e.g. [1, 2]. The interest to this topic was heated after P. Erdős posted problems comparing SPSN with the next its term $(N+1)^{k}$ [3, 4]. He demonstrated invalidity of the first impression that the difference $\sum_{j=1}^{N}j^{k}-\left({N+1}\right)^{k}$ should have the constant sign for all $N$ . The problem was investigated in several works and has been included in R. K. Guy’s book [5] as problem D7 together with the Bowen hypothesis (BH) of impossibility for both terms of the difference to be equal for a natural $k$ .

The divisibility properties of SPSN are the subject of the present paper abutting to Ref. [6] and keeping the stile supposedly understandable for university students mastered the initial course of algebra. The consideration is relying on the proved [6] combinatorial identity. The identity is based on the possibility of the analytical continuation of the explicit formula for second kind Stirling numbers $S(k,j)$ onto the zero upperright triangle of $S(k,j)$ matrix. Similar formula was derived in the J. Riordan classical monography [7, (chapter 2, formula 35, problem 15)], (and although mentioned in Wikipedia and Math World, but without mentioning the original source). The identity allowed to deduce a new version of the Faulhaber formula for $\sum_{j=1}^{N}j^{k}$ , which contains now natural Stirling numbers instead of rational Bernoulli numbers. The use of only natural numbers let to establish divisibility properties of $\sum_{j=1}^{N}j^{k}$ with a prime number of terms: $N=p$ [6].

The purpose of the present article is to extend description of the divisibility properties onto SPSN of any segment of the natural series. There is a direct link between SPSN divided by any prime divisor of $N$ and the BH. It is shown that the proof of the hypothesis requires to check SPSN with the only square-free natural number of terms, which have no other “eigen-remainders” (defined below) except 1. Since the maximal of such numbers is proved to be 42, and they all can be checked directly, this is equivalent to a proof of BH.

2. Divisibility of sum of successive powers (SPSN)

Our starting point is the Eq. (14) from Ref. [6] for sum of successive powers, which we reproduce from there without comments (the present paper is an addition to Ref. [6] and the reader is supposed already read Secs. 1–4 of Ref. [6]).

$\displaystyle\sum_{j=1}^{N}j^{k}=N\left({N+1}\right)\sum_{i=1}^{k}i^{k-1}\left% (-1\right)^{i}\left(\begin{array}[]{l}{N-1}\\ {i-1}\\ \end{array}\right)\sum_{l=i}^{k}\frac{(-1)^{l}}{l+1}\left(\begin{array}[]{c}{N% -i}\\ {l-i}\\ \end{array}\right).$ (1)

If $k<N$ and $N=p$ is prime then $N$ divides the sum since all factors in the expression Eq. (1) are integer and all denominators in the internal sum are smaller than $N$ .

This reasoning does not work in only one case when the exponent $k=p-1$ . Then the denominator in the last term of internal sum is $p$ . However, the “little” Fermat theorem works for division by $p$ of each term $j^{p-1}$ (beside the last one $p^{p-1})$ just in this case. Every addend contributes 1 to the remainder, giving the net value $p-1$ . The same remains valid if $k=a(p-1)$ since the contribution from $(j^{a})^{p-1}/p$ to the remainder is also 1 for any natural $a$ .

These assertions remain valid if the first term of the sum is any natural number. It is evident for $k=p-1$ since Fermat’s theorem is applied to each term separately. If $k\neq p-1$ it follows by induction: let it is valid for the sum with the first term $a^{k}$ (the last one is $(a+p-1)^{k})$ . To verify it for the sum with the first term $(a+1)^{k}$ we must add $(a+p)^{k}-a^{k}=p(a^{k-1}+\ldots+p^{k-1})$ , which is divisible by $p$ .

Let $p$ be a prime divisor of $N$ . Then joining $N/p$ successive sums of powers of integers, each of length $p$ , in a sum of $N$ successive powers $\sum_{\left(N\right)}j^{k}$ we conclude that sum is divisible by $p$ if $k$ is not divisible by $p-1$ . In the opposite case, the remainder of division by $p$ of the sum of $N/p$ identical portions each giving $p-1$ , is same as the remainder of $(p-1)N/p$ , denoted by $e_{N}(p)$ and named eigen-remainder of $N$ :

$\displaystyle e_{N}\left(p\right)=\left({N-N/p}\right)\left({\text{mod}\ p}% \right).$ (2)

(We use notation “mod $p$ ” from the theory of congruences with the equality sign “ $=$ ” (but not “ $\equiv$ ”) in the following sense: $r=A$ (mod $p$ ) is a remainder of $A$ divided by $p$ , and $0<r<p)$ .

It is convenient to distinguish the square free $N_{L}=p_{1}p_{1}\ldots p_{L}$ where $p_{j}$ are all $L$ prime divisors of $N_{L}$ , which we supply with subscript $L$ . Number $N$ without a subscript means that $N$ is not square free. Exactly $L$ eigen-remainders belong to $N_{L}$ . With Eq. (2) in mind we see that the following proposition gives almost a complete description of SPSN divisibility.

Proposition. The remainder of SPSN divided by any prime factor $p$ of $N$ , which contributes factor $p-1$ to $k$ is

$\displaystyle\sum_{\left(N\right)}j^{k}\left({{\text{mod}}\ p}\right)=e_{N}% \left(p\right).$ (3)

The same sum is divisible by the rest of $N$ prime factors. If $N$ is divisible by $p^{2}$ such $p$ divides the sum even if $p-1$ divides $k$ (since $e_{N}(p)=$ 0 then).

It seems, at the first glance, the proposition does not work if $N\leqslant k$ (because of the inequality mentioned after the basic Eq. (1)). However, it is easy to show its validity for $k=N,N+1,\ldots$ Indeed, if $N=k$ after changing the summation subscript $l\to l-i$ in the internal sum the result in Eq. (1) looks like:

$\displaystyle\sum_{i=1}^{k}i^{k}\left({{\begin{array}[]{c}k\hfill\\ l\hfill\\ \end{array}}}\right)\sum_{l=0}^{k-i}\frac{\left({k+1}\right)\left({-1}\right)^% {l}}{l+{i}+1}\left({{\begin{array}[]{c}{k-i}\\ l\\ \end{array}}}\right).$

The whole expression at $i^{k}$ equals to 1 as can be verified by induction (see Ref. [6]). Thus, the sum Eq. (1) keeps the initial form for $N=k$ , too. If so, the conclusion is true also for $k-1$ terms in sum since the last term $k^{k}$ term is, obviously, divisible by $k$ . Moving in this way step by step, we extend the assertion onto $N=k-1$ , $N=k-2$ , etc.

It is worthwhile to emphasize that the Proposition guarantees the existence of a prime factor of $N$ , which divides the left part of the equality $\sum_{j=1}^{N}j^{k}=\left({N+1}\right)^{k}$ in many cases, e.g., if $N$ is not free of squares or if $N$ is free and $k$ is simultaneously free of at least one factor $p-1$ with some prime divisor $p$ of $N$ . Such assertions are important since they mean the impossibility of the equality $\sum_{j=1}^{N}j^{k}-\left({N+1}\right)^{k}=0$ . Indeed, applying the admissible prime divisor to this difference we come to the contradiction: the remainder of the first term is 0 and that of the second term is 1 (due to the Newton binomial formula). Indeed integers $a_{1}$ and $a_{2}$ cannot be equal if the remainders of division of $a_{1}$ and $a_{2}$ by a natural $d$ are not equal. It follows from a sequence of inequalities: $a_{1}={dq}_{1}+r_{1}$ , $a_{2}={dq}_{2}+r_{2}$ , $0<r_{1}<d$ , $0<r_{2}<d$ , then $|r_{1}-r_{2}|<d$ , and from the impossibility of $a_{1}=a_{2}$ in the equality $a_{1}-a_{2}=d(q_{1}-q_{2})+r_{1}-r_{2}$ as a result of the strict inequality $|r_{1}-r_{2}|<d$ for $r_{1}-r_{2}\neq$ 0.

The proposition leaves only one chance for $\sum_{j=1}^{N}j^{k}-\left({N+1}\right)^{k}=0$ to be valid: the situation with no prime factors of $N$ dividing SPSN. It can happen when all primes from $N_{L}$ contribute $p-1$ to $k$ (i.e., $k$ contains the Euler function of $N_{L}$ ). Here the symmetric polynomial of order $L-1$ , built of all prime divisors of $N_{L}$ will be denoted as $P_{L-1}$ :

$\displaystyle P_{L-1}=N_{L}\sum_{j=1}^{L}1/p_{j}=\sum_{j=1}^{L}p_{1}p_{2}% \ldots p_{j-1}p_{j+1}\ldots p_{L}.$ (4)

It appears in the expression of eigen-remainder of $N_{L}$

$\displaystyle e_{N_{L}}\left({p_{j}}\right)=\left({N_{L}-P_{L-1}}\right)\left(% {\text{mod}\ p_{j}}\right)$ (5)

The product $p_{1}p_{2}\ldots p_{n-1}p_{n+1}\ldots p_{L}$ , which presents in definition (2), occurs also in $N_{L}$ and in $P_{L-1}$ . All other terms of $P_{L-1}$ as well as $N_{L}$ have factor $p_{n}$ and do not contribute to the remainder. Expression (5) is $\neq$ 0. These facts direct our attention to BH. Why the difference $\sum_{j=1}^{N}j^{k}-\left({N+1}\right)^{k}$ attracted Bowen’s attention we will see in the next Sec. Now we only mention that the examination of the condition (5) can decide whether the divisibility analysis alone can prove BH.

Our next comment connects Eq. (5) to the Chinese remainders theorem (CRT). In our situation, this theorem requires finding a number $W$ , which being divided by a set of divisors $p_{1}$ , $\ldots$ , $p_{n}$ , $\ldots$ , $p_{L}$ , leads to the same set of given remainders $e_{N_{L}}\left({p_{1}}\right),\ldots,e_{N_{L}}\left({p_{L}}\right)$ as the corresponding dividends from the given collection $(N_{L}-N_{L}/p_{1},\ldots,N_{L}-N_{L}/p_{L})$ . The required generating number is recognized in the right part of Eq. (5). $W=N_{L}-P_{L-1}<N_{L}$ . CRT and Eq. (3) suggest that $W$ might differ from SPSN by an integer number of products $p_{1}\cdot\ldots\cdot p_{L}=N_{L}$ , i e. by $N_{L}\cdot J$ . Hence, $N_{L}-P_{L-1}$ is itself a remainder of division of SPSN by $N_{L}$ . Since the ratio of SPSN to $N_{L}$ is the averaged $k$ -th power over any segment of natural series of length $N_{L}$ the averaging gives the remainder

$\displaystyle\sum_{\left({N_{L}}\right)}j^{k}\left({\text{mod}\ N_{L}}\right)=% N_{L}-P_{L-1}$ (6)

$k=a\Phi N_{L})$ , $\Phi(N_{L})$ is the Euler function of $N_{L}$ (the product of all $p-1$ ), $a$ is natural, and $P_{L-1}$ is defined in Eq. (4).

It is possible to extend the result Eq. (6) onto the general $N$ . An arbitrary $N$ can be presented as $N_{L}\cdot J$ where $N_{L}$ is a product of all prime divisors of $N$ , and $J$ accounts all their repetitions. Then we cut the sum with $N$ terms on $J$ pieces of length $N_{L}$ and apply Eq. (6) to each partial sum. Continuance of this repetition of reasoning, used above to obtain Eq. (3), leads to

$\displaystyle\sum_{\left(N\right)}j^{k}\left({\text{mod}\ N}\right)=N-N\mathop% {\sum}\limits_{p}1/p$ (7)

where summation is taken over prime divisors of $N$ contributing $p-1$ to $k$ (1/2 is always present in sum Eq. (7) for even $N$ ). If such prime divisors are absent, the remainder is 0 for odd $N$ and for $N=4n$ , the remainder is $n$ for $N=4n+2$ .

Similar formula has been discussed in Ref. [8] as deduced in 1994 by P. Moree from the von Staudt-Clausen theorem. The logic of the present derivation is opposite: the von Staudt-Clausen theorem itself has been simply derived in Ref. [6] together with Eq. (1) after consideration divisibility properties of the latter.

3. Divisibility and Erdős’ problem #4347

At the first glance the difference $D\left({N,k}\right)=\sum_{j=1}^{N}j^{k}-\left({N+1}\right)^{k}$ seems keeping the same sign for all $N$ , but this is not the case Erdős formulated “condition K”: $(1-1/(N+1))^{k}>1/2>(1-1/N)^{k}$ to clarify the situation. Erdős double inequalities can be rewritten in a “modulus form” $|(1-2^{-1/k})^{-1}-1/2-N|<1/2$ , which means that natural $N$ cannot deviate from real ( $2^{1/k}+1)/2(2^{1/k}-1)$ by more than 1/2. For large $N$ and $k$ , it means that points $N$ form a broken line, which follows along the strait asymptotic line $k/$ ln2 not deviating from the latter by more than 1/2 (as is clear after applying the L’Hôpital rule to ( $2^{1/k}+1)/2(2^{1/k}-1)$ ).

Erdős’ condition K is related to localizing the point of sign change of the first difference: $d(N+1k)=D(N+1k)-D(N,k)=2N^{k}-(N+1)^{k}$ . The real value $1/(2^{1/k}-1)$ of $N$ , which turns $d(N+1,k)$ into 0 is shifted from $N+1/2$ (up or down) not more than by 1/2. It means that the zero-point is between $N$ and $N+1$ where $N$ is the natural number referred to the given $k$ by P. Erdős’ condition K. Thus, the sign of $d(N+1k)$ changes just in the mentioned interval because the first derivative $\partial d(N+1k)/\partial N|=k(N+1)^{k-1}/(N-1)>0$ .

Similarly, P. Erdős’ conjectures L and M are related to localization of the zero-point for $D\left({N,k}\right)$ , which is a more difficult problem because the explicit solution of the equation $D\left({N,k}\right)=0$ for $k$ is not known independently of how to consider it: as Diophantine or as with $k$ real. P. Erdős’ conjectures mean that the zero-point is between $N+2$ and $N$ (defined by condition K). As mentioned by R. Guy [5], conjectures are proved in Ref. [9], and Ref. [10], respectively. The asymptote $k/$ ln2 passing through the middle between $N+2$ and $N+1$ , has been found in Ref [8]. The effective method of solving the equation $D\left({N,k}\right)=0$ for $k$ is invented by L. Moser [11], who found that natural solutions of $D\left({N,k}\right)=0$ for $k$ might exist only for natural $N>10^{u}$ , $u=10^{7}$ . Since then, this equation gained the name Erdős-Moser. A new technique to solve Erdős-Moser equation for real $k$ has been proposed in Ref. [8].

All these results make clear the behavior of function $D\left({N,k}\right)=\sum_{j=1}^{N}j^{k}-\left({N+1}\right)^{k}$ . In particular, natural numbers $N$ defined by condition K, become asymptotically close to the change-sign-point. However, this still does not answer whether a natural solution to the Diophantine equation $D(N,k)=0$ really exists. BH rejects the existence.

The results of the preceding Section demonstrate that BH is valid in the most of cases. We prove it completely using the divisibility technique. As was mentioned, the Proposition proves that at least one prime divisor of the left part of the Erdős-Moser equation exists for all not square free $N$ (and any $k$ ) and for square free $N_{L}$ and such $k$ , which do not contain even a single factor $p-1$ with a prime divisor $p$ of this $N_{L}$ . The BH is valid for these cases since the remainder of the right part of equation is 1 for the same divisors.

To complete the proof of BH it is necessary to consider the remaining cases ( $N_{L},k=a\Phi(N_{L})$ with neither of factors $p-1$ being absent in $k$ and to collect exotic pairs (EP), which have all eigen-remainders of $N_{L}$ equal to 1

$\displaystyle e_{N_{L}}(p_{j})=\left({N_{L}-P_{L-1}}\right)\left({\text{mod}\ % p_{j}}\right)=1,j=1,\ldots,L.$ (8)

Divisibility arguments do not work for EP because the divisibility criterion is only a necessary one, but not sufficient. It does not prove neither existence nor absence of solutions, and the further seeking of explicit solutions of the Erdős-Moser equation among EP is necessary.

4. Exotic pairs

Firstly, does such exotic pairs really exist? (If not BH would be already proved!). Unfortunately, yes, as even the first two primes show: for $N_{2}=2\cdot 3$ , Eq. (5) gives $e_{6}(2)=(6-6/2)(\text{mod2})=1$ and $e_{6}(3)=(6-6/3)(\text{mod3})=1$ . Examples justifying the term “exotic” also exist, e.g., $N_{4}=1806=2\cdot 3\cdot 7\cdot 43$ has $e_{1806}(2)=(1806-1806/2)(\text{mod2})=1$ , $e_{1806}(3)=(1806-602)(\text{mod3})=1$ , $e_{1806}(7)=(1806-258)(\text{mod7})=1$ , and $e_{1806}(43)=(1806-42)(\text{mod43})=1$ .

Because of 1(mod $d$ ) $=$ 1 for any natural $d$ , the equality

$\displaystyle N_{L}-P_{L-1}=1.$ (9)

is the sufficient condition for Eq. (8) to be valid for all primes $p_{j}$ . Let us show that this condition is also necessary if $N_{L}-P_{L}=W$ is treated as the CRT generating number $W$ for a set of unit remainders. Such $W$ is possible to calculate generalizing the technique of linear nondegenerate Diophantine equation proposed in Ref. [12] for the classic case of CRT with two divisors. The number of divisors in CRT is $L$ in the present case. This Diophantine equation in $L$ unknowns

$\displaystyle N_{L}\sum_{n=1}^{L}\frac{x_{n}}{p_{n}}=1.$ (10)

is solvable by means of the Euclidian algorithm repeated $L-1$ times [13]. Coefficients $N_{L}/p_{n}=p_{1}\ldots p_{n}\ldots p_{L}$ of this equation for $L$ auxiliary integer unknowns $x_{n}$ guarantee that the generating number $W$ can be determined in terms of products of $x_{n}$ and the given remainders $r_{1},\ldots,r_{n},\ldots,r_{L}$ .

$\displaystyle W=N_{L}\sum_{n=1}^{L}\frac{r_{n}x_{n}}{p_{n}}.$ (11)

Indeed: the remainder of division of $W$ by $p_{j}$ is $r_{j}\cdot N_{L}x_{j}/p_{j}$ , because the rest of $W$ terms contain $p_{j}$ : $N_{L}/p_{n}=p_{1}\ldots p_{j}\ldots p_{n-1}\cdot p_{n+1}\ldots p_{L}$ . The coefficient at $r_{j}$ in Eq. (11) differs from 1 by a number divisible by $p_{j}$ on account of Eq. (10): $N_{L}x_{j}/p_{j}=1-p_{j}\sum_{n\left({\neq j}\right)=1}^{L}x_{n}\ldots p_{n-1}% p_{n+1}\ldots p_{L}$ .

Therefore, if all remainders are identical, $W$ acquires the same value due to Eq. (10). In the present case $W$ should be 1 i.e., the condition (8) is also necessary. This has important consequences. The first odd $N_{2}=3\cdot 5=15$ corresponds to $W_{2}=15-3-5=7>1$ . Adding the smallest difference between primes, 2, to any factor in $N_{2}=p_{1}p_{2}$ increases $W_{2}=7$ by $2(p_{2}-1)$ and so on Similarly, $W_{3}$ , beginning from $34(N_{3}=3\cdot 5\cdot 7)$ goes up, etc. So, we have come to conclusion: all odd $N_{L}$ do not obey condition (8) that together with the restrictions discussed above reproduces the known result of L. Moser [11]: there is no solution to equation $D(N,k)=0$ with odd $N$ .

Thus, to prove BH it remains to consider only EP with even $N_{L}=2n_{L-1}$ . For such $N_{L}$ , requirement (8) leads to odd square-free $n_{L-1}$ obeying the condition $n_{L-1}-2P_{L-2}=1$ , which appears after substitution of $P_{L-1}=2P_{L-2}+n_{L}$ into Eq. (9). Note, Eq. (10) has the solution $x_{1}=1$ , $x_{j}=-1$ , $j=2,\ldots,L$ for $p_{1}=2$ , which transforms $N_{L}\sum_{j=1}^{L}x_{j}/p_{j}=1$ into

$\displaystyle n_{L-1}-2P_{L-2}=1.$ (12)

The next numbers are $n_{L}=n_{L-1}p_{L}$ and $P_{L-1}=P_{L-2}p_{L}+n_{L-1}$ by definition. Then $n_{L}-2P_{L-1}=(n_{L-1}-2P_{L-2})p_{L}-2n_{L-1}=p_{L}2n_{L-1}+1$ . If $p_{L}$ is chosen as $2n_{L-1}+1$ , then the next $n_{L}$ also obeying $n_{L}-2P_{L-1}=1$ , corresponds to EP and gives the recurrent relations

$\displaystyle N_{L+1}=N_{L}\left({N_{L}+1}\right),n_{L}=n_{L-1}\left({2n_{L-1}% +1}\right).$ (13)

Equation (13) gives rise to an infinite chain of numbers $N_{L}$ satisfying the condition (8). However, $p_{L+1}$ is not necessary a prime number. If it is not, the subscript $L$ is no more the number of prime divisors of $N_{L}$ because the last factor consists of more than one prime. Only their product $p_{L+1}$ , brings in $N_{L+1}$ the new unit eigen-remainder. So, we conclude: the equality $N_{L}-P_{L-1}=1$ is the necessary and sufficient for $(N_{L}k=m\Phi N_{L})$ to be EP; the proper $N_{L}$ can be found only among members of sequence (13).

The peculiarity of division by 4 allows to exclude half of $N_{L}$ from possible EP. As is known, division by 4 gives the remainder 1 for any odd dividend in even degree. Since $k$ is even (due to the factor $\Phi(N_{L}$ )) and $N_{L}+1$ is odd, the remainder of $(N_{L}+1)^{k}$ is 1 (4 is not assumed to be a divisor of $N_{L})$ . The remainder of SPSN is the same as the remainder of the number of odd terms $n_{L}$ in SPSN, $n_{L}$ is odd itself. Then all $n_{L}$ with remainders 3 contradict to right part $(N_{L}+1)^{k}$ heaving remainder 1. It can be verified that recurrent Eq. (13) leads to alternation of remainders of $n_{L}$ between 3 and 1 beginning from $n_{1}=3$ . Hence, the members of sequence Eq. (13) with even subscripts (marked *) do not belong to EP: the sequence begins from $N_{2\ast}=2\cdot 3=6$ , $N_{3}=2\cdot 3\cdot 7=42$ , $N_{4\ast}=2\cdot 3\cdot 7\cdot 43=1806$ , $N_{5}=2\cdot 3\cdot 7\cdot 43\cdot 1807=2\cdot 3\cdot 7\cdot 43\cdot 13\cdot 1% 39=3263442$ $N_{6\ast}=10650056950806=2\cdot 3\cdot 7\cdot 43\cdot 13\cdot 139\cdot 3263443$ , $N_{7}=113423713055421844361000442\ldots$

Both examples, we started from, found their subscripts (2 and 4) in this sequence (and should be ignored). The fifth member is a key for the further proof. Its fifth factor is not prime: $1807=13\cdot 139$ . Despite $e_{N_{5}}\left({1807}\right)=1$ , the new two prime divisors do not satisfy the condition (7): $e_{N_{5}}\left({13}\right)=9,e_{N_{5}}\left({139}\right)=13$ and can be used to produce the desired contradiction.

Nevertheless, Eq. (7) remains valid for $N_{5}$ and over all $N_{L}$ with $L>5$ , which guarantee the remainders of corresponding averaged powers to be 1. Therefore, Eq. (7) by itself cannot produce contradictions leading to the proof of BH.

Both factors 13 and 139 present in all subsequent members of the sequence Eq. (13), and their characteristic property: $e_{N_{L}}\left({13}\right)=9,e_{N_{L}}\left({139}\right)=13$ does not change due to the following identity.

$\displaystyle e_{N_{L+1}}\left(p\right)=\left({N_{L+1}-\frac{N_{L}}{p}}\right)% \left({\text{mod}\ p}\right)=\left({N_{L}\left({N_{L}+1}\right)-\frac{N_{L}^{2% }}{p}-\frac{N_{L}}{p}}\right)\left({\text{mod}\ p}\right)=e_{N_{L}}\left(p\right)$ (14)

This formula results from the presence of $p^{2}$ in $(N_{L})^{2}$ , leading to the multiplier $p$ in the term $(N_{L})^{2}/p$ . Hence $N_{L}/p$ determines all eigen-remainders of $N_{L+1}$ as well as of $N_{L}$ . Then all subsequent members keep factors leading to the required contradiction, e.g., $e_{N_{5}}\left({13}\right)=e_{N_{6}}\left({13}\right)=e_{N_{7}}\left({13}% \right)=9\ldots$

Elimination of the born “dangerous” primes from $N_{L}$ does not save from the contradiction because it violates the recurrent relation Eq. (13). As a result the eigen-remainders, produced by the rest “normal” primes, begin to differ from 1, e.g., $e_{N_{7}/1807}\left(7\right)=4$ , $e_{N_{7}/139}\left(7\right)=6$ , $e_{N_{7}/1807}\left({43}\right)=21$ , $e_{N_{7}/139}\left({43}\right)=13$ etc. Summarizing all the above let us to assert:

There is only a single square free number $N_{3}=$ 42, which does not produce a contradiction as a result of division of the Erdős-Moser equation by prime divisors of $N$ . However, 42 as well as 10^9.3u ( $u=$ 10⁶) other numbers were already excluded in classic Moser work [11] that transforms the preceding sentence into the final proof of BH.

5. Conclusion

The Riordan formula, extended onto upper-right triangle of the second kind Stirling matrix, let to find the results of division of SPSN by the prime factors of the number of SPSN terms. In most cases, the remainders are zero that makes Bowen’s hypothesis rather probable. The final proof results from demonstration that there is only a small number of “exotic pairs” $N$ , $k$ with all eigen-remainders equal 1, which were already verified as not satisfying Erdős-Moser equation.

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