On two combinatorial identities

Abstract

Two combinatorial identities are proved. The first one allows easy to establish connections between Bernoulli and Stirling numbers and values of zeta function of integer arguments and to find numerous identities, including the new ones, for these quantities. It also leads to a simple derivation of Abel’s analog of the Newton binomial formula, to the shortest proof of Staudt-Clausen theorem on Bernoulli numbers, and to a convenient summation formula for powers of natural numbers and series including such powers. The second identity presents the sum of two prime powers $x^{p}+y^{p}$ ( $p=2n+1=6m\pm 1$ ) of real or natural numbers as Chebyshev polynomial of $4{xy}/(x+y)^{2}$ . The same sum is expressed as a sum of $m$ subsequent powers of $[{xy}(x+y)]^{2}({xy}+x^{2}+y^{2})^{3}$ with coefficients $(n-j-1|2j)/(2j+1)$ .

Keywords

Bernoulli numbers Abel formula sum of natural powers generation function Chebyshev polynomial

1. Introduction

Two seemingly unknown identities for natural numbers which may be useful for some problems currently attracting attention will be deduced (Sections 2 and 10–13). The first one allows considering from a single viewpoint various connections between well-known mathematical objects such as Abel analog of Newton binomial formula (Section 3), modified Bernoulli summation formula (Section 4) and numerous identities (including new) for Bernoulli numbers (Sections 5 and 8), connection between sums of second kind Stirling numbers over the first and second argument (Section 6), etc. A simple proof of von Staudt-Clausen theorem (Section 4) as well as various identities for zeta-function (Sections 7–9) such as Hasse formula will be also derived. A highly accurate numerical connection between the fundamental dimensionless parameter of quantum electrodynamics, the fine structure constant, and a sum of zeta-functions is observed.

The second identity connecting $x^{p}+y^{p}$ to Chebyshev polynomial of the first kind is obtained and allows to present $x^{p}+y^{p}$ as a product of $n$ real factors (for $p=2n+1=6m\pm 1$ ). This leads for a prime $p$ to presentation of $x^{p}+y^{p}$ as a sum of $m$ products of powers of even squares and mutually-prime odd cubes (Sections 11–13).

This methodical article is mainly addressed to beginning researchers in mathematical physics. Therefore it is completely self-contained, and all calculations are included in text and explained in detail.

2. Definition of symbols $\bm{\{n,m\}}$ , recurrence relation and basic identity

Let us write down some relations for smallest natural numbers which may be named astonishing: $-2\cdot 1+2=0,-3\cdot 1+3\cdot 2-3=0,-3\cdot 1^{2}+3\cdot 2^{2}-9=0;-4\cdot 1+% 6\cdot 2-4\cdot 3+4=0,-4\cdot 1^{2}+6\cdot 2^{2}-4\cdot 3^{2}+4^{2}=0,-4\cdot 1% ^{3}+6\cdot 2^{3}-4\cdot 3^{3}+4^{3}=0$ ; etc. These equalities suggest introducing a following general notation where $(n|k)$ is a binomial coefficient:

$\displaystyle\{n,m\}=\sum\limits_{k=1}^{n}{(-1)^{k}(n|k)k^{m}}.$ (1)

To prove that $\{n,m\}$ is zero we derive a recurrent formula over the first argument.

$\displaystyle\{n+1,m\}=\sum\limits_{k=1}^{n+1}{(-1)^{k}(n+1|k)k^{m}}=\sum% \limits_{k=1}^{n+1}{(-1)^{k}(n|k-1)k^{m}}+\sum\limits_{k=1}^{n}{(-1)^{k}(n|k)k% ^{m}}=\{n,m\}+\sum\limits_{k=0}^{n}{(-1)^{k+1}(n|k)(k+1)^{m}}=\{n,m\}\!-\!1\!-% \!\sum\limits_{k=1}^{n}(-1)^{k}(n|k)\sum\limits_{p=0}^{m}{(m|p)k^{p}}=\{n,m\}-% 1-\sum\limits_{p=0}^{m}(m|p)\sum\limits_{k=1}^{n}{(-1)^{k}(n|k)k^{p}}=\{n,m\}+% \{n,0\}-\sum\limits_{p=0}^{m}{(m|p)\{n,p\}}.$

The Definition (1) has been used twice. Since from Eq. (1) it follows $\{n,0\}=-1$ two extreme terms of the sum disappear, and we finally have:

$\displaystyle\{n+1,m\}=-\sum\limits_{p=1}^{m-1}{(m|p)\{n,p\}}.$ (2)

Initial examples suggest that symbol $\{n,m\}$ is zero if $m\leqslant n-1$ . This is verified easily by induction and Eq. (2) since $p$ does not exceed $n-1$ for any $m\leqslant n$ . Therefore we have

$\displaystyle\{n,m\}=0,m<n,m=1,2,\ldots,n-1,n=2,3,\ldots;\{n,0\}=-1.$ (3)

Equation (3) expresses the first identity meant in the title of this paper.

Equation (2) is valid independently of the condition $m\leqslant n$ and can be used to define $\{n,m\}$ for $m\geqslant n$ . In such case, it may be interpreted as a reduction formula for the first parameter on account of Eq. (3) and including $\{n,m\}$ in the sum.

$\displaystyle\{n+1,m\}-\{n,m\}=\{n+1,m+1\}/(n+1)=-\sum\limits_{p=n}^{m}{(m|p)% \{n,p\}}$ (4)

The result of summation follows directly from definition (1) thanks to relation $(n|p)=(n+1|p)(n+1-\linebreak p)/(n+1)$ .

Some explicit expressions for nonzero symbols ensue from Eq. (4): $\{n+1,n+1\}=-(n+1)\{n,n\}\linebreak=(n+1)n\{n-1,n-1\}=\ldots=(n+1)n\ldots 2(-1% )^{n}\{1,1\}=(-1)^{n+1}(n+1)!$ since by the definition Eq. (1) $\{1,1\}=-1$ . Similarly transformation $\{n,n+1\}=-(n+1)(n\{n-1,n-1\}/2+\{n-1,n\})$ , if repeated $k$ times, gives $\{n,n+1\}=k(-1)^{n}(n+1)!/2+(-1)^{k}(n+1)\ldots(n-k+2)\{n-k,n-k+1\}=\ldots=n(n% +1)!(-1)^{n}/2$ and so on. These values are collected in Eq. (5), and $\{0,m\}=0$ by definition (1).

$\displaystyle\{n,0\}=-1,\{0,m\}=0,\{1,m\}=-1,\{2,m\}=2^{m}-2,\{n,n\}=(-1)^{n}n!,$ (5) $\displaystyle\{n,n+1\}=(-1)^{n}n(n+1)!/2,\ldots$

Simple results (5) lead to interesting conclusions when $n$ is a prime number, say, $p$ . Then $k$ in definition (1) of $\{p-1,m\}$ form a whole set of remainders of division by $p$ . For $m<p-1$ , all $\{p-1,m\}$ vanish due to identity (3). The fractional part of $\{p-1,p-1\}/p$ is $-1/p$ since $\{p-1,p-1\}/p=\sum_{k=1}^{p-1}(-1)^{k}(p-1|k)(k^{p-1}-1)/p+\sum_{k=1}^{p-1}(-1% )^{k}(p-1|k)/p$ , and $k^{p-1}-1$ is divisible by $p$ thanks to Fermat’s “little” theorem. Hence, the remainder of $(p-1)!/p$ according to Eq. (5) is $p-1$ (something similar to the “converse theorem” of Euclid’s poof of the infinite number of primes). For the next, $\{p-1,p\}/p$ and subsequent $\{p-1,m\}/p$ we present $k^{m}/p$ as $k^{m-p+1}(k^{p-1}-1)/p+k^{m-p+1}/p$ . This separates the fractional part as $\{p-1,m-p+1\}/p$ . The first argument exceeds the second one for all $m<2(p-1)$ , and $\{p-1,m-p+1\}$ vanishes due to identity (3) which means that $\{p-1,m\}$ is divisible by $p$ for all such $m$ . If $m=2(p-1)$ , or more generally, if $m$ is divisible by $p-1$ then the remainder is $p-1$ because of $k^{c(p-1)}/p=((k^{c})^{p-1}-1)/p+1/p$ . Thus, we have

Corollary 1. For prime $p$ and $m$ divisible by $p-1$ , the remainder of $\{p-1,m\}/p$ is $p-1$ , for all other $m,\{p-1,m\}$ is divisible by $p$ .

Further. The introduced symbols are nonzero for negative values of the second argument as well. Since $\{n,0\}=-1$ and Eq. (4) allows decreasing $m$ by one, applying it $n-1$ times leads to $\{0,-m\}=0$ while in all other symbols the second argument is $-m+1$ . As a result, $\{n,-1\}$ equals to the negative segment of the harmonic sum:

$\displaystyle\{n,-m\}=\{n-1,-m\}+\{n,-(m-1)\}/n=\{n-2,-m\}∼{}+\{n-1,-(m-1)\}/(% n-1)+\{n,-(m-1)\}/n=\ldots$

Finally

$\displaystyle\{n,-m\}=\sum\limits_{p=1}^{n}{\{p,-(m-1)\}/p}.$ (6)

The process can be continued starting from $\{n,-1\}$ to find $\{n,-2\}$ , $\{n,-3\}\ldots$ by means of the recurrent Eq. (6).

$\displaystyle\{n,-1\}=\sum\limits_{p=1}^{n}{\{p,0\}/p}=-\sum\limits_{p=1}^{n}{% 1/p},∼{}∼{}\{n,-2\}=\sum\limits_{p=1}^{n}{\frac{1}{p}\{p,-1\}}=-\sum\limits_{p% =1}^{n}{\frac{1}{p}\sum\limits_{q=1}^{p}{\frac{1}{q}}},$ $\displaystyle\{n,-3\}=\sum\limits_{p=1}^{n}{\frac{1}{p}\{p,-2\}}=-\sum\limits_% {p=1}^{n}{\frac{1}{p}}\sum\limits_{q=1}^{p}{\frac{1}{q}\sum\limits_{r=1}^{q}{% \frac{1}{r}}},\ldots$

and leading to

$\displaystyle\{n,-m\}=\sum\limits_{p=1}^{n}{(-1)^{p}(n|p)/p^{m}}=-\sum\limits_% {p_{1}=1}^{n}{1/p_{1}\sum\limits_{p_{2}=1}^{p_{1}}{1/p_{2}}\ldots}\sum\limits_% {p_{m}=1}^{p_{m-1}}{1/p_{m}}.$ (7)

Now symbol $\{n,m\}$ is defined for all natural $n$ and all integer $m$ . It is easy to prove by induction that the total number of terms in this $m$ -multiple sum is $(m+n-1|m)$ , i.e. it consists of all ordered m-products of $1,1/2,1/3,\ldots,1/n$ (with repetitions). In particular, it contains $n$ terms of the generalized finite harmonic sum: $H_{m}(n)=\sum_{p=1}^{n}1/p^{m}$ .

3. Abel’s formula

A generalization of symbol $\{n,m\}$ introducing a continuous parameter extends its possible applications.

$\displaystyle\{n,m|a\}=\sum\limits_{k=0}^{n}{(-1)^{k}(n|k)(k+a)^{m}.}$ (8)

Note a simultaneous change of the lower sum limit which, in fact, does not change the basic identity (3):

$\displaystyle\{n,m|a\}=0,∼{}0\leqslant m<n.$ (9)

Indeed, the interchange of the summation order and account of identities (3) and $\{n,0\}=-1$ leads to zero:

$\displaystyle\{n,m|a\}=a^{m}+\sum\limits_{k=1}^{n}(-1)^{k}(n|k)\sum\limits_{j=% 0}^{m}{(m|j)k^{j}a^{m-j}}=a^{m}+\sum\limits_{j=0}^{m}{(m|j)a^{m-j}\sum\limits_% {k=1}^{n}{(-1)^{k}(n|k)k^{j}}}=a^{m}-a^{m}=0.$

However, $\{n,0\}=-1$ itself is changed for the new quantity: $\{n,0|a\}=0,\{0,0|a\}=0$ , but still $\{n,$ $n|a\}=(-1)^{n}n!$ .

Let us add $b^{n}$ to the sum of quantities $\{n-m,n-m-1|a\}$ (which all are zero according to Eq. (9)) with the proper coefficients. After using the definition (8) for $\{n-m,n-m-1|a\}$ , changing summation order, and taking into account the property of binomial symbols: $(n|k)(n-k|m)=(n|m)(n-m|k)$ we have

$\displaystyle b^{n}=a\sum\limits_{m=0}^{n-1}{(-1)^{n-m}(n|m)b^{m}\{n-m,n-m-1|a% \}}+b^{n}=a\sum\limits_{m=0}^{n}(n|m)b^{m}\sum\limits_{k=0}^{n-m}{(-1)^{n-m-k}% (n-m|k)(k+a)^{n-m-1}}=a\sum\limits_{m=0}^{n}b^{m}\sum\limits_{k=0}^{n-m}(-1)^{% n-m-k}(n|k)(n-k|m)(k+a)^{n-m-1}=a\sum\limits_{k=0}^{n}{(n|k)\sum\limits_{m=0}^% {n-k}{(-1)^{n-k-m}b^{m}(n-k|m)(k+a)^{n-m-1}}}.$

We see that the term $k=0$ is the Newtonian expansion of the binomial $(b-a)^{n}$ while each of summands of the remaining sum can be prepared as the Newtonian expansion of binomial $(b-a-k)^{n}$ with coefficient $(n|k)(k+a)^{k-1}$ :

$\displaystyle b^{n}=(b-a)^{n}+a\sum\limits_{k=1}^{n}{(n|k)(k+a)^{k-1}\sum% \limits_{m=0}^{n-k}{(-1)^{n-k-m}b^{m}(n-k|m)(k+a)^{n-k-m}}}=(b-a)^{n}+a\sum% \limits_{k=1}^{n}{(n|k)(k+a)^{k-1}(b-k-a)^{n-k}}$

After substitution of $a+b$ for $b$ it is easy to recognize the Abel binomial formula in the last equation:

$\displaystyle(a+b)^{n}=b^{n}+a\sum\limits_{k=1}^{n}{(n|k)(a+k)^{k-1}(b-k)^{n-k% }}=a\sum\limits_{k=0}^{n}{(n|k)(a+k)^{k-1}(b-k)^{n-k}}$ (10)

It is instructive to compare the above six rows of transformations with the derivation of Abel’s formula on six pages of the basic source of binomial formulae [1, pp. 27–32].

4. Sums and series including powers of natural numbers

Calculation of sums of powers of natural numbers attracts attention from the Bernoulli times. These formulae in popular handbooks (e.g. [4, 5]) either require a preliminary computing Bernoulli numbers or use artificial transformations of recurrent type. The symbols $\{n,m\}$ lead to very simple and transparent expressions for these and other sums and series. Let $n>m>0$ . We start from the identity

$\displaystyle\sum\limits_{n=k}^{p}{(-1)^{n}(p|n)}(n|k)=(-1)^{k}\delta_{kp}$ (11)

It allows finding a sum of introduced symbols similar to recurrent Eq. (2), but over the first argument.

$\displaystyle\sum\limits_{n=1}^{p}{(-1)^{n}}(p|n)\{n,m\}=\sum\limits_{n=1}^{p}% {(-1)^{n}}(p|n)\sum\limits_{k=1}^{n}{(-1)^{k}(n|k)k^{m}}=\sum\limits_{k=1}^{p}% {(-1)^{k}k^{m}\sum\limits_{n=k}^{p}{(-1)^{n}(p|n)}(n|k)}=p^{m}.$ (12)

The power sum is shortened after transformation into a sum of symbols $\{n,m\}$ .

$\displaystyle\sum\limits_{p=1}^{q}p^{m}=\sum\limits_{p=1}^{q}\sum\limits_{n=1}% ^{p}{(-1)^{n}}(p|n)\{n,m\}=\sum\limits_{n=1}^{q}{(-1)^{n}\{n,m\}}\sum\limits_{% p=n}^{q}{(p|n)}=\sum\limits_{n=1}^{m}{(-1)^{n}\{n,m\}(q+1|n+1)}.$ (13)

Interchanging summation order leads to the internal sum of binomial coefficients which amounts to a single one. The main result is that most terms of the external sum vanish due to the basic identity Eq. (3) for symbols $\{n,m\}$ . It remains only to present $\{n,m\}$ using definition Eq. (1)

$\displaystyle\sum\limits_{p=1}^{q}p^{m}=\sum\limits_{p=1}^{m}{p^{m}}\sum% \limits_{n=0}^{m-p}{(-1)^{n}(n+p|p)(q+1|n+p+1)}=2(q+1|2)\sum\limits_{p=1}^{m}{% p^{m-1}}(q-1|p-1)\sum\limits_{n=0}^{m-p}{(-1)^{n}(q-p|n)/(p+n+1)}$ (14)

Thus the number of terms $p^{m}$ in the final result equals to the exponent. In particular, if $m=1$ or $m=2$ we have the well-known results: progression sum $q(q+1)/2$ and $(q+1|2)-2(q+1|3)+2^{2}(q+1|3))=(q+1|2)(2q+1)/3$ , etc.

Equation (14) is “self-consistent”: it suggests that the single way to calculate the sum if $q=m$ is to use its definition (13) directly. This results from the identity $\sum_{n=0}^{m-p}(-1)^{n}(n+p|p)(m+1|n+p+1)=1$ which in its turn follows from the identity $\sum_{n=0}^{m-p}(-1)^{n}(m-p|n)/(n+p+1)=1/[(m-p+1)(m+1|m-p+1)$ (see Ref. [4, p. 452, Table 1]).

Equation (14) leads to the following theorem. The remainder of division by a prime $q$ of the sum of $q$ any successive natural numbers, raised to any natural power divisible by $q-1$ , is $q-1$ . For any other natural powers, $q$ divides such sum (compare to corollary after Eq. (5)).

Indeed, sum (14) which is divisible by $q$ due to the presence of $(q+1|2)$ in the right-hand part differs from the sum of all remainders of division by $q$ only by $q^{m}$ (obviously divisible by $q$ ). This assertion has one exception $m=q-1$ when the denominator in Eq. (14) equals $q$ (in last term of internal sum). However, just for this exponent, $q-1$ , due to Fermat’s “little” theorem remainders of all powers are 1 (except of the last one of $q^{q-1}$ which is 0) giving the net value $q-1$ . The situation for sequence of $q$ arbitrary natural numbers is reduced to the above one.

The identity Eq. (3) “cuts off” not only power sums, but also series with a running term containing integer power. For instance, let us consider the following series which is absent in handbooks [2, 3]

$\displaystyle\sum\limits_{p=1}^{\infty}a^{p}p^{m}/p!=\sum\limits_{p=1}^{\infty% }{a^{p}/p!}\sum\limits_{n=1}^{p}{(-1)^{n}}(p|n)\{n,m\}=\sum\limits_{n=1}^{% \infty}{(-1)^{n}}\{n,m\}/n!\sum\limits_{p=n}^{\infty}a^{p}/(p-n)!=e^{a}\sum% \limits_{k=1}^{m}{k^{m}/k!}\sum\limits_{n=k}^{m}{(-1)^{n-k}a^{n}/(n-k)!}$ (15)

The final result is the $m$ -long first segment of the initial series. The $p$ -th coefficient consists of $m-p$ first terms of $e^{-a}$ -series multiplied by $e^{a}$ :

$\displaystyle\sum\limits_{p=1}^{\infty}a^{p}p^{m}/p!=\sum\limits_{p=1}^{m}a^{k% }p^{m}/p!\left[e^{a}\sum\limits_{n=0}^{m-p}{(-a)^{n}/n!}\right].$ (16)

There are some examples of a simpler series with $a=1$ in Ref. [2, #0.248], e.g. coefficient at $e$ in the final result for $m=5$ is 52, the same as in Eq. (16): $1/2-1/6+/24+16(1/2-1/6)+81/4+625/24=52$ .

On account of the equality $1\cdot 2..(k-1)/a^{k}+2\cdot 3..k/a^{k+1}+3\cdot 4..(k+1)/a^{k+2}\ldots=(k-1)!% /(a-1)^{k}$ resulting from $k-1$ -times differentiation of the convergent progression $\sum_{p=0}^{\infty}1/a^{p}=(a-1)^{-1}$ , which may be written as $a\sum{}_{p=n}^{\infty}(p|n)/a^{p+1}=a/(a-1)^{n+1}$ , and the property (3) of $\{n,m\}$ we can sum a more complicated series.

$\displaystyle\sum\limits_{p=1}^{\infty}p^{m}/a^{p}=\sum\limits_{p=1}^{\infty}{% a^{-p}}\sum\limits_{n=1}^{p}{(-1)^{n}}(p|n)\{n,m\}=\sum\limits_{n=1}^{\infty}{% (-1)^{n}}\{n,m\}\sum\limits_{p=n}^{\infty}{(p|n)/a^{p}}=\sum\limits_{n=1}^{m}(% -1)^{n}a/(a-1)^{n+1}\{n,m\}=a/(a-1)^{m+1}\sum\limits_{k=1}^{m}{k^{m}}\sum% \limits_{n=k}^{m}(-1)^{m-n}a^{n-k}(m+1|n+1)=a/(a-1)^{m+1}\sum\limits_{n=0}^{m-% 1}{a^{n}}\sum\limits_{k=1}^{m-n}{(-1)^{m-n-k}k^{m}(m+1|n+k+1)}.$ (17)

The Eq. (1), subsequent interchange of summation order and using Newton’s formula for $(a-1)^{m-n}$ and binomial identity $\sum^{m-p}_{n=k}(m-n|p)(n|k)=(m+1|k+p+1)$ are seen in the second row Eq. (17), and further change of the summation index $n\to n-k$ and summation order give the final result. This series defines a set of functions $A^{(m)}(a)$ for any $a$ outside of the interval $|a|\leqslant 1$ which can be put in a form similar to all other demonstrations of the above technique:

$\displaystyle\sum\limits_{p=1}^{\infty}p^{m}/a^{p}=\sum\limits_{p=1}^{m}{p^{m}% /a^{p}\sum\limits_{n=p}^{m}{(-1)^{m-n}(m+1|n+1)a^{n+1}/(a-1)^{m+1}}}=A^{(m)}(a)$ (18)

The first $A^{(m)}(a)$ are: $(a-1)^{2}A^{(1)}(a)=a,(a-1)^{3}A^{(2)}(a)=a(a+1),(a-1)^{4}A^{(3)}(a)=a(a^{2}+4% a+1),(a-1)^{5}A^{(4)}(a)=a(a^{3}+11a^{2}+11a+1),(a-1)^{6}A^{(5)}(a)=a(a^{4}+26% a^{3}+66a^{2}+26a+1)\ldots$ The roots of these $A^{(m)}(a)$ are: for $m=2,a=-1$ ; for $m=3$ , $a=-2\pm\surd 3$ ; for $m=4,a=-1,a=-5\pm 2\surd 6$ , etc. The first root does not correspond to a convergent series since $a=-1$ does not belong to the proper interval while the next ones do: $-(2+\surd 3),-(5+2\surd 6)$ . For $A^{(5)}(a)$ the regularities are similar: $a^{4}+26a^{3}+66a^{2}+26a+1=(a^{2}+2x_{1}a+1)(a^{2}+2x_{2}a+1)=0$ ; hence $x_{1}+x_{2}=13,x_{1}x_{2}=16$ , and the square equation for $x$ appears: $x^{2}-13x+16=0$ . Each root of this equation leads to one acceptable solution of $a^{2}+2xa+1=0$ : $a=-23.217\ldots$ and $-3.428\ldots$ guarantee the convergence of series (18) to 0, etc.

It can be proved that the properties demonstrated by these examples: the equality of coefficients which are equally distant from the polynomial ends (in particular, the first and the last are 1) are valid in the general case, too.

5. Sums and series of symbols

\bm{\{{n},{m}\}}

and Bernoulli numbers

Now we present some useful sums of symbols $\{n,m\}$ . Using the identity $\sum^{m}_{n=k}(n-1|k-1)=(m|k)$ we can transform the sum of $\{n,m\}$ into a binomial sum which admits the further simplification.

$\displaystyle\sum\limits_{n=1}^{m}\{n,m\}=\sum\limits_{n=1}^{m}{\sum\limits_{k% =1}^{n}{(-1)^{k}(n|k)k^{m}}}=\sum\limits_{k=1}^{m}{(-1)^{k}k^{m}\sum\limits_{n% =k}^{m}{(n|k)}}=\sum\limits_{k=1}^{m}{(-1)^{k}k^{m}(m+1|k+1)}=(-1)^{m}.$ (19)

The $(-1)^{m}$ appears thanks to $(m+1|k+1)=(m|k+1)+(m|k)$ , the definition Eq. (1) of $\{m,m\}=(-1)^{m}m!$ (see Eq. (5)), change of summation index $k\to k+1$ , and to the result of transformation (5) below of the appeared there second sum:

$\displaystyle\sum\limits_{k=1}^{m}{(-1)^{k}k^{m}(m+1|k+1)}=\{m,m\}-\sum\limits% _{k=2}^{m}{(-1)^{k}(k-1)^{m}(m|k)},$ $\displaystyle\sum\limits_{k=2}^{m}{(-1)^{k}(k-1)^{m}(m|k)}=(-1)^{m}m!-\sum% \limits_{k=1}^{m}{(-1)^{k}(m|k)}\sum\limits_{p=1}^{m}{(-1)^{p}k^{m-p}(m|p)}=(-% 1)^{m}m!-\sum\limits_{p=1}^{m}{(-1)^{p}(m|p)\{m,m-p\}}=(-1)^{m}m!-(-1)^{m}$ (20)

After separation of zero term of Newton’s formula for the rest of them we obtain sum of zero symbols except the last one which is equal to $(-1)^{m}$ .

These transformations suggest a method based on property (3) to obtain other similar sums.

$\displaystyle\sum\limits_{n=1}^{m}{\{n,m\}/n\!=\!\sum\limits_{n=1}^{m}\!{\sum% \limits_{k=1}^{n}\!{(-1)^{k}(n-1|k-1)k^{m-1}}}}\!=\!\!\sum\limits_{n=1}^{m}{(-% 1)^{k}(m|k)k^{m-1}}\!=\!\{m,m-1\}\!=\!0.$ (21)

If the denominator is decreased by 1 we can use Eq. (4) for $n\to n-1$ and after that Eqs (19) and (21). Repeating these operations until $\{2,2\}=2$ appears we come to

$\displaystyle\sum\limits_{n=2}^{m}\{n,m\}/(n-1)=\sum\limits_{n=2}^{m}n(\{n,m-1% \}-\{n-1,m-1\})/(n-1)=\sum\limits_{n=2}^{m-1}{\{n,m-1\}/(n-1)}+1=m.$ (22)

A sum, $B_{m}$ , with the increased by 1 denominator leads to a somewhat unexpected result.

$\displaystyle B_{m}=\sum\limits_{n=0}^{m}{\{n,m\}/(n+1).}$ (23)

The designation $B_{m}$ is justified below. Recurrent Eq. (2), values (5), and interchanging summation order gives

$\displaystyle B_{m}=-\left(\frac{1}{2}+\sum\limits_{n=2}^{m}\sum\limits_{p=n-1% }^{m-1}(m|p)\{n-1,p\}/(n+1)\right)=-\left(\frac{1}{2}+\sum\limits_{p=1}^{m-1}(% m|p)\sum\limits_{n=2}^{p+1}\{n-1,p\}/(n+1)\right).$ (24)

Applying the “lowering” Eq. (4) to the internal sum we transform it into initial quantities

$\displaystyle\sum\limits_{n=1}^{p+1}{\{n-1,p\}/(n+1)}=\sum\limits_{n=1}^{p+1}{% (\{n,p\}/(n+1)-\{n,p+1\}/n(n+1))}=B_{p}-\sum\limits_{n=1}^{p+1}\left(\{n,p+1\}% \left(\frac{1}{n}-\frac{1}{n+1}\right)\right)=B_{p}+B_{p+1}$

where Eq. (21) and $\{p+1,p\}=0$ demonstrated vanishing “unnecessary” sum and term while “necessary” term with $n=1$ which is zero: $\{0,p\}=0$ according to Eq. (5) has been initially added. Thus a recurrent formula has appeared.

$\displaystyle B_{m}=-\left(\frac{1}{2}+\sum\limits_{p=1}^{m-1}(m|p)(B_{p}+B_{p% +1})\right)=-\frac{1}{m+1}\left(1+\sum\limits_{p=1}^{m-1}(m+1|p)B_{p}\right).$ (25)

It works beginning from $m=1$ and gives $B_{1}=-1/2$ when the sum is absent. Then the next value $B_{2}=1/6$ suggests that $B_{m}$ are Bernoulli numbers, and it would be natural to put $B_{0}=1$ simultaneously complimenting the list (5) by $\{0,0\}=1$ .

To prove that $B_{m}$ are really Bernoulli numbers it should be noticed that Eq. (25) is a “real” formula corresponding to the usual symbolic definition $B_{m+1}=(B+1)^{m+1}$ of Bernoulli numbers [2, p. 1090].

Using $B_{1}=-1/2$ and $B_{2}=1/6$ from Eq. (25) we find $B_{3}=0$ . Then supposing that all odd $B_{m}$ vanish beginning from $m=3$ we see from Eq. (25) that this hypothesis is reproduced, and for even $B_{2m}$ the recurrent formula and new identity appear.

$\displaystyle B_{2m}=\frac{2m-1}{2(2m+1)}-\sum\limits_{p=1}^{m-1}{(2m|2p-1)B_{% 2p}/2p},∼{}∼{}\sum\limits_{p=1}^{m-1}(2m|2p)B_{2p}=m-1.$ (26)

The identity is not mentioned in Refs [2, 3] and has not been noticed in such sources as Ref. [5], but it is supported by presented there numerical values e.g.: $(10|2)B_{2}+(10|4)B_{4}+(10|6)B_{6}+(10|8)B_{8}=45/6-210/30+210/42-45/30=4.$ It is this identity which, in fact, “explains” vanishing odd Bernoulli numbers.

Thus we have demonstrated a similarity of the sum (23) for odd numbers $m$ to the sum (21): both are zero.

The immediate consequence of Eq. (23) for even subscript $2m$ is the von Staudt-Clausen theorem which states that the fraction part of Bernoulli number $B_{2m}$ is the negative sum of all $1/p$ where $p$ are such primes that $p-1$ divides $2m$ . Indeed, it is a direct application of the Corollary 1 of Eq. (5) to the sum (23) for even $m$ which states that each factor $p-1$ of $2m$ where $p$ is prime contributes to sum (23) the term $(p-1)/p$ , i.e. $-1/p$ to its fractional part.

6. Symbols

\bm{\{{n},{m}\}}

and second kind Stirling numbers

The connection of $\{n,m\}$ to the Stirling numbers $S(p,m)$ clarifies the nature of the identity Eq. (3). Since $S(p,m)$ $(S(0,0)=1)$ transforms $k^{m}$ into the “falling” power $k_{m}=k(k-1)(k-2)\ldots(k-m+1)$ it follows from the definition (1) for $k\geqslant m$

$\displaystyle\{n,m\}=\sum\limits_{k=1}^{n}(-1)^{k}(n|k)\sum\limits_{j=1}^{(m,k% )}{S(m,j)k_{j}}=\sum\limits_{j=1}^{(m,n)}S(m,j)\sum\limits_{k=j}^{n}{(-1)^{k}(% n|k)k_{j}}=\sum\limits_{j=1}^{(m,n)}S(m,j)\sum\limits_{k=j}^{n}{(-1)^{k}(n-j|k% -j)n_{j}}=\sum\limits_{j=1}^{(m,n)}(-1)^{j}S(m,j)n_{j}\sum\limits_{k=0}^{n-j}{% (-1)^{k}(n-j|k)}=\sum\limits_{j=1}^{(m,n)}(-1)^{j}S(m,j)n_{j}\delta_{nj}=(-1)^% {n}S(m,n)n!,m\geqslant n,n_{n}=n!.$ (27)

The “reduction” Eq. (4) is equivalent to the recurrent formula for Stirling numbers: $S(p+1,n)=S(p,n-1)+{nS}(p,n)$ . The symbol $\{n,m\}$ “mysterious” vanishing Eq. (3) if condition $m\geqslant n$ is not fulfilled now looks as analytic continuation of the Stirling numbers onto the upper zero triangle of their matrix.

$\displaystyle\{n,m\}={\begin{array}[]{@{}ccccl@{}}{n\backslash m}&0&1&2&3\\ 1&{-1}&{-1}&{-1}&{-1}\\ 2&{-1}&0&2&6\\ 3&{-1}&0&0&{-6}\\ 4&{-1}&0&0&0\ldots\\ \end{array}},$ $\displaystyle S(m,n)={\begin{array}[]{@{}ccccl@{}}{m\backslash n}&1&2&3&4\\ 1&1&0&0&0\\ 2&1&1&0&0\\ 3&1&3&1&0\\ 4&1&7&6&1\ldots\\ \end{array}}=\sum\limits_{k=1}^{n}{(-1)^{n-k}\frac{k^{m}}{k!(n-k)!}}.$ (28)

The used conditions $\{0,0\}=1$ and $\{0,m\}=0$ for $m\neq 0$ Eq. (5) now look natural, and the first one corresponds to Riordan’s proposition $S(0,0)=1$ [6, p. 43].

The established connection to Stirling numbers allows expressing sums of $\{n,m\}$ over the first argument through those over the second one using the relation $S(p+k,m+k)=S(p+k+1,m+k+1)-(m+k+1)S(p+k,m+k+1)$ . Let us find a sum of both sides of this equality in order to come to sums similar to Eq. (21) with a more complicated denominator. Considering $p$ and $k$ as free integers we define

$\sigma_{k}^{(p)}=\sum\limits_{m=0}^{p}{(-1)^{m}m!S(p+k,m+k)}$

and obtain a recurrent relation for it:

$\displaystyle\sigma_{k}^{(p)}=\sigma_{k+1}^{(p)}-\sum\limits_{m=0}^{p}{(-1)^{m% }(m+1)!S(p+k,m+k+1)}-k\sum\limits_{m=0}^{p}{(-1)^{m}m!S(p+k,m+k+1)}=\sigma_{k+% 1}^{(p)}+\sigma_{k}^{(p)}-k\sigma_{k+1}^{(p-1)}-S(p+k,k),$ $\displaystyle\sigma_{k+1}^{(p)}=k\sigma_{k+1}^{(p-1)}+S(p+k,k)=\sum\limits_{m=% 0}^{p}{k^{m}S(p+k-m,k)}.$

It is easy to recognize $\sigma^{(p)}_{(k)}$ , in the first sum after change the summation index $m+1\to m$ , including the zero term $S(p+k,k)$ , and accounting for $S(p+k,p+k+1)=0$ . The last term of the second sum disappears by the same reason and if the summand is written as $S(p-1+k+1,m+k+1)$ it is clear that this sum is $\sigma^{(p-1)}_{(k+1)}$ . The repeated use of the recurrent formula obtained beginning from $\sigma^{(0)}_{(k)}$ $=$ $S(k,k)=1$ gives the last equality above and desired identities:

$\displaystyle\sum\limits_{m=0}^{p}{k^{m}S(p+k-m,k)}=\sum\limits_{m=0}^{p}(-1)^% {m}m!S(p+k,m+k),$ $\displaystyle\sum\limits_{n=k+1}^{p}\{n,p\}/(n|k+1)=-(k+1)\sum\limits_{m=k}^{p% -1}{k^{p-m-1}\{k,m\}}.$ (29)

It seems this identity is absent in popular books [1, 6] as well as in Ref. [8]. Returning to $\{n,m\}$ according to Eq. (6) led to the second identity. It can be illustrated by $k=3,p=7$ case: the left side is $\{4,7\}+\{5,7\}/5+\{6,7\}/15+\{7,7\}/35=8400-3360+1008-144=5904$ , and the right one is $4(3^{3}\cdot 6+3^{2}\cdot 36+3\cdot 150+540)=5904$ .

7. Connection of symbols

\bm{\{{n},{m}\}}

to Riemann zeta-function of negative integer argument

It is worthwhile to remind that definition (1) admits $m$ to be a continuous real or even a complex parameter (to emphasize this we use letter $z$ instead of $m$ ) and does not violet the transformation of $\{n,z\}$ into power of $z$ Eq. (12). Parameter $n$ remains natural. Bearing this in mind let us consider series with denominator $2^{n}n$ as function of $z$ on account of $\sum_{n=k-1}^{\infty}(n-1|k-1)/2^{n}=1$ .

$\displaystyle\sum\limits_{n=1}^{\infty}{\{n,z\}/2^{n}n}=\sum\limits_{k=1}^{% \infty}(-1)^{k}/k^{1-z}=-\eta(1-z)=(2^{z}-1)\zeta(1-z),$ $\displaystyle(2^{1-z}-1)\zeta(z)=\sum\limits_{n=1}^{\infty}{\{n,1-z\}/2^{n}n}$ (30)

It was not difficult to recognize in Eq. (7) the Dirichlet eta-function – the alternating sign version of the Euler $\zeta(z)$ -series as is already noted in Eq. (7). On account of Eq. (4) for $n\to n-1$ , $m\to 1-z$ and change the summation index $n\to n+1$ in the second sum, we remove 1 from the argument, sum a progression using $\{n,0\}=-1$ , and return to $m>0$ .

$\displaystyle(2^{1-z}-1)\zeta(z)=\sum\limits_{n=1}^{\infty}{\{n,-z\}/2^{n+1}=% \sum\limits_{k=1}^{\infty}{(-1)^{k}/k^{z}}},∼{}\zeta(0)=-1/2,$ $\displaystyle(2^{m+1}-1)\zeta(-m)=\sum\limits_{n=1}^{m}{\{n,m\}/2^{n+1}}.$ (31)

Equation (7) give analytical continuation of Euler series onto zero and negative values of $\zeta$ -function arguments. It is necessary to emphasize that for integer negative values of argument in $\zeta(-m)$ the symbol $\{n,m\}$ depend on positive $m$ and as a result the identity (3) cuts the series leading to a finite sum Eq. (7). The first Eq. (7) also can, in principal, be used to find $\zeta(m)$ by means of expressions of type Eq. (7) (but this is not an effective way).

An equivalent method of analytical continuation of $\zeta(z)$ has been proposed by Hasse [8] who used Euler transformation of the Leibnitz series. After substitution the definition (1) into Eq. (7) we cast $(2^{1-z}-1)\zeta(z)$ in a form

$\displaystyle(2^{1-z}-1)\zeta(z)=\sum\limits_{n=1}^{\infty}{\sum\limits_{k=1}^% {n}{(-1)^{k}(n|k)/2^{n+1}k^{z}}},$ $\displaystyle(2^{1-z}-1)\zeta(z)=\sum\limits_{n=0}^{\infty}{\sum\limits_{k=0}^% {n}{(-1)^{k-1}(n|k)/2^{n+1}(k+1)^{z}}},$ (32)

which insignificantly differs from the Hasse formula written behind. Both follow from Eq. (7) where the internal sum is reduced to 1 after interchange of the summation order. The outer sum differs only by a shift $k\to k+1$ of the summation index. Nevertheless, their application is somewhat unlike, e.g. $\zeta(0)=-1/2$ follows from the first Eq. (7) after summation of progression while the second Eq. (7) leads to $\zeta(0)=-1/2$ due to vanishing all terms except $n=0$ if $z=0$ .

8. Connection of zeta-function of negative integer argument to Bernoulli numbers

Substitution of $\{n,m\}$ from the recurrent Eq. (2) into Eq. (7) after changing the summation order, separation zero term, using inside Eq. (7) once more, and increasing summation index $n-1\to n$ gives the recurrent formula for calculation $\zeta(-m)$ one by one.

$\displaystyle(2^{m+1}-1)\zeta(-m)=-\frac{1}{2}\left[\frac{1}{2}+\sum\limits_{p% =1}^{m-1}{(m|p)(2^{p+1}-1)}\zeta(-p)\right].$ (33)

If $m=1$ the sum is absent, and $\zeta(-1)=-1/12$ . Then we obtain: $\zeta(-2)=0$ , $\zeta(-3)=-(1/2-3\cdot 3/12)/(2\cdot 15)=1/120$ , $\zeta(-4)=0\ldots$ The situation very resembles the one for Bernoulli numbers. Omitting even terms in Eq. (33) we come to two formulae for $\zeta$ of odd argument; the third one expresses vanishing $\zeta(-2m)$ .

$\displaystyle(4^{m}-1)\zeta(-2m+1)=-\frac{1}{2}\left(\frac{1}{2}+\sum\limits_{% q=1}^{m-1}{(2m-1|2q-1)(4^{q}-1)\zeta(-2q+1)}\right)=-\frac{1}{2m}\left(\frac{1% }{2}+\sum\limits_{q=1}^{m-1}{(2m|2q-1)(4^{q}-1)\zeta(-2q+1)}\right)$ (34)

Both of them follow from Eq. (33) and the assumption $\zeta(-2m)=0$ as well as the identities Eq. (26) for Bernoulli numbers are derivable from the recurrent Eq. (25) and $B_{2m+1}=0$ . This similarity is strongly supported by calculated $\zeta(-2m+1):\zeta(-1)=-B_{2}/2$ , $\zeta(-3)=-B_{4}/4$ etc. However, if we would try to prove the hypothesis substituting $\zeta(-2m+1)=-B_{2m}/2m$ into recurrent Eq. (33) it is evident that recurrent formula for Bernoulli numbers Eq. (25) will not appear.

Nevertheless we can obtain the proof in a quite different way. A generating function for $\zeta(-2q+1)$ can be built from the starting Eq. (7) on account of the main identity Eq. (3).

$\displaystyle\sum\limits_{m=1}^{\infty}{(2^{m+1}-1)\zeta(-m)x^{m}/m!}=\sum% \limits_{n=1}^{\infty}{\sum\limits_{m=n}^{\infty}{\{n,m\}x^{m}/2^{n+1}m!}}=% \sum\limits_{n=1}^{\infty}(1-e^{x})^{n}/2^{n+1}=-1/2\operatorname{th}(x/2).$ (35)

The steps of transformation in Eq. (35) are: substitution of $\zeta(-m)$ by expression (7) and change of summation order, application of identity (3) in order to expand the internal summation from 0 (bearing in mind that $\{n,0\}=-1$ ), using the definition Eq. (1) of $\{n,m\}$ . The next steps: summation of the $m$ -series to obtain $e^{kx}$ and the finite sum to obtain $(1-e^{x})^{n}$ . The last summation of progression gives the final result (35).

Further, it can be noticed that the left part of Eq. (35) consists of two sums, and each of them can be expressed through the same function.

$\displaystyle\sum\limits_{q=1}^{\infty}{(4^{q}-1)\zeta(-2q+1)x^{2q-1}}/(2q-1)!% =2f(2x)-f(x)=-1/2\operatorname{th}(x/2),$ $\displaystyle f(x)=\sum\limits_{q=1}^{\infty}{\zeta(-2q+1)x^{2q-1}}/(2q-1)!$

This functional equation can be solved if we remember the identity for hyperbolic functions: $2\operatorname{cth}2x-\operatorname{cth}x=\operatorname{th}x$ which suggests the solution $f(x)=-1/2\operatorname{cth}x/2$ . Thus

$\displaystyle\sum\limits_{q=1}^{\infty}{\zeta(-2q+1)x^{2q}}/(2q-1)!=-x/2% \operatorname{cth}(x/2)=-(x/2+x/(e^{x}-1)).$ (36)

As is well known the second term from the right is the generating function for the Bernoulli numbers (see e.g. Ref. [1, p. 57]) and the first term emphasizes the specifics of $B_{1}$ . Now the theorem is completely proved. Hence, the observed numerical coincidences are not occasional.

The established connection between $\zeta(-2m+1)$ and $B_{2m}$ allows to transfer onto $B_{2m}$ all identities found for $\zeta(-2m+1)$ and vice-versa, in particular, identities (34) onto $B_{2m}$ and Eq. (26) on $\zeta(-2m+1)$ (We shall write here explicitly only latter).

$\displaystyle\sum\limits_{q=1}^{m}{(2m|2q-1)\zeta(-2q+1)}=-\frac{2m-1}{2(2m+1)% },∼{}\sum\limits_{q=1}^{m-1}{(2m-1|2q-1)}\zeta(-2q+1)=\frac{1-m}{2m}.$ (37)

A lot of this type identities can be obtained by means of a following technique which allows to prove that the second equality (34) is a consequence of the first one, or to answer “why odd Bernoulli numbers vanish”. If we denote for brevity $-2(4^{q}-1)\zeta(-2q+1)$ by $\tau_{q}$ then both Eq. (26) take a similar form: $P_{m}=0$ and $Q_{m}=0,m=1,\ldots$ where

$\displaystyle P_{m}\equiv\sum\limits_{q=1}^{m-1}{(2m-1|2q-1)\tau_{q}}+2\tau_{m% }-1,∼{}Q_{m}\equiv\sum\limits_{q=1}^{m-1}{(2m|2q-1)\tau_{q}}+2m\tau_{m}-1,$

As is seen $P_{1}\equiv Q_{1}$ . Therefore both formulae give the same $\tau_{1}=1/2=-6\zeta(-1)$ . All remaining $\tau_{m}$ have factor $\tau_{1}$ . All next $\tau_{m}$ obtained from both formulae by means of Eq. (37) can be equalized on each step by means of the corresponding factor $\alpha_{m}$ found from the requirement $P_{m}=Q_{m}$ . It means, e.g. that $3+2\alpha_{2}=4+4\alpha_{2}$ follows from the above equations, giving $\alpha_{2}=-1/2$ that delivers $\zeta(-3)=-1/2\cdot 1/2\cdot(-1/2)\cdot 1/15=1/120$ . Then using already obtained $\alpha_{2}$ we find $\alpha_{3}=1$ from $5-10/2+2\alpha_{3}=6-20/2+6\alpha_{3}$ which leads to $\zeta(-5)=-1/2\cdot 1/2\cdot 1/63=-1/252$ etc.

It is clear that plenty of other identities containing positive and negative powers of 4 is possible to build for $\tau_{q}$ in a similar way and hence for $\zeta(-2m+1)$ and $B_{2m}$ . This possibility arises thanks to the known sums of binomial coefficients. For example,

$\displaystyle 0=\sum\limits_{m=1}^{n-1}(2n-1|2m)Q_{m}=\sum\limits_{m=1}^{n-1}{% (2n-1|2m)}\sum\limits_{q=1}^{m}{(2m|2q-1)\tau_{q}}-4^{n-1}+1=\sum\limits_{m=1}% ^{n}{(2n-1|2m-1)}4^{n-q}\tau_{m}/2-4^{n-1}+1.$

Here the result of interchange of summation order, binomial coefficient property $(a|b)(b|c)=$ $(a|c)(a-c|b-c)$ , and the values of binomial sums:

$\displaystyle\sum\limits_{q=1}^{n-1}{(2n-1|2q-1)}\tau_{q}\sum\limits_{m=0}^{n-% q-1}{(2(n-q)|2m+1)},∼{}\sum\limits_{m=1}^{n-1}{(2n-1|2m)}=4^{n-1}-1,$ $\displaystyle\sum\limits_{m=1}^{n-q}{(2(n-q)|2m-1)}=4^{n-q}/2$

have been taken into account. The return from $\tau_{m}$ to $\zeta(-2m+1)$ allows to write this identity as

$\displaystyle\sum\limits_{m=1}^{n-1}{(2n-1|2m-1)}(4^{-m}-1)\zeta(-2m+1)=1/4-1/% 4^{n},$ $\displaystyle\sum\limits_{m=1}^{n}{(2n|2m-1)}(4^{-m}-1)\zeta(-2m+1)=1/4-1/(2% \cdot 4^{n}).$ (38)

Another identity obtained in a similar way from $P_{m}$ is shown next to it.

Combining these formulae with Eqs (34) and (37) allows to separate identities for $\zeta(-2m+1)$ which contain positive powers of 4 from those containing negative and zero powers.

$\displaystyle\sum\limits_{q=1}^{m}{(2m|2q-1)4^{q}\zeta(-2q+1)}=\frac{1}{2m+1}-1,$ $\displaystyle\sum\limits_{q=1}^{m}{[(2m-1|2q-1)+\delta_{qm}]}4^{q}\zeta(-2q+1)% =\frac{1}{2m}-1+2\zeta(-2m+1),$ $\displaystyle\sum\limits_{q=1}^{m}{(2m|2q-1)4^{-q}\zeta(-2q+1)}=\frac{1}{2m+1}% -\frac{1}{4}-4^{-m}/2,$ $\displaystyle\sum\limits_{q=1}^{m-1}{(2m-1|2q-1)}4^{-q}\zeta(-2q+1)=\frac{1}{2% m}-\frac{1}{4}-4^{-m}.$ (39)

Identities with more complicated expressions built of powers of 4 can be derived in a similar manner. For instance, taking half of sum of the first and third equalities Eq. (8) and subtracting the first one of Eq. (37) leads to

$\displaystyle\sum\limits_{q=1}^{m}{(2m|2q-1)}(4^{-q}-1)^{2}\zeta(-2q+1)=(3% \cdot 4^{m}-3^{2m}-1-4^{2m}/2)/4^{2m+1}$ (40)

All these identities can be rewritten in terms of Bernoulli numbers after substitution $\zeta(-2m+1)\to-B_{2m}/2m$ . It is worthwhile to mention that they are seemed absent in popular compendiums on Bernoulli numbers [5] and on zeta function or in sources quoted there.

9. Generating functions for zeta-function of positive integer argument

The calculated values of $\zeta(-m)$ as well as identities for them and generating functions thanks to the Riemann functional equation for zeta-function can be carried-over onto positive argument case. Only a preliminary modification of the equation by means of the complementary formula for the Euler $\Gamma(z)$ -function is needed:

$\displaystyle\zeta(z)=2^{z}\pi^{z-1}\sin(\pi z/2)\Gamma(1-z)\zeta(1-z),∼{}% \Gamma(1-z)=\pi/\Gamma(z)\sin\pi z,$ $\displaystyle\zeta(z)=2^{z-1}\pi^{z}\zeta(1-z)/\cos(\pi z/2)\Gamma(z),$ (41)

It immediately leads to the Euler formula for $\zeta(2m)$ if we take into account simplifications for integer $z=2m$ : $\cos(\pi m)=(-1)^{m},\Gamma(2m)=(2m-1)!$

$\displaystyle\zeta(2m)=(2\pi)^{2m}(-1)^{m}m\zeta(-2m+1)/(2m)!$ (42)

Using in Eq. (42) the results obtained above for $\zeta(-2m+1)$ we obtain various consequences for $\zeta(2m)$ . For example, $\zeta(-1)=-1/12$ turns into the Euler “Basel” formulae $\zeta(2)=\pi^{2}/6$ , and Eq. (37) gives a recurrent formula for $\zeta(2m)$ :

$\displaystyle\zeta(2m)=\frac{(2m-1)\pi^{2}(2\pi)^{2(m-1)}}{(-1)^{m-1}(2m+1)!}+% \sum\limits_{q=1}^{m-1}{(-1)^{q-1}(2\pi)^{2q}\zeta(2(m-q))/(2q+1)!},$ $\displaystyle\sum\limits_{q=1}^{m-1}{\frac{(-1)^{q}\zeta(2q)}{(2\pi)^{2q}(2(m-% q))!}}=\frac{1-m}{2(2m)!}.$ (43)

The first formula allows to find, e.g. $\zeta(4)$ from $\zeta(2)$ : $\zeta(4)=-12\pi^{4}/120+4\pi^{2}\zeta(2)/6=\pi^{4}/90$ , the second one generates somewhat unexpected identities, e.g. $\zeta(2)/3\pi^{2}-\zeta(4)/34\pi^{4}=1/45$ etc.

Substitution of $\zeta(-2m+1)$ from Eq. (35) into expansions Eqs (35) and (36) and simultaneous change of $x$ by $2\pi x$ lead to the generation functions for $\zeta(2m)$ .

$\displaystyle\sum\limits_{m=1}^{\infty}{(-1)^{m-1}(4^{m}-1)\zeta(2m)x^{2m}=(% \pi x/2)}\operatorname{th}(\pi x),$ $\displaystyle\sum\limits_{m=0}^{\infty}{(-1)^{m-1}\zeta(2m)x^{2m}=(\pi x/2)}% \operatorname{cth}(\pi x)$ (44)

It is convenient to use $(-1)^{m}=i^{2m}$ and $\operatorname{th}({ix})=i\operatorname{tg}x$ to pass from alternating series to monotone series:

$\displaystyle\sum\limits_{m=1}^{\infty}{(4^{m}-1)\zeta(2m)x^{2m}=(\pi x/2)}% \operatorname{tg}(\pi x),∼{}\sum\limits_{m=0}^{\infty}{\zeta(2m)x^{2m}=-(\pi x% /2)}\operatorname{ctg}(\pi x),$ $\displaystyle\sum\limits_{m=0}^{\infty}{\zeta(2m)/4^{m}=0,∼{}(x=1/2)}.$ (45)

It is possible to deduce a lot of similar expansions for zeta function from series Eqs (9) and (9). The second Eq. (9) is itself a consequence of the first one on account of the equality $2\operatorname{cth}2x-\operatorname{cth}x=\operatorname{th}x$ . The other way to obtain new series for $\zeta(2m)$ is the change of the expansion parameter, e.g. $x\to 1/x$ (bearing in mind simultaneous change of the convergence domain from $|x|<1$ to $|x|>1$ ). It is also possible to exclude singular points from the result. For instance, the singular point of cotangent is concentrated in the term $2x/\pi(x^{2}-1)\to\infty$ for $x\to 1$ as is known from its decomposition on partial fractions. It is removed after subtraction of the progression $2x\sum_{m=1}^{\infty}x^{-2m}/\pi$ . Almost all such type series from Ref. [9] can be obtained in this way, e.g.

$\displaystyle\sum\limits_{m=1}^{\infty}\zeta(2m)/x^{2m}=1/2-\pi\operatorname{% ctg}(\pi/x)/2x.$ (46)

Numerous identities can be obtained for Hurwitz zeta function in a similar way by using instead of $\{n,m\}$ the quantity $\{n,m|a\}$ introduced in Eq. (8) for derivation the Abel formula.

Perhaps, it is appropriate to accompany Eq. (9) by the following plot. At the “time of storm and onset” in physics, in twenties Einstein just after creation of the theory of gravitation tried to extract physics of elementary particles treating these as singularities of the set differential equations of gravity. His closest follower Eddington proposed to interpret the main dimensionless constant of microphysics, the fine structure parameter $1/\alpha=137$ , as the reciprocal of the number of independent components of the four-dimensional Riemann-Christoffel curvature tensor: 136 plus 1. This fantasy attracted Dirac. He proposed a different approach when experiments already gave non-integer value for $1/\alpha$ . Dirac equated the constant $\sigma$ in the Stefan-Boltzmann radiation low to $(4\pi e)^{-6}$ , the quantity of a proper dimensionality where $e$ is the elementary charge, and obtained $1/\alpha=(4\pi)^{2}(\pi^{2}/1\cdot 3\cdot 5)^{1/3}=137.358$ . Dirac’s motives remain incomprehensible (see the discussion in Ref. [10, p. 160]). The modern value of $1/\alpha$ is 137.0360.

The importance of primes, zeta function, equations of type Eq. (9), and coryphée’s efforts inspire to seek connections between $\zeta$ and $1/\alpha$ :

$\displaystyle 2\pi\operatorname{cth}\pi-\pi+\pi\operatorname{cth}\pi/(\pi% \operatorname{cth}^{2}\pi-\pi)=137.0376,∼{}\pi\operatorname{cth}\pi=2\sum% \limits_{m=0}^{\infty}{(-1)^{m-1}\zeta(2m)}.$ (47)

Maybe, indeed, there exists some reason why the constant which influences physics of the Universe is connected through $\zeta$ to primes, these “atoms” of mathematics, and to $\pi$ embodying the “simplicity and spherical beauty” of the Universe. Maybe, something “dark” as the dark matter is responsible for the remaining 0.001% $\alpha$ .

10. Sum of powers of two numbers

We proved above numerous identities. However, only the identity (3) of them was meant in the title of the article as was already mentioned. Now we shall prove the second identity kept in mind in the title. We mean transformation of the sum $x^{p}+y^{p}$ for natural and prime $p$ . Instead of $x, y$ , using variables $\sigma=x+y$ and $\rho={xy}$ we have

$\displaystyle x=(\sigma+\sqrt{\sigma^{2}-4\rho})/2,∼{}y=(\sigma-\sqrt{\sigma{}% ^{2}-4\rho})/2,$ $\displaystyle x^{p}+y^{p}=1/2^{p}\sum\limits_{i=0}^{p}{(p|i)\sigma^{i}(\sqrt{% \sigma^{2}-4\rho})^{p-i}[1+(-1)^{p-i}]}.$

If $p$ is odd: $p=2n+1$ all even terms with radicals annihilate, if $p$ is even: $p=2n$ the remained terms without radicals are even. The binomial coefficient is doubled:

$\displaystyle x^{2n+1}+y^{2n+1}=2\sigma/2^{p}\sum\limits_{i=0}^{n}{(2n+1|2i+1)% \sigma^{2i}(\sigma^{2}-4\rho)^{n-i}}=\sum\limits_{i=0}^{n}(p|2i+1)\sigma^{2i}% \sum\limits_{k=0}^{n-i}{(n-i|k)}\sigma^{2k+1}(-4\rho)^{n-i-k}/4^{n}=\sum% \limits_{k=0}^{n}{\sigma^{2k+1}(-4\rho)^{n-k}S_{n,k}/4^{n}};$ $\displaystyle x^{2n}+y^{2n}=2/4^{n}\sum\limits_{i=0}^{n}{(2n|2i)\sigma^{2i}(% \sigma^{2}-4\rho)^{n-i}}=2\sum\limits_{i=0}^{n}(2n|2i)\sigma^{2i}\sum\limits_{% k=0}^{n-i}{(n-i|k)}\sigma^{2k}(-4\rho)^{n-i-k}/4^{n}=2\sum\limits_{k=0}^{n}% \sigma^{2k}(-4\rho)^{n-k}T_{n,k}/4^{n}.$

Both introduced coefficients $S_{n,k}$ and $T_{n,k}$ can be calculated simultaneously.

$\displaystyle S_{n,k}=\sum\limits_{i=0}^{k}(2n+1|2i+1)(n-i|n-k),∼{}T_{n,k}=% \sum\limits_{i=0}^{k}(2n|2i)(n-i|n-k),$ $\displaystyle S_{n-1,k}=S_{n,k}-2T_{n,k},T_{n-1,k}=T_{n,k}-2S_{n-1,k-1}.$ (48)

A chain of simple transformations converts the right part of first formula into the left one:

$\displaystyle\sum\limits_{i=0}^{k}{[(2n+1|2i+1)-2(2n|2i)](n-i|n-k)}=\sum% \limits_{i=0}^{k}{[(2n|2i+1)-(2n|2i)](n-i|n-k)}$ $\displaystyle=\sum\limits_{i=0}^{k}(2n-1|2i+1)(n-i|n-k)+\sum\limits_{i=1}^{k}{% [(2n-1|2i)-(2n|2i)](n-i|n-k)]}$ $\displaystyle=\sum\limits_{i=0}^{k}{(2n-1|2i+1)(n-i|n-k)}-\sum\limits_{i=1}^{k% }{(2n-1|2i-1)](n-i|n-k)}$ $\displaystyle=\sum\limits_{i=1}^{k}{(2n-1|2i+1)[(n-i|n-k)-(n-i-1|n-k)]}+(2n-1|% 2k+1)$ $\displaystyle=\sum\limits_{i=0}^{k}(2n-1|2i+1)(n-i-1|n-k-1)=S_{n-1,k}.$

Similar transformations prove also the second relation:

$\displaystyle\sum\limits_{i=0}^{k}[(2n|2i)(n-i|n-k)-2\sum\limits_{i=1}^{k}{(2n% -1|2i-1)(n-i|n-k)}$ $\displaystyle=(n|k)+\sum\limits_{i=1}^{k}{[(2n-1|2i)-(2n-1|2i-1)](n-i|n-k)}$ $\displaystyle=(n|k)+\sum\limits_{i=1}^{k}[(2n-2|2i)+(2n-2|2i-1)-(2n-1|2i-1)(n-% i|n-k)$ $\displaystyle=(n|k)+\sum\limits_{i=1}^{k}{[(2n-2|2i)-(2n-2|2i-2)](n-i|n-k)}(n|k)$ $\displaystyle\quad∼{}+\sum\limits_{i=1}^{k}{(2n-2|2i)(n-i|n-k)-\sum\limits_{i=% 0}^{k-1}{(2n-2|2i)(n-i-1|n-k)}}$ $\displaystyle=\sum\limits_{i=0}^{k}{(2n-2|2i)[(n-i|n-k)-(n-i-1|n-k)]}$ $\displaystyle=\sum\limits_{i=0}^{k}{(2n-2|2i)(n-i-1|n-k-1)}=T_{n-1,k}.$

From the definition (10) directly, we find $S_{n,0}=2n+1,S_{n,1}=(2n+1)2n(n+1)/3,\ldots T_{n,0}=1,T_{n,1}=2n^{2}\ldots$ The further use of recurrent relations suggests

$\displaystyle S_{n,k}=(2n+1)(n+k|2k+1)4^{k}/(n-k),∼{}T_{n,k}=n(n+k-1|2k)4^{k}/% (n-k).$ (49)

These final results can be verified by substitution into the same recurrent relations and by comparison for $k=0$ with found $S_{n,0}$ and $T_{n,0}$ . The proved identities

$\displaystyle\sum\limits_{i=0}^{k}{(2n+1|2i+1)(n-i|n-k)}=(2n+1)(n+k|2k+1)4^{k}% /(n-k),$ $\displaystyle\sum\limits_{i=0}^{k}{(2n|2i)(n-i|n-k)=n(n+k-1|2k)4^{k}/(n-k)}.$ (50)

lead to the following identities for the sum of two identical natural powers of two real numbers.

$\displaystyle x^{2n+1}+y^{2n+1}=(2n+1)\sigma\rho^{n}\sum\limits_{k=0}^{n}{(-1)% ^{n-k}\frac{(n+k|2k+1)}{n-k}(\sigma^{2}/\rho)^{k}},$ $\displaystyle x^{2n}+y^{2n}=2n\rho^{n-1}\sum\limits_{k=0}^{n}{(-1)^{n-k}\frac{% (n+k-1|2k)}{n-k}(\sigma^{2}/\rho)^{k}}.$ (51)

It is seen that sums in Eq. (10) have a singularity for $k=n$ . If this term is calculated separately using in Eq. (10) the value of sum is $\sum_{i=0}^{n}(2n+1|2i+1)=4^{n}$ , and the peculiarity is removed. This result may be formulated as a prescription to replace the zero denominators by $0!=1$ which transforms both coefficients into unity: $(2n+1|2n+1)$ and $(2n|2n)$ .

The identity Eq. (10) also demonstrates that the integrity of the expression $(n+k|2k+1)/(n-k)$ is guaranteed by the primality of $2n+1$ which itself is not present explicitly in this expression. Indeed, $(n+k|2k+1)/(n-k)$ is generally an uncancellable fraction in spite of that nominator $(n+k)(n+k-1)\ldots(n-k)$ contains $n-k$ , e.g. $(20|7)/14=38760/7$ $(n=17,k=3)$ . Since the prime factor, $2n+1$ , of the nominator in the right-hand part of identity (10) is the largest one and not divisible by $n-k$ as well as by any other factor in the denominator, and the sum is integer, the remaining part $(n+k|2k+1)/(n-k)$ is also integer (in the mentioned example $2n+1=35$ is not a prime).

11. Connection to Chebyshev polynomials and factorization of

\bm{{x}^{p}+y^{p}}

There is an unexpected connection between sums in Eq. (10) and the first kind Chebyshev polynomials. By definition, the coefficient $L_{n-k}^{(2n+1)}$ of the polynomial $T^{(2n+1)}(\cos\varphi)$ appears at $\cos^{2k+1}\varphi$ in $\cos(2n+1)\varphi={\rm Re}(\cos\varphi+i\sin\varphi)^{2n+1}=\sum_{k=0}^{n}L_{n% -k}^{(2n+1)}\cos^{2k+1}\varphi=T^{(2n+1)}(\cos\varphi)$ . The numeration of coefficients is such that $L_{0}$ is at the highest degree of $\cos\varphi$ . Using Newton’s formula for the power expansion and for $(1-\cos^{2}\varphi)^{j}$ we have

$\displaystyle\cos(2n+1)\varphi=\sum\limits_{j=0}^{n}{(-1)^{j}(2n+1|2j)}\cos^{2% n-2j+1}\varphi\sum\limits_{k=0}^{j}{(-1)^{j}(j|k)}\cos^{2k}\varphi=\sum\limits% _{k=0}^{n}{(-1)^{n-k}\left[\sum\limits_{j=0}^{k}{(2n+1|2j+1)}(n-j|n-k)\right]% \cos^{2k+1}\varphi}.$

It is easy to recognize the sum from Eq. (10) in brackets which appeared after interchange of summation orders and $j\to n-j,k\to n-k$ . Comparison of this formulae with Eq. (10) gives $(p=2n+1)$

$\displaystyle\cos(2n+1)\varphi=\sum\limits_{k=0}^{n}{L_{n-k}^{(2n+1)}\cos^{2k+% 1}\varphi},∼{}L_{n-k}^{(2n+1)}=(-1)^{n-k}p\frac{4^{k}(n+k|2k+1)}{n-k},$ $\displaystyle x^{2n+1}+y^{2n+1}=\sigma\rho^{n}\sum\limits_{k=0}^{n}{L_{n-k}^{(% 2n+1)}(\sigma^{2}/4\rho)^{k}}$ (52)

Restoring standard terms order of the Chebyshev polynomial and using reciprocal variable of $\sigma^{2}/4\rho$ as an argument we finally obtain

$\displaystyle x^{2n+1}+y^{2n+1}=\sigma^{2n+1}\sum\limits_{k=0}^{n}{(L_{n-k}^{(% 2n+1)}/L_{0}^{(2n+1)})(4\rho/\sigma^{2})^{n}(\sigma^{2}/4\rho)^{k}}=\sigma^{2n% +1}/4^{n}\sum\limits_{k=0}^{n}{L_{k}^{(2n+1)}(4\rho/\sigma^{2})^{k}}.$ (53)

Because roots of the polynomial (53) are $\cos^{2}[(2k+1)\pi/2(2n+1)],k=0,1,\ldots,n-1$ and the coefficient at the new variable highest power is $L_{0}^{(2n+1)}=4^{n}$ we can reach our goal and cast the Eq. (53) in the factorized form.

$\displaystyle x^{2n+1}+y^{2n+1}=\sigma\prod\nolimits_{k=0}^{n-1}{\left(\sigma^% {2}-2\rho-2\rho\cos\frac{(2k+1)\pi}{2n+1}\right)}=\sigma^{2n+1}\prod\nolimits_% {k=0}^{n-1}{\left(1-4\frac{\rho}{\sigma^{2}}\cos^{2}\frac{(2k+1)\pi}{2(2n+1)}% \right)}.$ (54)

Since $\sigma/2$ is the arithmetic mean of $x$ and $y$ and $\rho$ is a square of their geometric mean the ratio $4\rho/\sigma^{2}<1$ , and each factor in Eq. (54) is greater than $\sin^{2}[(2k+1)\pi/2(2n+1)]$ and the whole product of sinus squares is equal to $1/4^{n}$ according to equation 1.392 from Ref. [2] we obtain the inequality $x^{2n+1}+y^{2n+1}\geqslant 2\sigma^{2n+1}/2^{2n+1}$ . This inequality supports factorization Eq. (53) because it is specific for power averages: $[(x^{2n+1}+y^{2n+1})/2]^{1/(2n+1)}>(x+y)/2$ .

The first Eq. (54) demonstrates that $x^{2n+1}+y^{2n+1}$ is the volume of $2n+1$ -dimetsional right-angled parallelepiped. Its edges are $2n$ chords of two concentric circles of radii $x$ and $y$ starting from a point on the internal circle and going to $2n$ points of intersection of radii with the outer circle. The angles between radius of the initial point and the rest ones are $\pm(2k+1)\pi/(2n+1)$ . The last edge, $x+y$ , corresponds to the angle $\pi$ and may be considered as a height while the parallelepiped bases consists of $n$ pairs of identical edges.

12. Parameter for

\bm{{x}^{p}+{y}^{p}}

expansion

A direct calculation of $x^{p}+y^{p}$ in terms of $\rho$ and $\sigma$ for few first primes $p=2n+1$ leads to: $x^{3}+y^{3}=\sigma^{3}-3\sigma\rho,x^{5}+y^{5}=\sigma^{5}-5\sigma\rho(\sigma^{% 2}-\rho),x^{7}+y^{7}=\sigma^{7}-7\sigma\rho(\sigma^{2}-\rho)^{2},x^{11}+y^{11}% =\sigma^{11}-11\sigma\rho(\sigma^{2}-\rho)[(\sigma^{2}-\rho)^{3}+\sigma^{2}% \rho^{2}],x^{13}+y^{13}=\sigma^{13}-13\sigma\rho(\sigma^{2}-\rho)^{2}[(\sigma^% {2}-\rho)^{3}+2\sigma^{2}\rho^{2}],x^{17}+y^{17}=\sigma^{17}-17\sigma\rho(% \sigma^{2}-\rho)[(\sigma^{2}-\rho)^{6}+5\sigma^{2}\rho^{2}(\sigma^{2}-\rho)^{3% }+\sigma^{4}\rho^{4}],x^{19}+y^{19}=\sigma^{19}-19\sigma\rho(\sigma^{2}-\rho)^% {2}[(\sigma^{2}-\rho)^{6}+7\sigma^{2}\rho^{2}(\sigma^{2}-\rho)^{3}+3\sigma^{4}% \rho^{4}]\ldots$ The sum in brackets, in fact, depends on one parameter through two arguments: $\mu=\sigma^{2}/\rho$ and $(\mu-1)^{3}$ after its division by $\rho^{n-1}$ or $\rho^{n-2}$ . We shall show that all these clearly visible regularities have general character. First we prove that after separation $\sigma^{p}$ the remaining sum is divisible by $\sigma^{2}-\rho$ .

The residue of division of $x^{2n+1}+y^{2n+1}$ by $\sigma^{2}-\rho$ appears according to Bezout theorem after substitution $\rho$ for $\sigma^{2}$ in Eq. (11)

$\displaystyle\sum\limits_{k=0}^{n}{L_{n-k}^{(2n+1)}/4^{k}}=2\cos[(2n+1)\pi/3]=% 2\cos[(6m\pm 1)\pi/3]=1,$ $\displaystyle\sum\limits_{k=0}^{n-1}{L_{n-k}^{(2n+1)}/4^{k}}=\sum\limits_{k=0}% ^{n-1}{\left.{(-1)^{k}\frac{(n+k|2k+1)}{n-k}}\right|_{n=3m-1,3m}}=0.$ (55)

This result is identical for both possible kinds of prime numbers (except 3): $p=6m\pm 1,n=3m,n=3m-1$ . The final term of the sum (12) $L_{0}^{2n+1}/4^{n}=1$ according to the prescription after Eq. (10). Therefore if the first term $\sigma^{2n+1}$ is separated from the sum Eq. (11) the rest of it is divisible by $\sigma^{2}-\rho$ , and we have on account of the sum of progression $\sigma{}^{2k}/\rho^{k}-1=(\sigma^{2}/\rho-1)\sum_{j=0}^{k-1}\sigma^{2j}/\rho^{j}$ :

$\displaystyle x^{p}+y^{p}=\sigma^{p}+\sigma\rho^{n}\sum\limits_{k=1}^{n-1}{L_{% n-k}^{(2n+1)}/4^{k}(\frac{\sigma^{2k}}{\rho^{k}}-1)}=\sigma^{p}-p\sigma\rho^{n% -1}(\sigma^{2}-\rho)\sum\limits_{k=1}^{n-1}{\left.{(-1)^{n-k+1}\frac{(n+k|2k+1% )}{n-k}}\right|_{n=3m-1,n=3m}\sum\limits_{j=0}^{k-1}{\frac{\sigma^{2j}}{\rho^{% j}}}},$ (56)

The interchange of summation order allows casting $x^{p}+y^{p}$ in a form of a power expansion over $\mu$

$\displaystyle x^{p}+y^{p}=\sigma^{p}-p\sigma\rho^{n-1}(\sigma^{2}-\rho)\sum% \limits_{k=0}^{n-2}{\mu^{k}U_{k}},$ $\displaystyle U_{k}=\sum\limits_{j=0}^{k}{\left.{(-1)^{n-j}\frac{(n+j|2j+1)}{n% -j}}\right|}_{n=3m-1,n=3m},4^{k}L_{n-k}^{(2n+1)}=(-1)^{n}(U_{k}-U_{k-1}).$ (57)

where the property (12) has been used.

The same technique works for attempt to separate one more factor $\sigma^{2}-\rho$ .

$\displaystyle x^{p}+y^{p}=\sigma^{p}+p\sigma\rho^{n-1}(\sigma^{2}-\rho)\sum% \limits_{k=1}^{n-1}(-1)^{n-k}\frac{(n+k|2k+1)}{n-k}\left[\sum\limits_{j=1}^{k-% 1}{\left(\frac{\sigma^{2j}}{\rho^{j}}-1\right)}+k\right]=\sigma^{p}+p\sigma% \rho^{n}\left(\frac{\sigma^{2}}{\rho}-1\right)^{2}\sum\limits_{k=1}^{n-1}{(-1)% ^{n-k}\frac{(n+k|2k+1)}{n-k}\sum\limits_{j=0}^{k-2}{(k-j-1)\frac{\sigma^{2}}{% \rho}}}\quad∼{}+p\sigma\rho^{n-1}(\sigma^{2}-\rho)\sum\limits_{k=1}^{n-1}{(-1)% ^{n-k}\frac{(n+k|2k+1)}{n-k}k}.x^{p}+y^{p}=\sigma^{p}-p\sigma\rho^{n-2}(\sigma% ^{2}-\rho)^{2}\sum\limits_{k=0}^{n-1}{(-1)^{n-k+1}\frac{(n+k|2k+1)}{n-k}\sum% \limits_{j=0}^{k-2}{(k-j-1)\left.{\frac{\sigma^{2j}}{\rho^{j}}}\right|}}_{n=3m}.$ (58)

If the first sum in the second row vanishes the attempt would be successful since the factor $\sigma^{2}-\rho$ is extracted from each term of the internal sum from the first row. Presenting $k$ from the key sum as $k-n+n$ (which does not change it due to identity (12)) we transform it into $\sum_{k=0}^{n-1}(-1)^{n-k}(n+k|2k+1)$ .

To establish whether this sum is zero consider the Chebyshev polynomial of the second kind. Following the same technique, but dealing with the imaginary part of binomial power $(\cos\varphi+i\sin\varphi)^{2n+1}$ we obtain the expression for Chebyshev polynomial of the second kind. After substitution of $\pi/3$ for $\varphi$ and $3m$ for $n$ , the desirable 0 appears only for the first kind primes: $p=6m+1$ . Indeed,

$\displaystyle\sin(2n\varphi)/\sin\varphi=2\sum\limits_{k=0}^{n-1}{(-1)^{n-k+1}% 4^{k}(n+k|2k+1)\cos^{2k+1}\varphi},\sin(2\cdot 3m\pi/3)/\sin(\pi/3)=\sum% \limits_{k=0}^{n-1}{\left.{(-1)^{n-k+1}(n+k|2k+1)}\right|_{n=3m}=0}.$ (59)

The last regularity seen from the examples is the dependence of the remaining sum on a single parameter $\mu=\sigma^{2}/\rho$ through two arguments $\mu$ and $(\mu-1)^{3}$ . It will be true if the following equation has a proper solution for constants $M_{j}$ . The left part is taken from Eq. (12), the right one is an assumption suggested by examples.

$\displaystyle\sum\limits_{k=0}^{n-2}{\mu^{k}U_{k}}=\sum\limits_{j=0}^{m-1}{M_{% j}\mu^{j}(\mu-1)^{3(m-1-j)}},\mu=\sigma^{2}/\rho.$ (60)

It is clear beforehand that this is not a trivial problem because there are $n-1$ powers of $\mu$ on the left side of Eq. (60) and only $m$ constants $M_{j}$ . Let us check the free term on the right $M_{0}(-1)^{3(m-1)}=M_{0}(-1)^{n}$ , and from Eq. (12) $U_{0}=(-1)^{n}$ , therefore $M_{0}=1$ . The rest of $M_{j}$ should be found from

$\displaystyle\sum\limits_{k=0}^{n-2}{\mu^{k}U_{k}}=\sum\limits_{k=0}^{m-1}{M_{% k}\mu^{k}\sum\limits_{j=0}^{3(m-1-k)}{(3(m-1-k)|j)(-1)^{n-k-j}\mu^{j}}}=\sum% \limits_{k=0}^{m-1}{M_{k}\sum\limits_{j=k}^{n-2-2k}{(n-2-3k|j-k)(-1)^{n-j}\mu^% {j}}}.$

Here Newton’s formula for $(\mu-1)^{3(m-1-j)}$ has been used, and internal summation index has been changed $j\to j+k$ .

Now summation in the plane $(k,j)$ is taken over a triangle restricted by $j$ -axis and two straight lines: $j=k$ and $j=3(m-1)-2k$ . It consists of two contributions from two triangles obtained from the initial one by division by the height parallel to the $k$ -axes:

$\displaystyle\sum\limits_{j=0}^{n-2}{\mu^{j}U_{j}}=\sum\limits_{j=0}^{m-1}{(-1% )^{n-j}\mu^{j}\sum\limits_{k=0}^{j}{(n-2-3k|j-k)M_{k}}}\quad∼{}+\sum\limits_{j% =m}^{3(m-1)}{(-1)^{n-j}\mu^{j}\sum\limits_{k=j}^{[(n-j-1)/2]}{(n-2-3k|j-k)M_{k% }}}.$

The first contribution gives $m$ equations, the second one adds $2(m-1)$ more, in total $n-2$ .

$\displaystyle U_{j}=(-1)^{n-j}\sum\limits_{k=0}^{j}{(3(m-k-1)|j-k)}M_{k},j=0,% \ldots,m-1;$ $\displaystyle U_{j}=(-1)^{n-j}\sum\limits_{k=0}^{[(n-j-1)/2]}{(3(m-1-k)|j-k)M_% {k}},j=m,m+1,\ldots,n-2.$

The first set of linear equations should be used to find $m$ coefficients $M_{k}$ , and it is necessary to prove that the second set is also satisfied by the found $M_{k}$ . Equations can be simplified: intermediate quantities $U_{j}$ can be removed and substitute by coefficients of Chebyshev polynomials. For this, the difference of the above two adjacent equations is formed, the rule of addition of binomial coefficients and Eq. (12) are used giving

$\displaystyle\sum\limits_{k=0}^{j}{(n-3k-1|j-k)}M_{k}=(n+j|2j+1)/(n-j),j=0\div m% -1.$ $\displaystyle\sum\limits_{k=0}^{[(n-j-1)/2]}(n-1-3k|j-k)M_{k}=(n+j|2j+1)/(n-j)% ,j=m\div n-2.$ (61)

The first set of equations may be considered as a system of recurrent formulas for $M_{j}$ beginning from $M_{0}=1$ . Then each next equation contains only one new unknown. In particular, $M_{1}=(n-2|2)/3$ , and so on. After finding each next $M_{k}$ , the second set, if used in the backward order, allows to check $M_{k}$ by each next two equations. For example, $M_{0}$ is verified by the last equation: $(n-1|0)=(2n-1|0)$ and the preceding one $(n-1|1)=(2n-2|2)$ . Similarly $M_{1}=(n-2|2)/3$ is checked by two next preceding equations: $(n-1|2)+(n-2)(n-3)/6=(2n-3)|2)/3$ and $(n-1|3)+(n-4|1)(n-2)(n-3)/6=(2(n-2)|3)/4$ , etc.

It is worthwhile to note that $M_{1}$ for both kinds of primes, $n=3m$ and $n=3m-1$ , is integer because it is equal to: $(m-1)^{2}+m(m-1)/2$ and $(m-1)(m-2)+m(m-1)/2$ , respectively. At the same time, the third possible kind of odd numbers $6m+3$ , e.g. $n=7$ ( $p=15$ ), does not guarantee the integrity of $M_{1}:M_{1}=10/3$ . Next $M_{2}$ also can be found directly from Eq. (12) and has a similar form: $M_{2}=(n-3|4)/5$ . It demonstrates that only primality of $p$ guarantees the integrity of $M_{2}$ , e.g. $p=35$ has a proper form $6\cdot 6-1$ , but $M_{2}=(14|4)/5=1001/5$ . A similar peculiarity was observed in Eq. (10).

The considered examples suggest that the general solution for $M_{j}$ should be

$\displaystyle M_{j}=(n-j-1|2j)/(2j+1)=(n-j|2j+1)/(n-j),j=0,\ldots,m-1.$ (62)

This hypothesis will be justified if we prove two binomial identities which result from substitution of supposed $M_{j}$ into Eq. (12) on account of the equality $(n-3k-1|j-k)(n-k|2k+1)/(n-k)=(n-j|2k+1)(n-k-1|j-k)/(n-j)$ .

$\displaystyle\sum\limits_{k=0}^{j}{(n-j|2k+1)}(n-k-1|n-j-1)=(n+j|2j+1),j=0\div m% -1;$ $\displaystyle\sum\limits_{k=0}^{[(n-j-1)/2]}{(n-j|2k+1)}(n-k-1|n-j-1)=(n+j|2j+% 1),j=m\div n-2.$ (63)

Only the domain of external parameter $j$ and the upper sum limit differ in the second identity.

13. Proof of two obtained binomial identities (63) and final form of

\bm{{x}^{p}+y^{p}}

We begin from consideration of the identity in the second domain by introducing new parameters $n-j=2l+1$ if $n-j$ is odd and $n-j=2l+2$ if it is even, and $0\leqslant l\leqslant m-1$ according to Eq. (12). We consider in detail the odd case (even one is quite similar). Here the second binomial coefficient in the sum Eq. (12) is $(n-k-1)\ldots(n-k-2l)/(2l)!$ Hence, $l$ subsequent factors $(n-l-1)(n-l-2)\ldots(n-2l)$ are present in the numerator of each term: these are the first $l$ factors of the last, $l$ -th, term and last $l$ factors of the zero term. It is convenient to use the “falling” powers defined above in connection with Stirling numbers (7): $k_{m}=k(k-1)(k-2)\ldots(k-m+1)$ . The second Eq. (12) written in reverse order in new notation looks like

$\displaystyle(2(n-l)-1|2l+1)=\sum\limits_{k=0}^{l}(2l+1|2k+1)(n-1-k|2l)=[n-l-1% ]_{l}/(2l)!\sum\limits_{k=0}^{l}{(2l+1|2k+1)}[n-k-1]_{l-k}[n-2l-1]_{k}.$

The first of falling powers remained in $k$ -th term was situated before the taken out product, the second one after it. The binomial coefficient in the left part of this equality can also be prepared in a similar way: $(2(n-l)-1|2l)=[2(n-l)-1]_{l}^{(2)}[n-l-1]_{l}/((2l-1)!!l!$ . Here we introduced a superscript ${}^{(2)}$ which indicates the value of “jump” in the falling power, viz: $k_{m}^{(2)}=k(k-2)(k-4)\ldots(k-2m)$ . The first symbol in $(2(n-l)-1|2l)$ contains all odd factors of its numerator. We see that the “traces” of even factors are the same as the first product on the right. This allows removing completely all even factors from the equality being proved. Thus it remains to prove the identity for odd factors:

$\displaystyle\sum\limits_{j=0}^{l}{(2l+1|2j+1)}[n-j-1]_{l-j}[n-2l-1]_{j}=2^{l}% [2(n-l)-1]_{l}^{(2)}$ (64)

This equation maybe considered as an equality of two polynomials of $n$ of order $l$ with identical coefficients $4^{l}$ at the highest terms. The latter is evident on the right side. On the left one, it is a consequence of the identity $\sum_{i=0}^{l}(2l+1|2i+1)=4^{l}$ mentioned after Eq. (10). Hence, the task is reduced to prove the identity of roots of both polynomials. Those of the right side are seen directly: $n_{i}=l+i+1/2,i=0,\ldots,l-1$ . Let us prove that these $n_{i}$ turn into 0 the left part, too.

In order not to deal with all roots we shall use induction over $l$ : it is evident that the first root for a given $l$ becomes the zero root for the next $l+1$ , the second one becomes the first one, etc., the last, $l-1$ -th, will be the $l-2$ -th for $l+1$ . Then if all roots for $l$ are already found the induction hypothesis requires proving that $2l+1/2$ and $2l+3/2$ are $l-1$ -th and $l$ -th roots for $l+1$ . This is true for initial $l=2$ as is clear from Eq. (64) where the left part is equal to the right one: $5(n-1)(n-2)+10(n-2)(n-5)+(n-5)(n-6)=4(4^{2}-24n+35)=2^{2}(2n-5)(2n-7)$ . In the general case, substitution of the first root gives

$\displaystyle\sum\limits_{j=0}^{l+1}{(2l+3|2j+1)}[n-j-1]_{l+1-j}[n-2l-3]_{j}|_% {n\to 2l+1/2}$ $\displaystyle\quad=2^{-l-1}\sum\limits_{j=0}^{l+1}(-1)^{j}(2l+3|2j+1)(2(2l-j)+% 1)_{l+1-j}^{(2)}(2j+3)!!/3$ (65)

Here all factors in the second falling power from the right are written in the reverse order to keep falling power decreasing since all of them are negative. The last multiplier from the falling power is $2l+1$ . If we place $(2l-1)(2l-3)\ldots(2j+5)$ between it and $(2j+3)!!$ the double factorial $(4l-2j+1)!!$ appears. The introduced factors can be taken from the binomial coefficient after it presented as $(2l+3|2j+1)=(l+1|j)(2l+3)(2l+1)(2l-1)!!/[(2j+1)!!(2(l-j)+1)!!]$ . These products are cancelled with both doubled factorials in the dominator, and Eq. (13) takes the form

$\displaystyle(2l+3)(2l+1)[(l+1)/2^{l}3\sum\limits_{j=0}^{l+1}{(-1)^{j}(l|j-1)(% 2(2l-j)+1)_{l}^{(2)}}$ $\displaystyle\quad+1/2^{l+1}\sum\limits_{j=0}^{l+1}{(-1)^{j}(l+1|j)(2(2l-j)+1)% _{l}^{(2)}}].$ (66)

Similarly it is possible to verify that the result of substitution of $n\to 2l+3/2$ into Eq. (64) coincides with second sum from Eq. (13).

All terms of sums Eq. (13) contain identical $l$ -th falling power of different numbers $2(2l-j)+1$ and admit a sequence of one-type transformations. Each step diminishes the degree of power by 1 and the base of power by 2 and consists of application of the recurrent formula for binomial coefficients and change of the summation index $j-1\to j$ . This leads to elimination of the two first factors of the falling power as is seen, e.g. for the second sum:

$\displaystyle\sum\limits_{j=0}^{l+1}{(-1)^{j}(l+1|j)(2(2l-j)+1)_{l}^{(2)}}$ $\displaystyle\quad=\sum\limits_{j=0}^{l}(-1)^{j}(l|j)(2(2(l-j)+1)_{l}^{(2)}+% \sum\limits_{j=1}^{l+1}(-1)^{j}(l|j-1)](2(2(l-j)+1)_{l}^{(2)}$ $\displaystyle\quad=\sum\limits_{j=0}^{l}(-1)^{j}(l|j)(2(2l-j)+1)_{l}^{(2)}% \left(1-\frac{2(l-j)-1}{2(2l-j)+1}\right)$ $\displaystyle\quad=2(l+1)\sum\limits_{j=0}^{l}{(-1)^{j}(l|j)(2(2l-j)-1)_{l-1}^% {(2)}}=\ldots$ $\displaystyle\quad=2^{l}(l+1)!\sum\limits_{j=0}^{1}{(-1)^{j}(1|j)(2(l-j)+1)_{0% }^{(2)}=0}.$

The exponent turns into 0 after $l$ steps, the falling power becomes independent of the summation index, and as a result the sum disappears. It was not necessary to reach 0 degree; we could stop at any degree and calculate the sum explicitly, e.g. if $l=2$ then $\sum_{j=0}^{2}(-1)^{j}(2|j)((2l-2j)+1)_{1}^{(2)}=2l+1-2(2l-1)+2l-3=0$ . It is easy to demonstrate in the same manner that the first sum in Eq. (13) is also 0. This completely proves induction hypothesis and equality Eq. (64) in the case of the odd difference $n-j$ in the second identity Eq. (12).

The poof of even case is almost the same. The total number of common factors $(n-l-1)(n-l-2)\ldots(n-2l-1)$ of all terms of sum Eq. (13) now is $2l+1$ . This only slightly modifies being proved Eq. (64):

$\sum\limits_{j=0}^{l}{(2l+2|2j+1)}[n-j-1]_{l-j}[n-2l-2]_{j}=2^{l+1}[2(n-l-1)-1% ]_{l}^{(2)},$

and requires to show that substitution $n\to 2l+1/2$ into it annuls the left side which differs from Eq. (64) only by a constant nonzero factor. Thus to complete the proof of theorem, it remains only to confirm the first identity Eq. (12).

It is clear that the above proof cannot be applied to this case since the powers of polynomials of $n$ in each term on the left part of identity differ from each other and also differ from the power of the right part polynomial. However, if we separate the last term of sum $(n-j|2j+1)$ , the identity can be written as

$\displaystyle(n-j)(n-j-1)\ldots(n-3j)/(2j+1)!=(n+j|2j+1)-\sum\limits_{k=0}^{j-1}$ $\displaystyle\quad{(n-j|2k+1)(n-k-1|j-k)}$

Let us verify that roots of the left part $n=j$ , $n=j+1,\ldots,n=3j$ turn the right one into 0 as well. It is clearly seen for $n=j$ : all terms of the sum vanish since the first symbol vanishes, and the last factor of the $(n+j|2j+1)$ numerator, $n-j$ , also vanishes. The second root $n=j+1$ transforms $(n+j|2j+1)$ into $(2j+1|2j+1)=1$ , and all terms of the sum vanish, besides the one with k=0, because of the first symbol $(1|2k+1)$ . This term is also $(1|1)(k|k)=1$ . The right hand part for any other root of type ${n=j+i}$ takes the form

$(2j+l|2j+1)-\sum\limits_{k=0}^{[(l-1)/2]}{(l|2k+1)(l+j-k-1|j-k)}=0$

The upper limit of the sum, if it does not exceed $j-1$ , guarantees that $(l|2k+1)\neq 0$ since $l<2j$ . Now, it should be noted that dealing with the second identity Eq. (12) we have proved something more than it was needed, viz. that it is fulfilled identically relative to n. Substituting $n$ by $l+j$ in second identity Eq. (12) we transform it into the desired equality.

Thus it is proved that the set of constants $M_{j}$ Eq. (62) satisfies both systems of Eq. (12). A peculiarity of proved for this combinatorial identities Eq. (12) should be mentioned. In contrast to the most combinatorial identities [1, 6], they are valid not for all values of parameter $n$ , but only for those which lead to natural constants $M_{j}$ . The proper $n$ should satisfy the additional condition: $p=2n+1$ must be prime.

At last, we may present the second identity meant in the title of the article. Collecting Eqs (12), (60) and (62), we have after elementary transformations

$\displaystyle x^{p}+y^{p}=\sigma^{p}-p\sigma\rho\sum^{m-1}_{j=0}\frac{(n-j|2j+% 1)}{n-j}(\rho\sigma)^{2j}(\sigma^{2}-\rho)^{n-1-3j},$ (67)

Identity (67) looks identical for both kinds of prime numbers $p=2n+1=6m\pm 1$ ; only connection between $n$ and $m$ differs: $n=3m$ and $n=3m-1$ . As compared to Eq. (53), the number of terms in it is three times fewer.

14. Conclusion

Various connections between known mathematical quantities are considered. Some of them are apparently new. The most of these probably are: the summation Eqs (14), (18), identities for Bernoulli and for Stirling numbers Eqs (26) and (6), modified Hasse Eq. (7) and identities Eqs (37)–(40), (9) for zeta function of integer argument, factorization Eq. (54) based on Chebyshev polynomials, and binomial identities Eqs (12) and (67). Simple proofs of Abel binomial formula and von Staudt-Clausen theorem are given.

References

Riordan

, Combinatorial Identities, John Wiley, New York, 1968.

Gradshtein

I.S.

and Ryzhik

I.M.

, Tables of Integrals, Sums, Series, and Products, FM, Moscow, 1962.

Beyer

W.H.

, CRC Standard Mathematical Tables, Florida, 1981.

Mestechkin

M.M.

, JCMSE 13 (2013), 395.

Weisstein

E.W.

, Bernoulli number from a wolfram, Math World.

Riordan

, Introduction to Combinatorial Analysis, John Wiley, New York, 1958.

Weisstein

E.W.

, Stirling number of the second kind from a wolfram, Math World.

Hasse

, Math Zeitst 32 (1930), 458.

Borwein

J.M.

Bradley

D.M.

and Crandall

R.E.

, J Compt Appl Math 121 (2000), 247.

10.

Rumer

I.B.

and Ryvkin

M.S.

, Thermodynamics, Statistical Physics, and Kinetics, Science, Moscow: Mir, 1980.