Modified Newton identities

Abstract

The sums of powers with identical exponents of natural, real, or complex numbers, considered as roots of algebraic equation, are expressed directly through the products of coefficients of that equation, starting from the well-known Newton identities. The final Eq. (6) includes the same power of sum of all numbers $\pm$ a sum over all partitions of the exponent. Each term of the last sum is the equation coefficients product with the net power keeping the “dimensionality” of the exponent and having a numerical factor, equal to a proper polynomial coefficient, built of exponents of equation coefficients entering the product. The revers Eq. (43) for equation coefficients is also a sum over all partitions of the same exponent with known numerical coefficients. The entering products are built of “commutators-anticommutators of power of sum and sum of powers” (C-A) of the initial sum addends. The numerous identities Eq. (44) for a C-A with an exponent, exceeding the number of C-A sum terms by 2, and similar C-A-s with lower exponents are established.

Keywords

Finite sum of powers polynomial coefficient Newton’s identities

1. Introduction

This presentation continues a series of methodical articles addressed merely to beginning researchers in mathematical physics. We considered some mathematical aspects of the known problems, usually not attracting much attention, like magnetic field of ring current, dispersion of molecular higher hyperpolarizabilities and crystal susceptibilities [1], or mathematical problems of density matrix $N$ -representability. We also did not avoid more specific mathematical topics like linear Diophantine equations with $n$ unknowns, which have physical applications, chain fractions, and algorithm of solution of Pell equation without a random search, etc. [2]. Simultaneously we tried to add something new to these problems. This article follows the same line.

Classical Newton’s formulae for sums of powers of natural, real, or complex numbers, which are roots of some algebraic equation, are an important tool of algebra, number and matrix theory, and have numerous applications in physics. They are more flexible than, e.g., various formulas of the Faulhaber type [3] for sums of sequential integer powers expressed in terms of Bernoulli numbers [4], which could be removed from the final result [2]. Newton’s formulae could be applied in consideration of several unresolved conjectures in number theory like Euler’s sum of powers conjecture, Lander et al. conjecture [5], etc. Some inconvenience of Newton’s formulae for large exponents is the necessity to find sums with all preceding exponents to obtain the sum needed. The purpose of this article is to present the Newton formulae in a form, which does not require such preliminary computations. This goal may be achieved by a method used to obtain the identity, which has been deduced in Ref. [2] for integer $x$ , $y$ , and prime $p=2n+1$ with $n=3m$ or $n=3m-1$ :

$\displaystyle x^{p}+y^{p}={(x+y)}^{p}-\textit{pxy}(x+y)\sum^{m-1}_{j=0}\binom{% n-j}{2j+1}\frac{{(x^{2}y+xy^{2})}^{2j}({x^{2}+y^{2}+xy)}^{n-3j-1}}{n-j}$ (1)

The requirement for $p$ to be prime is essential because the sum should be integer. It has been proved that primality of $p$ guarantees coefficients’ integrity. This requirement is not superfluous, as it may seem on the first glance, since $p$ is explicitly not present in $\frac{\binom{n-j}{2j+1}}{n-j}$ . Indeed, e.g. $\frac{{\binom{7-1}{2\cdot 1+1}}}{7-1}=\frac{10}{3}:2\cdot 7+1=15$ is not prime.

Thus, the first problem to be solved is to remove these restrictions. The second task is more essential and difficult: to include into the left part of Eq. (1) any number of variables. We solve the second task in three steps: at first, we introduce three variables, and then few more, and finally we extend the left sum in Eq. (1) to any number of terms. We consider a lot of numerical examples in accordance with this deduction method used. All calculation details are shown explicitly in accordance with our methodical intensions.

2. Sum of natural powers of two real numbers

In fact, Eq. (1) suggests itself the natural way to generalize it for real $x$ and $y$ and any natural $p$ . If we introduce the following notation $\lambda=xy(x+y)$ , $\mu=x^{2}+y^{2}+xy$ , and $\sigma=x+y$ it is clear that $\sigma$ is connected to $\lambda$ and $\mu$ directly.

$\displaystyle{\sigma}^{3}-\mu\sigma-\lambda=0.$ (2)

It is important that this cubic equation has all three roots real, and these are $\sigma$ , $-x$ , and $-y$ . Therefore, the sum Eq. (1) may be treated as the sum of powers of roots of Eq. (2) and expressed through $\lambda$ and $\mu$ by means of Newton’s formulae. The symmetric sums of roots products $P_{1}=\sigma+(-x)+(-y)=0$ , $P_{2}=\sigma(-x)+\sigma(-y)+(-x)(-y)=-\sigma^{2}+\rho=-\mu$ , $P_{3}=\sigma(-x)(-y)=\sigma\rho=\lambda$ , allow to write the sum of a natural power of all roots:

$\displaystyle S_{t}\equiv{\sigma}^{t}+(-1)^{t}(x^{t}+y^{t})=\mu S_{t-2}+% \lambda S_{t-3}.$ (3)

Starting from $S_{1}=\sigma-x-y=0$ , $S_{2}=\sigma^{2}+x^{2}+y^{2}=2\sigma^{2}-2\rho=2\mu$ , $S_{3}=\sigma^{3}-x^{3}-y^{3}=\sigma^{3}-(x+y)^{3}+3\sigma\rho=3\lambda$ , we obtain by means of Eq. (3) $S_{4}=2\mu^{2}$ , $S_{5}=3\lambda\mu+2\lambda\mu=5\lambda\mu$ , $S_{6}=2\mu^{3}+3\lambda^{2}$ , $S_{7}=5\lambda\mu^{2}+2\lambda\mu^{2}=7\lambda\mu^{2}$ , $S_{8}=2\mu^{4}+8\mu\lambda^{2}$ , $S_{9}=9\lambda\mu^{3}+3\lambda^{3}$ , $S_{10}=2\mu^{5}+15\mu^{2}\lambda^{2}$ , $S_{11}=11\lambda\mu(\mu^{3}+\lambda^{2})$ , $S_{12}=2\mu^{6}+24\mu^{3}\lambda^{2}+3\lambda^{4}$ , $S_{13}=13\lambda\mu^{2}(\mu^{3}+2\lambda^{2})$ , etc. Numerical coefficients for prime $t$ are the same, which Eq. (1) gives. To finalize the general case it remains to find in it the expression for numerical coefficients and verify that those in Eq. (1) are particular cases for prime $t$ .

It is clear that the net degree of all $\lambda^{f}\mu^{g}$ -products in $S_{t}$ should guarantee the total degree of $x y$ -products to be equal to $t$ . It is convenient to consider the even $2t$ and odd $2t+1$ case separately. The Eq. (1) gives an example for the latter case. However, the last term in Eq. (3) mixes even and odd cases. Since the order of $\lambda$ is 3 and of $\mu$ is 2 the order of $\lambda^{f}\mu^{g}$ is $3f+2g=2t$ in the even case and $3f+2g=2t+1$ in the odd one. Proper monomials have forms $\lambda^{2j}\mu^{t-3j}$ and $\lambda^{2j+1}\mu^{t-3j-1}$ , respectively, because $f=2j$ , $g=t-3j$ mean $3f+2g=6j+2t-2\cdot 3j=2t$ and $f=2j+1$ , $g=t-3j-1$ mean $3f+2g=6j+3+2t-2\cdot 3j-2=2t+1$ . Thus, we can accept

$\displaystyle S_{2t}=\sum^{[t/3]}_{j=0}A^{(t)}_{j}{\lambda}^{2j}{\mu}^{t-3j},S% _{2t+1}=\sum^{[t/3]}_{j=0}B^{(t)}_{j}{\lambda}^{2j+1}{\mu}^{t-3j-1};$

summation runs over integers $j\geqslant 0$ , which guarantee the non-negativity of the exponent of $\mu$ . Substitution of these expressions into Eq. (3) gives the recurrent formulae for coefficients after equating coefficients at identical products. The expressions considered should be compared only with the sum in Eq. (1) since $\sigma^{t}$ satisfies Eq. (3) due to Eq. (2). Therefore, we have

$\displaystyle A^{(t)}_{j}=A^{(t-1)}_{j}+B^{(t-2)}_{j-1},B^{(t)}_{j}=B^{(t-1)}_% {j}+A^{(t-1)}_{j}.$ (4)

The above examples e.g. $A^{(4)}_{1}=8,A^{(5)}_{1}=15,A^{(6)}_{1}=24,B^{(4)}_{0}=9,B^{(5)}_{1}=11$ satisfy Eq. (4): $A^{(6)}_{1}=A^{(5)}_{1}{+B}^{(4)}_{0}$ , etc.

We can suppose ${B}^{(t)}_{j}=(2t+1)\binom{t-j}{2j+1}/(t-j)$ , which is also confirmed by the examples. Then the second recurrent formula gives a similar expression of

$\displaystyle A^{(t)}_{j}=B^{(t+1)}_{j}-B^{(t)}_{j}=(2t+3)\binom{t-j+1}{2j+1}/% (t+1-j)-\frac{(2t+1)\binom{t-j}{2j+1}}{t-j}=\binom{t-j}{2j+1}\left[\frac{2t+3}% {t-3j}-\frac{2t+1}{t-j}\right]=\binom{t-j}{2j+1}\frac{2t(2j+1)}{(t-3j)(t-j)}=2% t\binom{t-j}{2j}/(t-j).$

It is easy to verify that substitution of this formula for $A$ together with the accepted one for $B$ into the first Eq. (4) turns it into identity. Thus, we have extended Eq. (1) onto any real numbers $x$ and $y$ and onto all natural exponents:

$\displaystyle S_{2t}=2t\sum^{[t/3]}_{j=0}\frac{\binom{t-j}{2j}}{t-j}{\lambda}^% {2j}{\mu}^{t-3j},S_{2t+1}=(2t+1)\sum^{[t/3]}_{j=0}\frac{\binom{t-j}{2j+1}}{t-j% }{\lambda}^{2j+1}{\mu}^{t-1-3j}.$ (5)

The former more cumbersome and elaborated proof [2] of the particular case Eq. (1) remains essential as the guarantee to write the prime subscript of $S_{2t+1}$ as a factor before the sum, while only the products with inner coefficients of sums become integer for other odd (and even) multipliers.

3. Generalization onto sum of three powers

The fourth-degree equation (instead of Eq. (2)) lets to test the idea of using the basic parameters and their sum as the roots of equation as the initial point. Simultaneously we introduce subscripts to simplify the notation in a multi-parameter case. Thus, the equation and parameters are now:

$\displaystyle{\xi}^{4}-{\mu}_{2}{\xi}^{2}-{\mu}_{3}\xi-{\mu}_{4}=0,{\mu}_{2}=% \sigma^{2}-\tau,{\mu}_{3}=\sigma\tau-\rho,{\mu}_{4}=\sigma\rho,$

(6) $\displaystyle\sigma=x+y+z,\tau=xy+xz+yz.$

If $p_{j}$ are symmetric polynomials of $x$ , $y$ , $z$ then $p_{0}=1$ , $p_{1}=\sigma$ , $p_{2}=\tau$ , $p_{3}=\rho=\textit{xyz}$ . Note, $\xi^{3}$ is absent in Eq. (3). It is clear that $\sigma$ and $-x$ , $-y$ , $-z$ are the roots of Eq. (3). The coefficients $\mu_{j}$ are symmetric functions of all roots of Eq. (3) including $\sigma$ . Therefore, the Newton equations give

$\displaystyle S_{t}\equiv{\sigma}^{t}+(-1)^{t}(x^{t}+y^{t}+z^{t})={\mu}_{2}S_{% t-2}+{\mu}_{3}S_{t-3}+{\mu}_{4}S_{t-4}.$ (7)

Beginning from $t=1$ we have $S_{1}=\sigma-x-y-z=0$ , $S_{2}=\sigma^{2}+x^{2}+y^{2}+z^{2}=2\sigma^{2}-2\tau=2\mu_{2}$ , $S_{3}=\sigma^{3}-x^{3}-y^{3}-z^{3}=3\sigma\tau-3\rho+x^{3}+y^{3}+z^{3}-x^{3}-y% ^{3}-z^{3}=3\mu_{2}$ since $\sigma^{3}=(x+y+z)(x^{2}+y^{2}+z^{2}+2\tau)=x^{3}+y^{3}+z^{3}+2\sigma\tau+x(xy% +xz+yz)-\rho+y\tau-\rho+z\tau-\rho=x^{3}+y^{3}+z^{3}+3\sigma\tau-3\rho$ and also $S_{4}=\sigma^{4}+x^{4}+y^{4}+z^{4}=4\mu_{4}+2\mu_{2}^{2}$ is a result of $\sigma^{4}=(x^{2}+y^{2}+z^{2}+2\tau)^{2}=x^{4}+y^{4}+z^{4}+4\tau^{2}+4\tau(x^{% 2}+y^{2}+z^{2})+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})=x^{4}+y^{4}+z^{4}+4\tau^{2% }+4\tau(\sigma^{2}-2\tau)+2(\tau^{2}-2\sigma\rho)$ . We move further using Eq. (7):

$\displaystyle S_{5}=\mu_{2}3\mu_{3}+2\mu_{2}\mu_{3}=5\mu_{2}\mu_{3},S_{6}=6\mu% _{2}\mu_{4}+2\mu_{2}^{3}+3\mu_{3}^{2},$ $\displaystyle S_{7}=7\mu_{3}\mu_{2}^{2}+7\mu_{3}\mu_{4}=7\mu_{3}(\mu_{4}+\mu_{% 2}^{2}),S_{8}=4\mu_{4}^{2}+8\mu_{4}\mu_{2}^{2}+2\mu_{2}^{4}+8\mu_{2}\mu_{3}^{2},$ $\displaystyle S_{9}=18\mu_{2}\mu_{3}\mu_{4}+9\mu_{3}\mu_{2}^{3}+3\mu_{3}^{3},S% _{10}=10\mu_{3}^{2}\mu_{4}+10\mu_{2}\mu_{4}^{2}+10\mu_{4}\mu_{2}^{3}+2\mu_{2}^% {5}+15\mu_{2}^{2}\mu_{3}^{2},$ $\displaystyle S_{11}=11\mu_{3}\mu_{4}^{2}+33\mu_{3}\mu_{2}^{2}\mu_{4}+11\mu_{3% }\mu_{2}^{4}+11\mu_{2}\mu_{3}^{3},$ $\displaystyle S_{12}=4\mu_{4}^{3}+36\mu_{2}\mu_{3}^{2}\mu_{4}+18\mu_{2}^{2}\mu% _{4}^{2}+12\mu_{4}\mu_{2}^{4}+2\mu_{2}^{6}+24\mu_{2}^{3}\mu_{3}^{2}+3\mu_{3}^{% 4},$ $\displaystyle S_{13}=13\mu_{3}^{3}\mu_{4}+39\mu_{2}\mu_{3}\mu_{4}^{2}+52\mu_{3% }\mu_{2}^{3}\mu_{4}+13\mu_{3}\mu_{2}^{5}+26\mu_{2}^{2}\mu_{3}^{3},$

etc. This amount of samples is sufficient for attempting to construct a formula for numerical coefficients by checking various hypotheses. The solved two-parameter case and the Eq. (5) suggest that the sough-for formula should have a form of a sum over all partitions of $t$ .

$\displaystyle S_{t}(3)=\sum_{P,t=2i+3j+4k}{A_{P}}{\mu}^{i}_{2}{\mu}^{j}_{3}{% \mu}^{k}_{4},$ (8)

with $\mu_{j}$ defined in Eq. (3) as functions of $x$ , $y$ , and $z$ . To find the explicit expression of $A_{P}$ , it is reasonably to give before a general formulation of the problem.
4. Commutator of power of sum and sum of powers of integers with identical exponents

In the general case, we rewrite Eq. (7) as

$\displaystyle(\xi+x_{1})(\xi+x_{2})(\xi+x_{3})\dots(\xi+x_{n})(\xi-\sigma)$

(9) $\displaystyle\quad=\sum^{n+1}_{j=0}{\xi}^{n-j+1}(p_{j}-\sigma p_{j-1})=\xi^{n+% 1}-\sum^{n+1}_{j=2}{\mu}_{j}{\xi}^{n-j+1}=0,$

with notation ${\mu}_{j}=\sigma p_{j-1}-p_{j}$ , ${\mu}_{1}=0$ , $\sigma p_{n}={\mu}_{n+1}$ . The subscript $j$ at $\mu_{j}$ is the “dimensionality” of $\mu_{j}$ if that of each $x_{k}$ is 1. The total number of $\mu_{j}$ in Eq. (4) is $n$ ; $p_{j}$ are symmetric polynomials built of basic parameters $x_{k}$ .

The generalization of the Eq. (7) is

$\displaystyle S_{t}\equiv{\sigma}^{t}+(-1)^{t}{\sigma}_{t},{\sigma}_{t}=\sum^{% n}_{k=1}x^{t}_{k},{\sigma}_{1}\equiv\sigma,S_{1}=0.$ (10)

The title of this section is connected to the odd subscript case since then $S_{t}={\sigma}^{t}-{\sigma}_{t}$ . The term “anticommutator” may be used in the even case. We mean the sequence of two operations addition and involution acting on a set of $x_{k}$ .

To express $S_{t}$ with $t>n$ in terms of $\mu_{j}$ we apply Newton’s formulae related to Eq. (4) and entirely exclude $p_{j}$ due to their connection to $\mu_{j}$ and to the family of $S_{u}$ with $u\leqslant n$ by means of

$\displaystyle{\sigma}_{t}=\sum^{t-1}_{u=1}(-1)^{u-1}p_{u}{\sigma}_{t-u}+{(-1)}% ^{t-1}tp_{t},p_{u}={\sigma}^{u}-\sum^{u}_{j=2}{\mu}_{j}{\sigma}^{u-j},{\sigma}% _{u}=(-1)^{u}(S_{u}-{\sigma}^{u}).$ (11)

Now we can rewrite Eq. (11) leaving only $S_{u}$ and $\mu_{j}$ after insertion of the last two Eq. (11) into the first one.

$\displaystyle S_{t}={\sigma}^{t}-\sum^{t-1}_{u=1}\left({\sigma}^{t-u}-\sum^{t-% u}_{v=2}{\mu}_{v}{\sigma}^{t-u-v}\right)(S_{u}-{\sigma}^{u})-t\left({\sigma}^{% u}-\sum^{t}_{u=2}{\mu}_{u}{\sigma}^{t-u}\right)={\sigma}^{t}\!-\!\sum^{t-1}_{u% =1}{\sigma}^{t-u}S_{u}\!+\!\sum^{t-1}_{u=1}{\sigma}^{t}\!+\!\sum^{t-1}_{u=1}{S% _{u}}\sum^{t-u}_{v=2}{\mu}_{v}{\sigma}^{t-u-v}-\sum^{t-1}_{u=1}\sum^{t-u}_{v=2% }{\mu}_{v}{\sigma}^{t-v}-t\left({\sigma}^{u}-\sum^{t}_{u=2}{\mu}_{u}{\sigma}^{% t-u}\right)=\sum^{t-1}_{u=2}(t\mu_{u}-S_{u}){\sigma}^{t-u}+t{\mu}_{t}+\sum^{t-% 1}_{u=1}{S_{u}}\sum^{t-u}_{v=2}{\mu}_{v}{\sigma}^{t-u-v}-\sum^{t-2}_{u=1}{u\mu% }_{t-u}{\sigma}^{u}.$

We take into account that $S_{1}=0$ , which follows from the Eq. (10), and the identity

$\displaystyle\sum^{t-1}_{u=1}{S_{u}}\sum^{t-u}_{v=2}{\mu}_{v}{\sigma}^{t-u-v}=% \sum^{t}_{u=4}{\sigma}^{t-u}\sum^{u-2}_{v=2}{\mu}_{v}S_{u-v}$

appearing after changes of summation order and notation of subscripts. Finally, we came to the following equation with two first terms of the first sum separated.

$\displaystyle S_{t}=\sum^{t-1}_{u=4}{\sigma}^{t-u}\left(u{\mu}_{u}-S_{u}+\sum^% {u-2}_{v=2}{\mu}_{v}S_{u-v}\right){}+\sum^{t-2}_{u=2}{\mu}_{u}S_{t-u}+t{\mu}_{% t}+{\sigma}^{t-2}(2\mu_{2}-S_{2})+{\sigma}^{t-3}(3\mu_{3}-S_{3}).$ (12)

Just only the first term of these, $2\mu_{2}+(2\mu_{2}-S_{2})$ , remains from the whole expression for $t=2$ , and equation becomes “self-consistent” if $S_{2}=2\mu_{2}$ . Similarly, $S_{3}=3\mu_{3}$ on account of the last result and $S_{3}=3\mu_{3}+(2\mu_{3}-S_{3})$ . The sums appear in the equation starting from $t=4$ along with disappearing of both last terms if preceding results are used:

$\displaystyle S_{t}=\sum^{t-1}_{u=4}{\sigma}^{t-u}\left(u{\mu}_{u}-S_{u}+\sum^% {u-2}_{v=2}{\mu}_{v}S_{u-v}\right)+\sum^{t-2}_{u=2}{\mu}_{u}S_{t-u}+t{\mu}_{t}.$ (13)

Only the last sum remains for $t=4$ giving $S_{4}=\mu_{2}S_{2}+4\mu_{4}=2(\mu_{2})^{2}+4\mu_{4}$ . The first sum begins to work from $t=5:\sigma(4\mu_{4}-S_{4}+\mu_{2}S_{2})+\mu_{2}S_{3}+\mu_{3}S_{2}+5\mu_{5}=0$ and accounting the last three results gives $S_{5}=2\mu_{2}\mu_{3}+3\mu_{3}\mu_{2}+5\mu_{5}=5\mu_{2}\mu_{3}+5\mu_{5}$ .

The requirement of self-consistency simplifies the result in the general case, too. If we assume for all preceding $t$ cases $r=6,7,\ldots,t-1$ that

$\displaystyle S_{r}=\sum^{r-2}_{j=2}{\mu}_{j}S_{r-j}+r{\mu}_{r},$ (14)

then for $t$ it becomes valid, too, since for $t=2,3,4$ it has been already verified, and the induction hypothesis leads the first sum to vanish for the rest values of $t$ including $t$ itself. Then $S_{2}=2\mu_{2}$ , $S_{3}=3\mu_{3}$ , $S_{4}=2(\mu_{2})^{2}+4\mu_{4}$ independently of the number ( $n$ ) of basic parameters $x_{j}$ . With these values, Eq. (14) begins to work for $t\geqslant 4$ and allows continuing the set $S_{t}$ , which looks identically for any $n$ , as well as $\mu_{t}$ :

$\displaystyle S_{1}=0,S_{2}=2\mu_{2},S_{3}=3\mu_{3},S_{4}=2\mu_{2}^{2}+4\mu_{4% },S_{5}=5\mu_{2}\mu_{3}+5\mu_{5},$ $\displaystyle S_{6}=\mu_{2}S_{4}+\mu_{3}S_{3}+\mu_{4}S_{2}+6\mu_{6}=2(\mu_{2})% ^{3}+3(\mu_{3})^{2}+6\mu_{4}\mu_{2}+6\mu_{6},$ $\displaystyle S_{7}=\mu_{2}S_{5}+\mu_{3}S_{4}+\mu_{4}S_{3}+\mu_{5}S_{2}+7\mu_{% 7}=7(\mu_{2})^{2}\mu_{3}+7\mu_{3}\mu_{4}+7\mu_{2}\mu_{5}+7\mu_{7},$ $\displaystyle S_{8}=\mu_{2}S_{6}+\mu_{3}S_{5}+\mu_{4}S_{4}+\mu_{5}S_{3}+\mu_{6% }S_{2}+8\mu_{8}=2(\mu_{2})^{4}+8\mu_{2}(\mu_{3})^{2}+8\mu_{4}(\mu_{2})^{2}{}+8% \mu_{2}\mu_{6}+8\mu_{3}\mu_{5}+4(\mu_{4})^{2}+8\mu_{8},$ $\displaystyle 4\mu_{4}=S_{4}-(S_{2})^{2}/2,5\mu_{5}=S_{5}-5S_{2}S_{3},6\mu_{6}% =S_{6}+S_{2}^{3}/8-S_{3}^{2}/3-3S_{4}S_{2}/4,$ $\displaystyle 7\mu_{7}=S_{7}+7S_{2}^{2}S_{3}/24-7S_{3}S_{4}/12-7S_{2}S_{5}/10,$ $\displaystyle 8\mu_{8}=S_{8}+2S_{2}S_{3}^{2}/9+S_{4}S_{2}^{2}/4-2S_{2}S_{6}/3-% 8S_{3}S_{5}/15-S_{4}^{2}/4-S_{2}^{4}/3,$ (15)

etc. The direct and reverse formulae correspond to the same partition of the subscript at $S_{t}$ or $\mu_{t}$ .

Thus, the simplification of Newton’s formulae is due to inclusion of the sum of basic parameters into the set of roots of the initial equation, on which the definition of commutators-anticommutators (C-A) is based. The recurrent formulae are similar in both intervals of the exponent $t:n\geqslant t$ and $n<t$ . Only the last term $t\mu_{t}$ of Eq. (14) is absent and the total number of terms ( $n$ ) remains constant for all $t$ in the second interval.

The same Eq. (14) can be used either for determination of $S_{t}$ or of $\mu_{t}$ as is shown in Eq. (4). The first possibility should be chosen for construction of $\mu_{u}$ using $S_{u}$ built by definition from the given initial parameters $x_{j}$ Eq. (10). The second one allows to express C-A for $t>n$ through coefficients $\mu_{u}$ of Eq. (4) even if $x_{j}$ are not known.

$\displaystyle u{\mu}_{u}=S_{u}-\sum^{u-2}_{r=2}{\mu}_{r}S_{u-r},u=2,\dots,n+1;% S_{t}(n)=\sum^{n+1}_{u=2}{\mu}_{u}S_{t-u}(n),t=n+2,\dots(n\geqslant 2).$ (16)

The total number of nonzero $\mu_{u}$ for a problem based on $n$ parameters $x_{k}$ is $n:\mu_{2},\mu_{3},\ldots,\mu_{n+1}$ . The proper C-A is convenient to denote by $S_{t}(n)$ . The second equation is consistent with all $\mu_{t}=0$ if $t>n+1$ . The missed in the set Eq. (16) for $t=n+1$ follows directly from Eq. (4) and delivers the test identity ${S}_{n+1}(n)=\sum^{n+1}_{j=2}{\mu}_{j}S_{n-j+1},(S_{0}=n+1)$ .

There is one-to-one correspondence between partitions and terms of the sum in Eq. (4). A partition $2e_{2}+3e_{3}+\ldots+(n+1)e_{n+1}=t$ leads to a typical product ${\mu}^{e_{2}}_{2}{\mu}^{e_{3}}_{3}\dots{\mu}^{e_{n+1}}_{n+1}$ . As some $e_{j}$ could be zero, the real number of factors may be lower than $n$ in some terms. For example, solutions of equation $2e_{2}+3e_{3}+4e_{4}+5e_{5}+6e_{6}=8$ are 1) $e_{2}=4$ , 2) $e_{2}=1$ , $e_{3}=2$ , 3) $e_{2}=2$ , $e_{4}=1$ , 4) $e_{2}=e_{6}=1$ , 5) $e_{3}=e_{5}=1$ , 6) $e_{4}=2$ that corresponds to the 4-th row of Eq. (4). Clearly, two integers: $n$ , the quantity of the basic parameters $x_{k}$ , and the exponent $t$ determine C-A, and our final goal is to obtain these dependencies in the explicit form.
5. More examples

“The conservation of dimensionality” remains two free integers in the expansion Eq. (8) of $S_{t}(3)$ .

$\displaystyle S_{2t}(3)=\sum^{[t/3]}_{j=0}\sum^{[t/2]}_{k=0}A^{(t)}_{j,k}{\mu}% ^{t-3j-2k}_{2}{\mu}^{2j}_{3}{\mu}^{k}_{4},$

(17) $\displaystyle S_{2t+1}(3)=\sum^{[t/3]}_{j=0}\sum^{[t/2]}_{k=0}B^{(t)}_{j,k}{% \mu}^{t-3j-2k-1}_{2}{\mu}^{2j+1}_{3}{\mu}^{k}_{4},t=2,\dots$

The result Eq. (5) of the two-parameter case suggests a separate consideration of the even and odd exponents. Substituting these expressions into the second recurrent Eq. (16), which looks like

$\displaystyle S_{2t}(3)={\mu}_{2}S_{2(t-1)}(3)+{\mu}_{3}S_{2(t-2)+1}(3)+{\mu}_% {4}S_{2(t-2)}(3),$ (18) $\displaystyle S_{2t+1}(3)={\mu}_{2}S_{2(t-1)+1}(3)+{\mu}_{3}S_{2(t-1)}(3)+{\mu% }_{4}S_{2(t-2)+1}(3),$

and comparing the factors at identical monomials in both parts of Eq. (5) gives recurrent relations for coefficients

$\displaystyle A^{(t)}_{j,k}=A^{(t-1)}_{j,k}+B^{(t-2)}_{j-1,k}+A^{(t-2)}_{j,k-1% },B^{(t)}_{j,k}=B^{(t-1)}_{j,k}+A^{(t-1)}_{j,k}+B^{(t-2)}_{j,k-1}.$ (19)

The known explicit expressions Eq. (5) of $A^{(t)}_{j,0}=2t\binom{t-j}{2j}/(t-j)$ and $B^{(t)}_{j,0}=(2t+1)\binom{t-j}{2j+1}/(t-j)$ are particular cases of similar double-subscript quantities. The above after Eq. (7) several explicit formulae of $S_{2t}(3)$ and $S_{2t+1}(3)$ give some coefficients with $k\neq 0$ , to which we adjoined few more:

$\displaystyle B^{(3)}_{0,1}=7,A^{(4)}_{0,1}=8,A^{(4)}_{1,1}=8,A^{(4)}_{0,2}=4,% B^{(4)}_{0,1}=18,A^{(5)}_{2,1}=10,A^{(5)}_{0,2}=10,A^{(5)}_{0,1}=10,$ $\displaystyle B^{(5)}_{0,1}=33,B^{(5)}_{0,2}=11,A^{(6)}_{0,1}=12,A^{(6)}_{0,2}% =60,{A}^{(9)}_{2,1}=90,{A}^{(9)}_{0,4}=18,A^{(9)}_{1,3}=36,B^{(9)}_{0,2}=285,$ $\displaystyle B^{(9)}_{0,1}=133,B^{(9)}_{1,1}=380,B^{(9)}_{0,3}=190,B^{(9)}_{1% ,2}=190,B^{(9)}_{2,1}=19,$

etc.

Looking at these results we feel the presence of the same factors $2t$ and $2t+1$ as in the case $k=0$ . Then excluding these factors we have

$\displaystyle A^{(4)}_{0,1}/8=1,A^{(4)}_{0,2}/8=1,A^{(5)}_{0,1}/10=1,A^{(5)}_{% 0,2}/10=1,A^{(6)}_{0,1}/12=1,A^{(6)}_{0,2}/12=3/2,A^{(6)}_{1,1}/12=3,$ $\displaystyle A^{(6)}_{0,3}/12=1/3,A^{(9)}_{0,1}/18=1,A^{(9)}_{1,2}/18=15,A^{(% 9)}_{0,2}/18=3,A^{(9)}_{2,1}/18=15,$

etc. The remaining quantity for $j>0$ we can describe by $\binom{t-j-k}{2j+k}/(t-j-k)$ corrected by an additional multiplier, following the similarity with $k=0$ case:

$\displaystyle S_{2t}(3)=2t\sum^{[t/2]}_{k=0}\sum^{[t/3]}_{j=0}\binom{2j+k}{k}% \binom{t-j-k}{2j+k}{\mu}^{t-3j-2k}_{2}{\mu}^{2j}_{3}\mu^{k}_{4}/(t-j-k),$ (20)

Indeed,

$\displaystyle\frac{A^{(9)}_{2,1}}{18}=\frac{\binom{2\cdot 2+1}{1}\binom{9-1-1}% {2\cdot 2+1}}{9-1-1}=5\cdot\frac{21}{7}=15,\frac{A^{(6)}_{1,1}}{12}=\frac{% \binom{2\cdot 1+1}{1}\binom{6-1-1}{2\cdot 1+1}}{6-1-1}=\frac{3\cdot 4}{4}=3,$ $\displaystyle\frac{A^{(6)}_{0,3}}{12}=\frac{\binom{6-3}{3}}{6-3}=\frac{1}{3},% \frac{A^{(9)}_{0,1}}{18}=\frac{\binom{9-1}{1}}{9-1}=1,$

etc.

The situation for $B$ -coefficients is similar

$\displaystyle S_{2t+1}(3)=(2t+1)\sum^{[t/2]}_{k=0}\sum^{[t/3]}_{j=0}\binom{2j+% k+1}{k}\binom{t-j-k}{2j+k+1}{\mu}^{t-3j-2k-1}_{2}{\mu}^{2j+1}_{3}\mu^{k}_{4}/(% t-j-k),$ (21)

and the examples

$\displaystyle\frac{B^{(4)}_{0,1}}{9}=\frac{\binom{1+1}{1}\binom{4-1}{2}}{4-1}=% \frac{2\cdot 3}{3}=2,\frac{B^{(6)}_{1,1}}{13}=\frac{\binom{2\cdot 1+1+1}{1}% \binom{6-1-1}{4}}{6-1-1}=\frac{4\cdot 4}{4}=4,$ $\displaystyle\frac{B^{(5)}_{0,1}}{11}=\frac{\binom{1+1}{1}\binom{5-1}{2}}{5-1}% =\frac{2\cdot 6}{4}=3,\frac{B^{(9)}_{0,1}}{19}=\frac{\binom{1+1}{1}\binom{9-1}% {2}}{9-1}=\frac{2\cdot 28}{8}=7$

also show the complete agreement with the above list.

To give a proof of Eqs (20) and (21) it is necessary to demonstrate that coefficients

$\displaystyle A^{(t)}_{j,k}=2t\binom{2j+k}{k}\binom{t-j-k}{2j+k}/(t-j-k),$ (22) $\displaystyle B^{(t)}_{j,k}=(2t+1)\binom{2j+k+1}{k}\binom{t-j-k}{2j+k+1}/(t-j-k)$

obey the recurrent relations Eq. (19). For this, we simply calculate the right-hand part.

$\displaystyle A^{(t-2)}_{j,k-1}+B^{(t-2)}_{j-1,k}+A^{(t-1)}_{j,k}=\frac{2(t-2)% \binom{2j+k-1}{k-1}\binom{t-j-k-1}{2j+k-1}}{t-j-k-1}+\frac{(2t-3)\binom{2j+k-1% }{k}\binom{t-j-k-1}{2j-1+k}}{t-j-k-1}$ $\displaystyle\quad+\frac{2(t-1)\binom{2j+k}{k}\binom{t-j-k-1}{2j+k}}{t-j-k-1}=% \binom{t-j-k-1}{2j+k-1}\cdot\binom{2j+k-1}{k-1}$ $\displaystyle\qquad\frac{2(t-2)k+(2t-3)2j+2(t-1)(t-3j-2k)}{(t-j-k-1)k}\frac{% \binom{t-j-k-1}{2j+k-1}\binom{2j+k-1}{k-1}2t}{k}$ $\displaystyle\quad=\frac{2t\binom{t-j-k}{2j+k}\binom{2j+k}{k}}{t-j-k}=A^{(t)}_% {j,k}.$

The proof for $B$ -coefficients does not differ from that for $A$ .

$\displaystyle B^{(t-2)}_{j,k-1}+A^{(t-1)}_{j,k}+B^{(t-1)}_{j,k}=\frac{(2t-3)% \binom{2j+k}{k-1}\binom{t-j-k-1}{2j+k}}{t-j-k-1}+\frac{(2t-2)\binom{2j+k}{k}% \binom{t-j-k-1}{2j+k}}{t-j-k-1}$ $\displaystyle\quad+\frac{(2t-1)\binom{2j+k+1}{k}\binom{t-j-k-1}{2j+k+1}}{t-j-k% -1}=\frac{\binom{2j+k}{k-1}\binom{t-j-k-1}{2j+k}}{t-j-k-1}$ $\displaystyle\qquad\left[(2t-3)+(2t-2)\frac{2j+1}{k}+(2t-1)\frac{t-3j-2k-1}{k}\right]$ $\displaystyle\quad=\frac{\binom{2j+k}{k-1}\binom{t-j-k-1}{2j+k}}{k(t-j-k-1)}(2% t+1)(t-j-k-1)=B^{(t)}_{j,k}.$

To comprehend complications caused by the increase of the number of defining C-A parameters and to find the form of higher coefficients in the general recurrent problem Eq. (16) we need to overcome few next steps. Similarly to Eq. (5), we write for $n=4$ .

$\displaystyle S_{2t}(4)=\sum^{[t/5]}_{l=0}\sum^{[t/4]}_{k=0}\sum^{[t/3]}_{j=0}% A^{(t)}_{l,k,j}{\mu}^{l}_{5}{\mu}^{k}_{4}{\mu}^{2j-l}_{3}{\mu}^{t-3j-2k-l}_{2},$ (23) $\displaystyle S_{2t+1}(4)=\sum^{[t/5]}_{l=0}\sum^{[t/4]}_{k=0}\sum^{[t/3]}_{j=% 0}B^{(t)}_{l,k,j}{\mu}^{l}_{5}{\mu}^{k}_{4}{\mu}^{2j-l+1}_{3}{\mu}^{t-3j-2k-l-% 1}_{2}$

The recurrent relation Eq. (5) now includes $S_{t-5}$ and takes the form

$\displaystyle S_{2t}(4)={\mu}_{2}S_{2(t-1)}(4)+{\mu}_{3}S_{2(t-2)+1}(4)+{\mu}_% {4}S_{2(t-2)}(4)+{\mu}_{5}S_{2(t-3)+1}(4)$ (24) $\displaystyle S_{2t+1}(4)={\mu}_{2}S_{2(t-1)+1}(4)+{\mu}_{3}S_{2(t-1)}(4)+{\mu% }_{4}S_{2(t-2)+1}(4)+{\mu}_{5}S_{2(t-2)}(4)$

The corresponding recurrent equations in terms of coefficients look like

$\displaystyle A^{(t)}_{j,k,l}=A^{(t-1)}_{j,k,l}+B^{(t-2)}_{j-1,k,l}+A^{(t-2)}_% {j,k-1,l}+B^{(t-3)}_{j-1,k,l-1},$ (25) $\displaystyle B^{(t)}_{j,k,l}=B^{(t-1)}_{j,k,l}+A^{(t-1)}_{j,k,l}+B^{(t-2)}_{j% ,k-1,l}+A^{(t-2)}_{j,k,l-1}$

It is necessary to bear in mind that the third subscript $j$ is not the exponent $e_{3}$ of $\mu_{3}$ , being connected to it by $e_{3}=2j-l$ in the even order case and by $e_{3}=2j-l+1$ in the odd case, where $l$ is the exponent of $\mu_{5}$ . We also notice that exponent of $\mu_{2}$ is not present among the subscripts in Eq. (5).

The review of the equations for $S_{t}(2)$ Eq. (5) and $S_{t}(3)$ Eqs (20) and (21) seems suggesting a hypothesis that the appearance of each new $\mu$ -factor is accompanied by subtracting its exponent from the upper symbol of the main binomial coefficient and from the denominator and also by joining one more binomial coefficient. The upper symbol of the last binomial coefficient is the same as the lower symbol of the preceding binomial coefficient, which is the sum of all $\mu$ -exponents, except of that of $\mu_{2}$ . The lower symbol equals to the sum of all remaining exponents. Thus, the specific form of this supposition is

$\displaystyle S_{2t}(4)=2t\sum^{[t/3]}_{j=0}\sum^{[t/2]}_{k=0}\sum^{[t/5]}_{l=% 0}\binom{t-j-k-l}{2j+k}\binom{2j+k}{k+l}\binom{k+l}{l}{\mu}^{t-3j-2k-l}_{2}{% \mu}^{2j-l}_{3}\mu^{k}_{4}{\mu}^{l}_{5}/(t-j-k-l),$ (26) $\displaystyle S_{2t+1}(4)=(2t+1)\sum^{[t/3]}_{j=0}\sum^{[t/2]}_{k=0}\sum^{[t/5% ]}_{l=0}\binom{t-j-k-l}{2j+k+1}\binom{2j+k+1}{l+k}\binom{l+k}{l}{\mu}^{t-3j-2k% -l-1}_{2}{\mu}^{2j+1-l}_{3}\mu^{k}_{4}{\mu}^{l}_{5}/(t-j-k-l)$ (27)

In terms of coefficients, it means

$\displaystyle A^{(t)}_{j,k,l}=2t\binom{t-j-k-l}{2j+k}\binom{2j+k}{k+l}\binom{k% +l}{l}/(t-j-k-l)=2t\frac{(t-j-k-l|2j-l,k,l)}{t-j-k-l}=\frac{2t(t-j-k-l-1)!}{(t% -3j-2k-l)!(2j-l)!k!l!},$ (28) $\displaystyle B^{(t)}_{j,k,l}=(2t+1)\binom{t-j-k-l}{2j+k+1}\binom{2j+k+1}{k+l}% \binom{k+l}{l}/(t-j-k-l)=(2t+1)\frac{(t-j-k-l|2j-l+1,k,l)}{t-j-k-l}=\frac{(2t+% 1)(t-j-k-l-1)!}{(t-3j-2k-l-1)!(2j-l+1)!k!l!}$

To estimate this hypothesis validity we, as before, verity directly by means of Eq. (5) that

$\displaystyle S_{15}(4)=S_{15}(3)+15\mu_{2}^{5}\mu_{5}+60\mu_{2}^{3}\mu_{4}\mu% _{5}+45\mu_{2}\mu_{3}\mu_{5}^{2}+45\mu_{2}\mu_{4}^{2}\mu_{5}+45\mu_{3}^{2}\mu_% {4}\mu_{5}+5\mu_{5}^{3},$ $\displaystyle S_{16}(4)=S_{16}(3)+80\mu_{2}^{4}\mu_{3}\mu_{5}+192\mu_{2}^{2}% \mu_{3}\mu_{4}\mu_{5}+24\mu_{3}^{2}\mu_{5}^{2}+32\mu_{2}^{3}\mu_{5}^{2}+64\mu_% {2}\mu_{3}^{3}\mu_{5}{}+48\mu_{2}\mu_{4}\mu_{5}^{2}+48\mu_{3}\mu_{4}^{2}\mu_{5},$ $\displaystyle S_{17}(4)=S_{17}(3)+17\mu_{2}^{6}\mu_{5}+85\mu_{2}^{4}\mu_{4}\mu% _{5}+170\mu_{2}^{3}\mu_{3}^{2}\mu_{5}+102\mu_{2}^{3}\mu_{3}\mu_{5}^{2}+102\mu_% {2}^{2}\mu_{4}^{2}\mu_{5}+17\mu_{3}^{4}\mu_{5}{}+204\mu_{2}\mu_{3}^{2}\mu_{4}% \mu_{5}+17\mu_{2}\mu_{5}^{3}+51\mu_{3}\mu_{4}\mu_{5}^{2}+17\mu_{4}^{3}\mu_{5}.$

These data give some coefficients Eq. (28) at $\mu_{5}$ :

$\displaystyle A^{(8)}_{1,0,1}=80,{A}^{(8)}_{1,1,1}=192,A^{(8)}_{1,0,2}=32,A^{(% 8)}_{2,0,1}=64,A^{(8)}_{1,1,2}=48,A^{(8)}_{1,2,1}=48,A^{(8)}_{2,0,2}=24;$ $\displaystyle B^{(7)}_{0,0,1}=15,B^{(7)}_{0,1.1}=60,B^{(7)}_{1,0,2}=45,B^{(7)}% _{0,2,1}=45,B^{(7)}_{1,1,1}=45,B^{(7)}_{1,0,3}=5;$ $\displaystyle B^{(8)}_{0,0,1}=17,B^{(8)}_{0,1,1}=85,B^{(8)}_{1,0,1}=170,B^{(8)% }_{1,0,2}=102,B^{(8)}_{2,0,1}=17,B^{(8)}_{1,1,1}=204,B^{(8)}_{1,0,3}=17,$ $\displaystyle B^{(8)}_{1,1,2}=51,B^{(8)}_{0,3,1}=17,$

which remain good chances for the hypothesis:

$\displaystyle A^{(8)}_{1,0,1}=16\frac{15\cdot 2\cdot 1}{6}=80,A^{(8)}_{1,1,2}=% 16\frac{4\cdot 1\cdot 3}{4}=48,A^{(8)}_{1,2,1}=16\frac{1\cdot 4\cdot 3}{4}=48,$ $\displaystyle{A}^{(8)}_{2,0,2}=16\frac{1\cdot 6\cdot 1}{4}=24,A^{(8)}_{1,0,2}=% 16\frac{10\cdot 1\cdot 1}{5}=32,A^{(8)}_{1,1,1}=16\frac{10\cdot 3\cdot 2}{5}=1% 92,$ $\displaystyle A^{(8)}_{2,0,1}=16\frac{5\cdot 4\cdot 1}{5}=64,B^{(7)}_{0,0,1}=1% 5\frac{6\cdot 1\cdot 1}{6}=15,B^{(7)}_{0,1,1}=15\frac{10\cdot 1\cdot 2}{5}=60,$ $\displaystyle B^{(7)}_{1,0,2}=15\frac{4\cdot 3\cdot 1}{4}=45,B^{(7)}_{0,2,1}=1% 5\frac{4\cdot 1\cdot 3}{4}=45,B^{(7)}_{1,1,1}=15\frac{1\cdot 6\cdot 2}{4}=45,$ $\displaystyle B^{(7)}_{1,0,3}=15\frac{1\cdot 1\cdot 1}{3}=5,B^{(8)}_{0,0,1}=17% \frac{7\cdot 1\cdot 1}{7}=17,B^{(8)}_{0,1,1}=17\frac{15\cdot 1\cdot 2}{6}=85,$ $\displaystyle B^{(8)}_{1,0,1}=17\frac{20\cdot 3\cdot 1}{6}=170,B^{(8)}_{1,0,2}% =17\frac{10\cdot 3\cdot 1}{5}=102,B^{(8)}_{2,0,1}=17\frac{1\cdot 5\cdot 1}{5}=% 17,$ $\displaystyle{B}^{(8)}_{1,1,1}=17\frac{5\cdot 6\cdot 2}{5}=204,{B}^{(8)}_{1,0,% 3}=17\frac{4\cdot 1\cdot 1}{4}=17,B^{(8)}_{1,1,2}=17\frac{1\cdot 4\cdot 3}{4}=% 51,$

etc.

Moreover, since the product of these binomial coefficients is a polynomial coefficient, built of all exponents, the simplicity of the final form makes it more plausible. The symbol $(a|b,c,\dots,f)=\frac{a!}{b!c!\dots f!(a-b-c-\dots-f)!}$ for polynomial coefficient was used in Eq. (28). The difference $a-b-c-\ldots-f$ is present in the factorial denominator although it is not present in the symbol. Computing the right-hand part of Eq. (5) gives

$\displaystyle 2(t-1)\frac{(t-j-k-l-1|2j-l,k,l)}{t-j-k-l-1}+(2t-3)\frac{(t-j-k-% l-1|2j-l-1,k,l)}{t-j-k-l-1}$ $\displaystyle\quad{}+2(t-2)\frac{(t-j-k-l-1|2j-l,k-1,l)}{t-j-k-l-1}+(2t-5)% \frac{(t-j-k-l-1|2j-l,k,l-1)}{t-j-k-l-1}$ $\displaystyle\quad=\frac{(t-j-k-l-2)!}{(t-3j-2k-l)!(2j-l)!k!l!}$ $\displaystyle\qquad(2(t-1)(t-3j-2k-l)+(2t-3)(2j-l)+2(t-2)k+(2t-5)l)$ $\displaystyle\quad=\frac{(t-j-k-l-2)!}{(t-3j-2k-l)!(2j-l)!k!l!}(2t^{2}-2tj-2tk% -2tl-2t)$ $\displaystyle\quad=\frac{2t(t-j-k-l)!}{(t-j-k-l)(t-3j-2k-l)!(2j-l)!k!l!}=2t% \frac{(t-j-k-l|2j-l,k,l)}{t-j-k-l}=A^{(t)}_{j,k,l}.$

Similarly we confirm the expression of $B(28)$ :

$\displaystyle(2t-1)\frac{(t-j-k-l-1|2j-l+1,k,l)}{t-j-k-l-1}+(2t-2)\frac{(t-j-k% -l-1|2j-l,k,l)}{t-j-k-l-1}$ $\displaystyle\quad{}+(2t-3)\frac{(t-j-k-l-1|2j-l+1,k-1,l)}{t-j-k-l-1}$ $\displaystyle\quad{}+(2t-4)\frac{(t-j-k-l-1|2j-l+1,k,l)}{t-j-k-l-1}$ $\displaystyle\quad=\frac{(t-j-k-l-2)!}{(t-3j-2k-l-1)!(2j-l+1)!k!l!}$ $\displaystyle\qquad[(2t-1)(t-3j-2k-l-1)+(2t-2)(2j-l+1)+(2t-3)k+(2t-4)l]$ $\displaystyle\quad=\frac{(t-j-k-l-2)!}{(t-3j-2k-l-1)!(2j-l+1)!k!l!}[2t(t-j-k-l% -1)+t-j-k-l-1]$ $\displaystyle\quad=\frac{(2t+1)(t-j-k-l-1)!}{(t-3j-2k-l-1)!(2j-l+1)!k!l!}.$

The procedure detail becomes clearer after the next step. The specific of all preceding cases is the presence of factorials of all exponents in the denominator, and hence of the factorial of their sum in the nominator. It suggests

$\displaystyle S_{2t}(5)=2t\sum^{[t/3]}_{j=0}\sum^{[t/2]}_{k=0}\sum^{[t/5]}_{l=% 0}\sum^{[t/6]}_{m=0}\frac{(t-j-k-l-2m|2j-l,k,l,m)}{t-j-k-l-2m}{\mu}^{t-3j-2k-l% -3m}_{2}{\mu}^{2j-l}_{3}\mu^{k}_{4}{\mu}^{l}_{5}{\mu}^{m}_{6},$ (29) $\displaystyle S_{2t+1}(5)=(2t+1)\sum^{[t/3]}_{j=0}\sum^{[t/2]}_{k=0}\sum^{[t/5% ]}_{l=0}\sum^{[t/6]}_{m=0}\frac{(t-j-k-l-2m|2j-l+1,k,l,m)}{t-j-k-l-2m}{\mu}^{t% -3j-2k-l-3m-1}_{2}{\mu}^{2j+1-l}_{3}\mu^{k}_{4}{\mu}^{l}_{5}{\mu}^{m}_{6}$ (30)

Summations go over all nonnegative integers $j, k, l, m$ , which belong to the hyperplane $2i^{\prime}+3j^{\prime}+4k+5l+6m=2t$ (or $2t+1$ for Eq. (30)), in which $j^{\prime}$ is substituted by the exponent of $\mu_{3}$ and $i^{\prime}$ by the exponent of $\mu_{2}$ from Eqs (29) and (30). Coefficients in Eqs (29) and (30)

$\displaystyle A^{(t)}_{j,k,l,m}=\frac{2t(t-j-k-l-2m|2j-l,k,l,m)}{t-j-k-l-2m}=% \frac{2t(t-j-k-l-2m-1)!}{(t-3j-2k-l-3m)!(2j-l)!k!l!m!},$ $\displaystyle B^{(t)}_{j,k,l,m}=\frac{(2t+1)(t-j-k-l-2m|2j-l+1,k,l,m)}{t-j-k-l% -2m}=\frac{(2t+1)(t-j-k-l-2m-1)!}{(t-3j-2k-l-3m-1)!(2j-l+1)!k!l!m!}$ (31)

must obey recurrent relations Eq. (5), which follow from properly prolonged Eq. (5).

$\displaystyle A^{(t)}_{j,k,l,m}=A^{(t-1)}_{j,k,l,m}+B^{(t-2)}_{j-1,k,l,m}+A^{(% t-2)}_{j,k-1,l,m}+B^{(t-3)}_{j-1,k,l-1,m}+A^{(t-3)}_{j,k,l,m-1},$ (32) $\displaystyle B^{(t)}_{j,k,l,m}=B^{(t-1)}_{j,k,l,m}+A^{(t-1)}_{j,k,l,m}+B^{(t-% 2)}_{j,k-1,l,m}+A^{(t-2)}_{j,k,l-1,m}+B^{(t-3)}_{j,k,l,m-1}$

Indeed, substituting Eqs (29) and (30) into basic recurrent Eq. (16), we should find in the right-hand internal sum all terms with the same monomial as in Eqs (29) and (30). The factor, multiplied by a proper $\mu$ , should have a coefficient with subscript reduced by 1. Then the net exponent will be the same as in the left-hand part. It is important that the exponent of $\mu_{2}$ automatically will be identical to that on the left at $A$ since the decrease of $t$ by 1 is compensated by multiplying by $\mu_{2}$ , the decrease of $t$ by 2 is compensated by multiplying by $\mu_{4}$ , the decrease of $t$ by 3 is compensated by multiplying by $\mu_{6}$ , (and $j$ remains unchanged). The picture for $B$ is somewhat more complicated: decrease $t$ by 2 and $j$ by 1 simultaneously transforms the exponents of $\mu_{2}$ and $\mu_{3}$ at $B$ into the corresponding ones at $A$ . A similar change of $j$ is required if odd terms are multiplied by $\mu_{6}$ , when $l$ is reduced by 1 and $t$ by 3.

The proof requires verifying that the right-hand part of Eq. (5) is equal to ${A}^{(t)}_{j,k,l,m}$ . It is really so:

$\displaystyle\frac{(2t-2)(t-j-k-l-2m-2)!}{(t-3j-2k-l-3m-1)!(2j-l)!k!l!m!}+% \frac{(2t-3)(t-j-k-l-2m-2)!(2j-l)}{(t-3j-2k-l-3m)!(2j-l)!k!l!m!}$ $\displaystyle\quad+\frac{(2t-4)(t-j-k-l-2m-2)!k}{(t-3j-2k-l-3m)!(2j-l)!k!l!m!}% +\frac{(2t-5)(t-j-k-l-2m-2)!l}{(t-3j-2k-l-3m)!(2j-l)!k!l!m!}$ (33) $\displaystyle\quad+\frac{(2t-6)(t-j-k-l-2m-2)!m}{(t-3j-2k-l-3m)!(2j-l)!k!l!m!}$ $\displaystyle\quad=\frac{(t-j-k-l-2m-2)![2t(t-j-k-l-2m)-2t]}{(t-3j-2k-l-3m)!(2% j-l)!k!l!m!}=A^{(t)}_{j,k,l,m}$
6. General case

Let us stop moving ahead gropingly from one case to another. Trying to discern the general features of expressions Eqs (29) and (30), we compare the first factor, denominator and the first binomial coefficient of the initial two-dimensional case Eq. (5) to those of the three- Eq. (5), four- Eq. (28), and five-dimensional cases Eq. (5). It seems initially that the sum of all subscripts is standing in place of one in the upper symbol of binomial coefficient and the denominator. However, this impression is refuted by the fifth example Eq. (5). In fact, the initial assumption should be replaced by a more natural one that sum of all exponents of all $\mu$ -factors is the upper symbol of the polynomial coefficient. The exponents of $\mu_{2}$ and $\mu_{3}$ are in fact determined to fulfill the “dimensionality conservation law”. Each new even factor, ${\mu}^{e_{2i}}_{2i}$ , contributes ${-ie}_{2i}$ into the exponent of ${\mu}_{2}$ as well as each odd factor, ${\mu}^{e_{2i+1}}_{2i+1}$ , contributes $-(i-1)e_{2i+1}$ into the exponent of ${\mu}_{2}$ and ${-e}_{i}$ into the exponent of ${\mu}_{3}$ . The expression of $S_{2t}(2n)$ contains a product ${\mu}^{e^{\prime}_{2}}_{2}\mu^{e^{\prime}_{3}}_{3}{\mu}^{e_{4}}_{4}{\mu}^{e_{5% }}_{5}{\mu}^{e_{6}}_{6}\dots{\mu}^{e_{2n+1}}_{n+1}$ with exponents satisfying the dimensionality condition $2e_{2}^{\prime}+3e_{3}^{\prime}+\cdots+2ne_{2n}+(2n+1)e_{2n+1}=2t,{e}_{j}\geqslant 0$ . The number of terms in $S_{2t}(2n)$ equals to the number of all nonnegative solutions for $e_{3},e_{4},\ldots,e_{2n+1}$ of this linear nondegenerate Diophantine equation with $2n-1$ unknowns [2] and $e^{\prime}_{2}$ , $e^{\prime}_{3}$ expressed through these as mentioned above. The contribution, $e$ , from exponents to the upper symbol of polynomial coefficient in $S_{2t}(2n)$ is determined from

$\displaystyle e^{\prime}_{2}+e^{\prime}_{3}+\sum^{2n+1}_{i=4}{e_{i}}=t-e,e=e_{% 3}+\sum^{n}_{i=2}{(i-1)(e_{2i}+e_{2i+1})}.$

(34) $\displaystyle e^{\prime}_{2}=t-3e_{3}-\sum^{n}_{i=2}{(ie_{2i}+(i-1)e_{2i+1})},% e^{\prime}_{3}=2e_{3}-\sum^{n}_{i=2}{e_{2i+1}},$

In particular, for $S_{2t}(4)e=t-e_{3}-e_{4}-e_{5}$ that coincides with the first binomial coefficient upper symbol in Eqs (26) and (27). We can verify that the correct dimensionality of the modified product ${\mu}^{e^{\prime}_{2}}_{2}\mu^{e^{\prime}_{3}}_{3}{\mu}^{e_{4}}_{4}{\mu}^{e_{5% }}_{5}{\mu}^{e_{6}}_{6}{\mu}^{e_{7}}_{7}\dots{\mu}^{e_{2n+1}}_{n+1}$ is consistent with the definitions in Eq. (6).

$\displaystyle 2e^{\prime}_{2}+3e^{\prime}_{3}+\sum^{2n+1}_{i=4}ie_{i}=2t-3e_{3% }-\sum^{n}_{i=2}(ie_{2i}+(i-1)e_{2i+1})+3\left(2e_{3}-\sum^{n}_{i=2}{e_{2i+1}}% \right)+\sum^{2n+1}_{i=4}ie_{i}=2t$ (35)

It is clear that adding 1 to $e^{\prime}_{3}$ and $-$ 1 to $e^{\prime}_{2}$ reduces the balance of exponents to $2t+1$ . The difference with the case of odd number of variables $S_{2t}(2n+1)$ is only in the number of last term $(n+1)e_{2n+2}$ of the $\mu$ -product. Thus, we can deal below only with $S_{2t}(2n)$ , bearing in mind these differences for the rest of cases. The anticipated general form of $S_{2t}(2n)$ is

$\displaystyle S_{2t}(2n)=\sum A^{(t)}_{e_{3},\cdots,e_{2n+1}}{\mu}^{e^{\prime}% _{2}}_{2}{\mu}^{e^{\prime}_{3}}_{3}{\mu}^{e_{4}}_{4}{\mu}^{e_{5}}_{5}\dots{\mu% }^{e_{2n}}_{2n}{\mu}^{e_{2n+1}}_{2n+1},S_{2t+1}(2n)=\sum B^{(t)}_{e_{3},\cdots% ,e_{2n+1}}{\mu}^{e^{\prime}_{2}-1}_{2}{\mu}^{e^{\prime}_{3}+1}_{3}{\mu}^{e_{4}% }_{4}{\mu}^{e_{5}}_{5}\dots{\mu}^{e_{2n}}_{2n}{\mu}^{e_{2n+1}}_{2n+1}$ (36)

Summation is taken over all sets of nonnegative integers $e_{3}\ldots,e_{2n+1}$ , which satisfy the dimensionality Eq. (35).

Generalizing Eq. (5) of the case $S_{2t}(5)$ , we suppose that

$\displaystyle A^{(t)}_{e_{3},\cdots,e_{2n+1}}=\frac{2t(t-e|e^{\prime}_{3},e_{4% },\dots,e_{2n+1})}{t-e}=\frac{2t(t-e-1)!}{e^{\prime}_{2}!e^{\prime}_{3}!e_{4}!% \dots e_{2n+1}!},$ (37) $\displaystyle B^{(t)}_{e_{3},\cdots,e_{2n+1}}=\frac{(2t+1)(t-e|e^{\prime}_{3}+% 1,e_{4},\dots,e_{2n+1})}{t-e}=\frac{(2t+1)(t-e-1)!}{(e^{\prime}_{2}-1)!(e^{% \prime}_{3}+1)!e_{4}!\dots e_{2n+1}!}$

We substitute expression Eq. (36) into the recurrent Eq. (16) to obtain the general recurrent formulae for coefficients. Both commutators and anticommutators enter each formula.

$\displaystyle S_{2t}(2n)=\sum^{2n+1}_{u=2}{\mu}_{u}S_{2t-u}(2n)=\sum^{n}_{u=1}% ({\mu}_{2u}S_{2(t-u)}(2n)+{\mu}_{2u+1}S_{2(t-u-1)+1}(2n))$ (38)

The procedure how to pass from this equation to equation for coefficients is described in detail in the case $n=5$ above. All steps remain the same. It should be emphasized only that the decrease of any subscript by 1, as in that case, is accompanied by a proper change of $e^{\prime}_{2}$ and $e^{\prime}_{3}$ , which make them equal to similar exponents in ${A}^{(t)}_{e_{3},\cdots,e_{2n+1}}$ . Indeed, the changes of individual exponents are compensated by suitable decrease of $t$ and $e$ . For instance, decrease of $e_{2u}$ by 1 produces increase of $e^{\prime}_{2}$ by $u$ according to Eq. (6), which is annulled by decrease of $t$ by the same amount, etc. As a result, the following recurrent equations for coefficients appear.

$\displaystyle A^{(t)}_{e_{3},\cdots,e_{2n+1}}=A^{(t-1)}_{e_{3},\cdots,e_{2n+1}% }+B^{(t-2)}_{e_{3}-1,e_{4},\cdots,e_{2n+1}}{}+\sum^{n}_{u=2}\left(A^{(t-u)}_{e% _{3},\cdots,e_{2u}-1,\cdots,e_{2n+1}}+B^{(t-u-1)}_{e_{3}-1,\cdots,e_{2u+1}-1,% \cdots,e_{2n+1}}\right)$ (39) $\displaystyle B^{(t)}_{e_{3},\cdots,e_{2n+1}}=A^{(t-1)}_{e_{3},\cdots,e_{2n+1}% }+B^{(t-1)}_{e_{3},\cdots,e_{2n+1}}{}+\sum^{n}_{u=2}\left(A^{(t-u)}_{e_{3},% \cdots,e_{2u+1}-1,\cdots,e_{2n+1}}+B^{(t-u)}_{e_{3},\cdots,e_{2u}-1,\cdots,e_{% 2n+1}}\right)$

The first terms with no changed subscripts are separated from both sums.

To simplify the check of our hypothesis we prepare $A^{(t-u)}_{\cdots e_{2u}-1\cdots}$ and ${B}^{(t-u-1)}_{e_{3}-1,\cdots,e_{2u+1}-1,\cdots}$ beforehand. We have to bear in mind that in the case of $A$ -coefficient, only even subscripts changed, hence $e^{\prime}_{3}$ remains unchanged. Moreover, $e^{\prime}_{2}$ remains unchanged, too, according to Eq. (6). If neither of subscripts changed, $e^{\prime}_{2}\to e^{\prime}_{2}-1$ in $A^{(t-1)}_{\cdots e_{2u}\cdots}$ . At the same time, $e$ is reduced by $u-1$ , if $e_{2u}\to e_{2u}-1$ while $t$ is reduced by $u$ , and $t-e\to t-e-1$ , but $e^{\prime}_{2}\to e^{\prime}_{2}$ . In the $B$ -case, $e$ is reduced by 1 by definition, and $e^{\prime}_{3}$ is decreased by 1 (as a result of subtraction of 1 from $e_{3}$ and $e_{2u+1}$ ), while $e^{\prime}_{2}$ is increased by 1 due to addition of 3 (from $e_{3}-1$ ) and the difference $-u-1-(u-1)$ in Eq. (6), which is annulled by $-$ 1, already present in Eq. (39). Therefore,

$\displaystyle A^{(t-u)}_{e_{3},\cdots,e_{2u}-1,\cdots,e_{2n+1}}=\frac{2(t-u)(t% -2-e)!e_{2u}}{e^{\prime}_{2}!e^{\prime}_{3}!e_{4}!\dots e_{2n+1}!},u\geqslant 2% ;{A}^{(t-1)}_{e_{3},\cdots,e_{2u},\cdots,e_{2n+1}}=\frac{2(t-1)(t-2-e)!e^{% \prime}_{2}}{e^{\prime}_{2}!e^{\prime}_{3}!e_{4}!\dots e_{2n+1}!},$ (40) $\displaystyle B^{(t-u-1)}_{e_{3}-1,\cdots,e_{2u+1}-1,\cdots,e_{2n+1}}=\frac{(2% (t-u)-1)(t-2-e)!e_{2u+1}}{e^{\prime}_{2}!e^{\prime}_{3}!e_{4}!\dots e_{2n+1}!}% ,{B}^{(t-2)}_{e_{3}-1,\cdots,e_{2u+1},\cdots,e_{2n+1}}=\frac{(2(t-1)-1)(t-2-e)% !e^{\prime}_{3}}{e^{\prime}_{2}!e^{\prime}_{3}!e_{4}!\dots e_{2n+1}!}$

The reduced subscripts absent in $A$ for $u=1$ , and 1 subtracted only from $e_{3}$ in $B$ . The specific feature of Eq. (40) is a very simple dependence of $u$ , which remained only in one factor. We see that these equations include all situations met in the particular case $S_{2t}(5)$ (Eq. (5)) and coincide with those results if we put $e_{3}=j,e_{4}=k,e_{5}=l,e_{6}=m,e_{u}=0$ ( $u>6$ ), $e^{\prime}_{3}=2j-l,e^{\prime}_{2}=t-3j-2k-l-3m,e=j+k+l+2m$ .

Now we can insert expressions Eq. (40) into the right part of Eq. (39).

$\displaystyle\frac{(t-2-e)![2(t-1)e^{\prime}_{2}+(2(t-1)-1)e^{\prime}_{3}]}{e^% {\prime}_{2}!e^{\prime}_{3}!e_{4}!\dots e_{2n+1}!}+\frac{(t-2-e)!}{e^{\prime}_% {2}!e^{\prime}_{3}!e_{4}!\dots e_{2n+1}!}$ $\displaystyle\sum^{n}_{i=2}[2(t-i)e_{2i}+(2(t-i)-1)e_{2i}]=\frac{(t-2-e)!}{e^{% \prime}_{2}!e^{\prime}_{3}!e_{4}!\dots e_{2n+1}!}$ $\displaystyle\left[2(t-1)\left(t-3e_{3}-\sum^{n}_{i=2}(ie_{2i}+(i-1)e_{2i+1})% \right)+(2(t-1)-1)\left(2e_{3}-\sum^{n}_{i=2}{e_{2i+1}}\right)\right.$ $\displaystyle\left.+\sum^{n}_{u=2}[2(t-i)e_{2i}+(2(t-i)-1)e_{2i+1}]\right]$

The proof requires calculating the expression in brackets and the sum that gives

$\displaystyle 2t(t-1)-2te_{3}-2(t-1)\sum^{n}_{i=2}(ie_{2i}+(i-1)e_{2i+1})-(2t-% 3)\sum^{n}_{i=2}e_{2i+1}$ $\displaystyle\quad+\sum^{n}_{i=2}[2(t-i)e_{2i}+(2(t-i)-1)e_{2i+1}]$ $\displaystyle\quad=2t(t-1-e_{3})-2t\sum^{n}_{i=2}(ie_{2i}+ie_{2i+1}-e_{2i}-e_{% 2i+1})$ $\displaystyle\quad+\sum^{n}_{i=2}[2(ie_{2i}+(i-1)e_{2i+1})+3e_{2i+1}-2ie_{2i}-% (2i+1)e_{2i+1}]$ $\displaystyle\quad=2t\left(t-1-e_{3}-\sum^{n}_{i=2}(i-1)(e_{2i}+e_{2i+1})% \right)=2t(t-1-e)$

We used the Eq. (6) on the last step that has given $e$ . Collecting all these results we obtain $A^{(t)}_{e_{3},\cdots,e_{2n+1}}$ as it defined in Eq. (6). The proof of the formula for $B$ -coefficient is similar.

Thus, we have proved that recurrent Eq. (16), based on the Newton theorem, leads to the following identities for “normalized C-A” $R_{u}=S_{u}/u$ with even $2t$ and odd $2t+1$ subscripts and $e^{\prime}_{2}$ , $e^{\prime}_{3}$ defined in Eq. (6).

$\displaystyle R_{2t}(n)=\sum\frac{(t-e-1)!}{e^{\prime}_{2}!e^{\prime}_{3}!e_{4% }!\dots e_{n+1}!}{\mu}^{e_{2}^{\prime}}_{2}{\mu}^{e_{3}^{\prime}}_{3}{\mu}^{e_% {4}}_{4}{\mu}^{e_{5}}_{5}\dots{\mu}^{e_{n}}_{n}{(\rho\sigma)}^{e_{n+1}},\quad R% _{2t}(n)\equiv\frac{{\sigma}^{2t}+{\sigma}_{2t}}{2t},$ $\displaystyle R_{2t+1}(n)=\sum\frac{(t-e-1)!}{(e^{\prime}_{2}-1)!(e^{\prime}_{% 3}+1)!e_{4}!\dots e_{n+1}!}{\mu}^{e^{\prime}_{2}-1}_{2}{\mu}^{e^{\prime}_{3}+1% }_{3}{\mu}^{e_{4}}_{4}{\mu}^{e_{5}}_{5}\dots{\mu}^{e_{n}}_{n}{(\rho\sigma)}^{e% _{n+1}},$ (41) $\displaystyle R_{2t+1}(n)\equiv\frac{\sigma^{2t+1}-{\sigma}_{2t+1}}{2t+1}.$

Equations for the odd number of basic parameters, included in Eq. (6), can be written in the same way as in the example of $2n+1=5$ Eqs (29) and (30). In each case, the last $\mu$ -factor number is greater by 1 than the number of basic parameters: $\mu_{n+1}=\rho\sigma$ where $\sigma$ is a sum and $\rho$ is a product of all basic parameters. The normalized C-A with a prime subscript only is guaranteed to be integer as explained in Sec. 1. In the other cases, it may have a denominator equal to a divisor of the subscript of $R_{u}$ . Equation (6) is the main result of the present work.

It is worthwhile now to repeat that $\mu_{k}$ appear in the equation ${\xi}^{n+1}-\sum^{n+1}_{j=2}{\mu}_{j}{\xi}^{n-j+1}=0$ , the roots $-x_{j}$ of this equation are the arguments of $R_{t}$ (through $\sigma$ and $\sigma_{t}$ in Eq. (10)), which may be natural, real or complex numbers or even commuting matrices. Summation in Eq. (6) is taken over all $n-1$ nonnegative integers $e_{3},e_{4},\ldots,e_{n+1}$ , which obey nondegenerate linear Diophantine equation $2e^{\prime}_{2}+3e^{\prime}_{3}+4e_{4}+\ldots+(n+1)e_{n+1}=2t$ (or $2t+1$ ). The latter condition is automatically fulfilled if linear combinations $e^{\prime}_{2}$ and $e^{\prime}_{3}$ , defined by Eq. (6), are nonnegative, too. Each variable $e_{k}$ can acquire integer values between 0 and [ $t/k$ ].
7. Other presentation of results and comments

So, identities Eq. (6) give a possibility to express sums of $n$ arbitrary numbers in high ( $\gg n$ ) degrees, $t$ , in terms of $n$ parameters $\mu_{k}$ . Some problems, which deal with such sums of natural numbers, are mentioned in Introduction. For example, the Euler sum of powers conjecture means if $kR_{k}(n)=\sigma^{k}(n)\pm a^{k}$ ( $a$ is natural) then $n\geqslant k$ . After computer finding of a counterexample for $k=5,n=4$ , Lander et al. [5] reduced by 1 the right side of Euler’s inequality. Other directions of investigation concern the interrelation between such $\sigma_{t}(n)$ with different $t$ of sequential natural numbers [6] or of the general type [5, 7].

Equation (6) allows finding these type connections between different $R_{k}$ . For this, it is necessary to express $\mu_{u}$ through $R_{k}$ . In fact, we already have examples of such Eq. (4). Now we shall find the general form of this connection. Bearing in mind that $\mu_{2}=R_{2}$ , $\mu_{3}=R_{3}$ we can rewrite Eq. (4) through $R_{u}$ beginning from $\mu_{4}$ , viz $\mu_{4}=R_{4}-\mu_{2}2R_{2}/4=R_{4}-R_{2}^{2}/2,\mu_{5}=R_{5}-2\mu_{3}R_{2}/5-% 3\mu_{2}R_{3}/5=R_{5}-R_{2}R_{3},\mu_{6}=R_{6}+R_{2}^{3}/3!-R_{3}^{2}/2!-R_{4}% R_{2},\mu_{7}=R_{7}+R_{2}^{2}R_{3}/2!-R_{3}R_{4}-R_{2}R_{5},\mu_{8}=R_{8}+R_{2% }R_{3}^{2}/2+R_{4}R_{2}^{2}/2-R_{2}R_{6}-R_{3}R_{5}-R_{4}^{2}/2-R_{2}^{4}/4!,% \mu_{9}=R_{9}+R_{2}R_{3}R_{4}-R_{2}R_{7}-R_{3}R_{6}-R_{4}R_{5}-R_{2}^{3}R_{3}/% 3!+R_{2}^{2}R_{5}/2+R_{3}^{3}/3!\ldots$ All of these may be obtained directly from Eq. (14) written in terms of $R_{u}$ .

$\displaystyle{\mu}_{t}=R_{t}-\sum^{t-2}_{u=2}u\mu_{t-u}R_{u}/t.$ (42)

Some peculiarities of the above formulae worthwhile to mention: the coefficients in denominators are artificially presented as factorials of the exponents of some factors, which, therefore, may be considered “normalized”. The sign of any term is “ $+$ ” if the net number of factors is odd, and “ $-$ ” in the opposite case. The coefficients at products, which are free of C-A squares, are 1. The noticed features may be proved general.

Indeed, a product $R_{i}R_{j}\ldots R_{k}R_{u}$ appears in Eq. (42) from those $\mu_{t-u}$ , which contain similar products with one factor omitted, i.e. from $i\mu_{t-i}R_{i}/t+j\mu_{t-j}R_{j}/t+\ldots+k\mu_{t-k}R_{k}/t+u\mu_{t-u}R_{u}/t$ . This contribution, $-(iR_{i}R_{j}\ldots R_{k}R_{u}/t+jR_{i}R_{j}\ldots R_{k}R_{u}/t+kR_{i}R_{j}% \ldots R_{k}R_{u}/t+uR_{i}R_{j}\ldots R_{k}R_{u})/t=(i+j+\ldots+k+u)R_{i}R_{j}% \ldots R_{k}R_{u}/t=R_{i}R_{j}\ldots R_{k}R_{u}$ , results from the “dimensionality conservation law” and from $i+j+\ldots+k+u=t$ . The number of cofactors in a term determines the sign before it. A term of type $R_{u}^{k}$ could appear in $\mu_{uk}$ only from $\mu_{uk-u}$ in Eq. (42). If we assume that it had there the coefficient $[(k-1)!]^{-1}$ then Eq. (42) gives $-(uR_{u}^{k-1}/(k-1)!)R_{u}/(uk)=R_{u}^{k}/k!$ . The hypothesis turned up to be self-consistent. The presence of another factor does not influence on this conclusion: $R_{j}R_{u}^{k}/k!$ appears from $R_{j}R_{u}^{k-1}/(k-1)!$ of $\mu_{uk-u+j}$ and from $R_{u}^{k}/k!$ of $\mu_{uk}$ , viz $-(uR_{j}R_{u}^{k-1}/(k-1)!)R_{u}/(uk+j)-(jR_{u}^{k}R_{j}/k!)/(uk+j)=-R_{u}^{k}% /k!$ .

Thus, the peculiarities, manifested in the above examples, are proved to be general, and the structure of $\mu_{t}$ looks like

$\displaystyle{\mu}_{t}=R_{t}+\sum_{P,t=i+j+ak+u+\dots+v}(-1)^{m-1}R_{i}(n)R_{j% }(n)[R_{k}(n)^{a}/a!]R_{u}(n)\dots R_{v}(n).$ (43)

Summation is taken over possible partitions of $t$ on all possible ( $\neq$ 1) nonzero parts (including equal “normalized” parts); $m$ is the number of all factors counting each one from the repeated ones. A specific “partition” on a single part is separated from the sum.

If the number of initial parameters $x_{j}$ is $n$ there are no $\mu-s$ with subscripts higher than $n+1$ , i.e. $\mu_{t}=0$ if $t>n+1$ . This has a going far consequence: there exist identities between the higher $R_{t}$ and lower ones, $R_{k}$ , with $k\leqslant n+1$ forming possible products corresponding to partitions of $t$ according to the equation, which follows from Eq. (43).

$\displaystyle R_{t}(n)=\sum_{P,t=i+j+ak+\dots+u}(-1)^{m}R_{i}(n)R_{j}(n)(R^{a}% _{k}(n)/a!)\dots R_{u}(n),t>n+1.$ (44)

The lower C-A will appear in the right side of Eq. (44) after sequential application of the same Eq. (44) to any factor $R_{q}$ . A lot of identities may be obtained in this way since we can stop at any moment and express the remaining $R_{q}$ through basic parameters directly by the Eq. (10). Thus, Eq. (44) is as important as Eq. (42).

Few examples demonstrate this result (notation of ( $n$ ) is omitted) beginning from $R_{5}=R_{2}R_{3}$ : $x^{5}+y^{5}+z^{5}=({x+y+z})^{5}-5[({x+y+z})^{2}+x^{2}+y^{2}+z^{2}][({x+y+z})^{% 3}-(x^{3}+y^{3}+z^{3})]/6$ . To be short we omit below the detailed form: ${R}_{6}={R}_{3}^{2}/2!+{R}_{4}{R}_{2}-{R}_{2}^{3}/3!,{R}_{7}={R}_{3}{R}_{4}+{R% }_{2}{R}_{5}-{R}_{2}^{2}{R}_{3}/2!={R}_{3}{R}_{4}+{R}_{2}^{2}{R}_{3}/2!,{R}_{8% }={R}_{2}{R}_{6}+{R}_{3}{R}_{5}+{R}_{4}^{2}/2+{R}_{2}^{4}/4!-{R}_{2}{R}_{3}^{2% }/2-R_{4}{R}_{2}^{2}/2={R}_{2}{R}_{3}^{2}/2!+{R}_{4}{R}_{2}^{2}-{R}_{2}^{4}/3!% +{R}_{2}{R}_{3}^{2}+R_{4}^{2}/2+{R}_{2}^{4}/4!-{R}_{2}{R}_{3}^{2}/2{-R}_{4}{R}% _{2}^{2}/2={R}_{2}{R}_{3}^{2}+{R}_{4}{R}_{2}^{2}/2+{R}_{4}^{2}/2-{R}_{2}^{4}/8% ,{R}_{9}={R}_{2}{R}_{7}+{R}_{3}{R}_{6}+{R}_{4}{R}_{5}+{R}_{2}^{3}{R}_{3}/3!-{R% }_{2}{R}_{3}{R}_{4}-{R}_{2}^{2}{R}_{5}/2-{R}_{3}^{3}/3!={R}_{2}{R}_{3}{R}_{4}+% {R}_{2}^{3}{R}_{3}/2!+{R}_{3}^{3}/2!+{R}_{4}{R}_{3}{R}_{2}-{R}_{2}^{3}{R}_{3}/% 3!+{R}_{2}^{3}{R}_{3}/3!-{R}_{2}^{2}{R}_{5}/2-{R}_{3}^{3}/3!=2{R}_{2}{R}_{3}{R% }_{4}+{R}_{3}^{3}/3$ , etc. It should be reminded that above identities are valid only for $n\leqslant 3$ .

It is worthwhile to note that the inequality in the Lander-Parkin-Selfridge conjecture [5], in fact, exactly opposite to the condition Eq. (44). That means the above conjecture admits substituting $\sigma_{k}(n)$ by the $k$ -th power of some natural number only in the domain on the exponent $k$ -axes where there are no identities between $R_{k}(n)$ and lower normalized C-A of the above type and also that Eq. (4) may be useful in a search of such numbers.

We mention at the end (remember: the above results cover also real numbers) some similarity between Eq. (6) for C-A and equation for the “power indicator” from the hyperpolarizability theory, which allows to establish connection between the latter and the highest susceptibilities of molecules and solids [8] and also to revers power series. By definition, power indicator is

$\displaystyle X^{(E)}_{t}=\sum_{t=2e_{2}+3e_{3}+\dots;e_{2}+e_{3}+\dots=E}% \frac{E!}{e_{2}!e_{3}!e_{4}!e_{5}!\dots}{\mu}^{e_{2}}_{2}{\mu}^{e_{3}}_{3}{\mu% }^{e_{4}}_{4}{\mu}^{e_{5}}_{5}\dots.$ (45)

Here, $\mu_{j}$ are coefficients of a power series. If the expansion parameter of this series is the external electric field strength, and series starts from $j=2$ (as in the Eq. (4)) then $\mu_{j}=\chi_{j}$ are nonlinear susceptibilities; if parameter is the local field then $\mu_{j}=\gamma_{j}$ are hyperpolarizabilities. The main difference between Eqs (6) and (45) is the additional restrictions of the summation domain. It requires the number of multipliers in a product of $\mu$ -s to be fixed with free $e_{2}$ and $e_{3}$ in the latter, while in the former such restriction is absent, but $e^{\prime}_{2}$ and $e^{\prime}_{3}$ are expressed through the rest of $e_{j}$ . This difference causes the important distinctions in properties of $R_{t}$ and $X_{t}^{({E})}$ . In particular, convolutions of the latter over the superscript are expressed through the same quantities. As a result, the power indicator allows to find formula for the reciprocal power series and due to it to connect explicitly the higher susceptibilities of a crystal with hyperpolarizabilities of its molecules [8].

8. Conclusion

Newton’s identities for sum of powers with identical exponents of natural, real, or complex numbers are presented in a form, not using similar sums with all lower exponents, directly through the coefficients of the corresponding equation with numerical polynomial coefficients, built of properly modified intermediate exponents. Numerous identities between such sums with the exponents, which are greater than the number of addends by more than 2, are established. The identities found are not linear, in contrast to the identities between sums of sequential integers [6], but remain valid only for a restricted number of sum terms.

This article is dedicated to the memory of Professor Michel Deza.

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