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Received December 4, 1930.
Mr. Lagrange, in a paper published in "Acta Mathematica" ( ¹ ), studied sequences of numbers from an algebraic point of view. He applied these to certain sequences of polynomials. In the present work, we will complement the "algebra of sequences of numbers" with an "algebra of sequences of polynomials." We will address the applications in other papers.

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We only provide definitions and results. The proofs often lead to somewhat lengthy calculations, but these present no real difficulty.

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1. 1.

   Definition of a sequence of polynomials. Consider a sequence of polynomials in$x$

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$$\mathrm{P}_{0},\mathrm{P}_{1},\ldots\mathrm{P}_{n},\ldots$$

(1)

taken in a specific order. This order is characterized by the number$n$which is the index or rank of the polynomial$\mathrm{P}_{n}$. Let us designate by$p(n)$the degree of the polynomial$\mathrm{P}_{n}$. If$\mathrm{P}_{n}$is identically zero, we assume that$p(n)$has a negative value as large as we want. The difference

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$$p(n)-n$$

(2)

is the order of the polynomial$\mathrm{P}_{n}$. If$p(n)<0$, so if$\mathrm{P}_{n}$is identically zero we say that it is of order$-\infty$.

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If the orders of all the elements of a sequence are equal to$-\infty$We say that this sequence is the zero sequence. For all other sequences, the order of the elements has an upper bound.$m$Finite or infinite. If$m$is finished, we say that the sequence is of finite order$m$. In

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In this case, the sequence (1) has at least one element of order$m$We call the smallest value of the characteristic index$n$for which

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$$p(n)-n=m$$

We say that a sequence is complete if all its elements have the same order. A complete sequence is necessarily of finite order and its characteristic index is 0.

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If$m$is infinite, the sequence is of infinite order. Such a sequence cannot be complete and does not admit a characteristic index. We call the class of the sequence (1) the number$k$such as

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$$p(0)<0,\quad p(1)<0,\ldots\quad p(k-1)<0,\quad p(k)\geqslant 0.$$

A sequence of negative order$m$is at least class$-m$We
ask

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$$\binom{i}{j}=\frac{i(i-1)\ldots(i-j+1)}{1.2.3\ldots j}\quad;\quad\binom{i}{j}=0,j>i.$$

Since accents denote derivatives, we say that a class sequence$k$and order$-k$is normal if

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	$$\displaystyle\sum_{j=0}^{i}\binom{i}{j}\mathrm{P}_{k+j}^{(j)}\neq 0$$		(3)
	$$\displaystyle i=0,1,2,\ldots$$

We see that the first members are numbers.
We denote by$[\mathrm{P}]$the sequence (1). When considering several sequences$[\mathrm{P}],[\mathrm{Q}],\ldots$we designate by$p(n),q(n),...$the degree of polynomials$\mathrm{P}_{n},Q_{n},\ldots$

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All the relationships we write between several polynomials are verified identically with respect to$x$
2. The algebra of sequences of polynomials . $1^{0}$We denote by [0] the zero sequence.
$2^{0}$The sequel

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$$\mathrm{P}_{0}=1;\quad\mathrm{P}_{n}=0,n>0$$

is the unit sequence and will be denoted by [1]. It is normal and of class 0.

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The sequel

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$$\mathrm{P}_{k}=1,\quad\mathrm{P}_{n}=0,\quad n\neq k$$

is the unit sequence of class$k,[1]_{k}$30.
Two suites$[\mathrm{P}],[\mathrm{Q}]$are equal if, and only if

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$$\begin{gathered}\mathrm{P}_{n}=\mathrm{Q}_{n}\\ n=0,1,2,\ldots\end{gathered}$$

We write

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$$[\mathrm{P}]=[\mathrm{Q}]$$

$4^{0}$The product of a sequence ^[P] by a number$\lambda$is by definition a new sequel$[Q]$given by

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We write

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$$\begin{gathered}Q_{n}=\lambda P_{n}\\ n=0,1,2,\ldots\\ \lambda\cdot\left[P_{j}=[\lambda P]\right.\end{gathered}$$

$-\mathrm{P}]$is the opposite of$[\mathrm{P}]$and is equal to$-[\mathrm{P}]$.
$5^{0}$The sum of two sequences$[\mathrm{P}],[\mathrm{Q}]$is a new sequel$[\mathrm{R}]$defined by

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We write

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$$\begin{gathered}\mathrm{R}_{n}=\mathrm{P}_{n}+\mathrm{Q}_{n}\\ n=0,1,2,\ldots\end{gathered}$$

1. 60.

   The elementary product of$[\mathrm{P}]$by$[\mathrm{Q}]$is a new sequence [R] defined by the equalities

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$$\mathrm{R}_{n}=\sum_{i=0}^{\infty}\mathrm{Q}_{i}\left[\sum_{j=0}^{n}\binom{i}{j}\mathrm{P}_{nj}^{(ij)}\right],\quad n=0,1,2,\ldots$$

(4)

It is easily verified that the right-hand side contains a finite number of terms. For sequences of negative or zero order, we can write the condensed formulas

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$$\mathrm{R}_{n}=\sum_{i=0}^{n}Q\left[\sum_{j=0}^{i}\binom{i}{j}\mathrm{P}_{nj}^{(ij)}\right],\quad n=0,1,2,\ldots$$

(5)

We write

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$$[\mathrm{Q}]\cdot[\mathrm{P}]=[\mathrm{R}]$$

The definitions$4^{0},5^{0}$allow us to find the difference between two sequences.

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The equality and addition of sequences enjoys all the properties of ordinary equality and addition.

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Elementary multiplication is associative and distributive with respect to addition, but it is not generally commutative. If

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$$[\mathrm{Q}]\cdot[\mathrm{P}]=[\mathrm{P}]\cdot[\mathrm{Q}]$$

we say that the sequels$[\mathrm{P}],[\mathrm{Q}]$are interchangeable. The unit sequence is interchangeable with any sequence.

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Multiplication by a non-zero number does not change the order, class, normality, or permutability of a sequence.

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The order of a sum is at most equal to the highest order of the sums added, and its class is at least equal to the lowest
class of the sums added. In the formula

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$$[\mathrm{R}]=[\mathrm{P}]+[\mathrm{Q}]$$

on a effect

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$$r(n)\leq\max[p(n),q(n)]$$

Whatever happens next$[\mathrm{P}]$we have

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$$[\mathrm{P}]+[0]=[\mathrm{P}]$$

The order of a product is at most equal to the sum of the orders of its factors, and its class is at least equal to the class of the sequence being multiplied. Indeed, we have
$r(n)\leq\max\left[q(0)+p_{1}(n),\quad q(1)+p_{1}(n)-1,\quad q(2)+p_{1}(n)-2,\ldots q\left(p_{1}(n)\right)\right]$Or

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$$p_{1}(n)=\max[p(n),p(n-1)+1,p(n-2)+2,\ldots,p(0)+n]$$

If one of the factors is [0], the product is equal to [0].
For normal sequences of class 0, the converse is true, but: in the general case, the product of two non-zero sequences can be zero; for example, for the sequences
$[P]\quad x,0,0,\ldots 0;\ldots$

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$$[Q]\quad 1,-x,\frac{x^{2}}{2;},\ldots\frac{(-1)^{n}x^{n}}{n!},\ldots$$

$$[Q]\cdot[P]=[0]^{1}$$

Knowing the product of two sequences, we can calculate any positive integer power. We will denote by$[\mathrm{P}]^{m}$, Or$[m\mathrm{P}]$there$m^{\text{èma }}$power of the suite$[\mathrm{P}]$We posit by definition:

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$$[P]^{0}=\left[{}_{0}P\right]=[1]$$

We then have for$m,k$positive integers or zero-

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$$[\mathrm{P}]^{m}\cdot[\mathrm{P}]^{k}=[\mathrm{P}]^{m+k};\left\{[\mathrm{P}]^{m}\right\}^{k}=[\mathrm{P}]^{mk}.$$

We must say a few more words about the elementary division of: sequences.

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$$[\mathrm{P}]=[\mathrm{Q}]\cdot[\mathrm{R}].$$

(6)

Equation (6) is not always possible in [ R ] and if it is possible the solution is not always unique.

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If [R] is the sequence [1], we say that [P] has a left inverse. Let us denote by$[\mathrm{P}]_{\mathrm{g}}^{-1}$this invention. We have,

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$$[\mathbb{P}]\cdot[\mathrm{P}]_{g}^{-1}=[\mathrm{I}]$$

From this we deduce that the following$[\mathrm{P}]$is divisible from the left by any sequence that has a left inverse.

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Similarly, we define right division and right division.$[\mathrm{P}]_{d}^{-1}$So that$[\mathrm{P}]_{d}^{-1}$exists, the sequel must follow$[\mathrm{P}]$either of class 0.

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If a sequel$[P]$has a left inverse and a right inverse net if

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$$[\mathrm{P}]_{g}^{-1}=[\mathrm{P}]_{d}^{-1}$$

We say that it is reversible. We then designate by$[\mathrm{P}]^{-1}$the inverse of [P]. All integer powers of such a sequence are determined.
3. We will point out some properties of normal sequences.

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The product of two normal sequences is a normal sequence of class -equal to the sum of the classes of the factors

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Posing

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$$[Q]\cdot[P]=[R]$$

we have

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$$\begin{gathered}\sum_{j=0}^{i}\binom{i}{j}\mathrm{R}_{k+k^{\prime}+j}^{(j)}=\left\{\sum_{j=0}^{i}\binom{i}{j}Q_{k^{\prime}+i}^{(j)}\right\}\left\{\sum_{j=0}^{k^{\prime}+i}\binom{k^{\prime}+i}{j}\mathrm{P}_{k+i}^{(j)}\right\}\\ i=0,1;2,\ldots\end{gathered}$$

$k,k^{\prime}$being the classes of$[\mathrm{P}]$And$[\mathrm{Q}]$It
follows that the product of two normal sequences is always different from [0].

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Every normal sequence of class 0 is invertible and its inverse is also a normal sequence of class 0.

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Either$\left[{}_{-1}\mathrm{P}\right]$the inverse sequence; we have

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$$\begin{gathered}\left\{\sum_{j=0}^{i}\binom{i}{j}-1\mathrm{P}_{j}^{(j)}\left\{\left\lvert\,\sum_{j=0}^{i}\binom{i}{j}\mathrm{P}_{j}^{(j)}\right.\right\}=1\right.\\ i=0,1,2,\ldots\end{gathered}$$

The normal sequences of class zero form a group. This group is not permutable, but it contains permutable subgroups.

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We already know any integer power of a normal sequence of class 0.

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Let's ask

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$$\mathrm{U}_{i}^{(j)}=\sum_{s=0}^{j}\binom{j}{s}\mathrm{P}_{s+i}^{(s)}$$

and let us introduce the following notation

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$$\left[a_{1},a_{2},\ldots a_{k}\right]=\left|\begin{array}[]{cccccc}1&a_{1}&a_{1}^{2}&\ldots&a_{1}^{k-2}&a_{1}^{m}\\ 1&a_{2}&a_{2}^{2}&\ldots&a_{2}^{k-2}&a_{2}^{m}\\ \cdot&\cdot&\cdot&\cdot&\cdot&\cdot\\ 1&a_{k}&a_{k}^{2}&\ldots&a_{k}^{k-2}&a_{k}^{m}\end{array}\right|:\left|\begin{array}[]{ccccc}1&a_{1}&a_{1}^{2}&\ldots&a_{1}^{k-1}\\ 1&a_{2}&a_{2}^{2}&\ldots&a_{2}^{k-1}\\ \cdot&\cdot&\cdot&\cdot&\cdot\\ 1&a_{k}&a_{k}^{2}&\ldots&a_{k}^{k-1}\end{array}\right|$$

Let us now consider the following$[m\mathrm{P}]$defined by the relationships

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$$\begin{gathered}{}_{m}\mathrm{P}_{n}=\sum_{n_{1},n_{2},\ldots n_{k}}\mathrm{U}_{n_{1}}^{(0)}\cdot\mathrm{U}_{n_{2}}^{\left(n_{1}\right)}\cdot\mathrm{U}_{n_{3}}^{\left(n_{1}+n_{2}\right)}\cdots\mathrm{U}_{n_{k}}^{\left(n_{1}+n_{2}+\cdots+n_{k-1}\right)}\\ \cdot\left[\mathrm{U}_{0}^{(0)},\mathrm{U}_{0}^{\left(n_{1}\right)},\mathrm{U}_{0}^{\left(n_{1}+n_{2}\right)},\ldots\mathrm{U}_{0}^{\left(n_{1}+n_{2}+\cdots+n_{k}\right)^{\prime}}\right.\\ n=0,1,2,\ldots\end{gathered}$$

the summation being extended to positive values ​​of$n_{1},n_{2},\ldots n_{\text{fin }}$ verifying equality

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$$n_{1}+n_{2}+\cdots+n_{k}=n$$

$\boldsymbol{k}$taking all possible values.
It can be shown that if$m$is an integer
(7)

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$$\left[{}_{m}\mathrm{P}\right]=[\mathrm{P}]^{m}$$

It is then demonstrated that

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$$\begin{gathered}{\left[m_{\mathrm{m}}\mathrm{P}\right]\cdot\left[m^{\prime}\mathrm{P}\right]=\left[m^{\prime}\mathrm{P}\right]\cdot\left[{}_{m}\mathrm{P}\right]=\left[m_{+m^{\prime}}\mathrm{P}\right]}\\ {\left[m^{\prime}[m\mathrm{P}]=\left[m\left[m^{\prime}\mathrm{P}\right]\right]=\left[mm^{\prime}\mathrm{P}\right]\right.}\end{gathered}$$

We can therefore keep equality (7) as defining a power: any of the sequence$[\mathrm{P}]$, normal and class 0.

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We have

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$$\sum_{j=0}^{i}\binom{i}{j}{}_{m}\mathrm{P}_{j}^{(j)}=\left(\sum_{j=0}^{i}\binom{i}{j}\mathrm{P}_{j}^{(\Omega)}\right)^{m}$$

For example, for the normal binomial sequence of class 0,
we have

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$$\begin{gathered}\mathrm{P}_{0},\mathrm{P}_{1},0,0,\ldots 0,\ldots\\ {}_{m}\mathrm{P}_{n}=\mathrm{P}_{-1}^{n}\cdot\left[\mathrm{U}_{0}^{(0)},\mathrm{U}_{0}^{(1)},\mathrm{U}_{0}^{(2)},\ldots\mathrm{U}_{0}^{(n)}\right]\end{gathered}$$

The number$\left[\mathrm{U}_{0}^{(0)},\mathrm{U}_{0}^{(1)},\ldots\mathrm{U}_{0}^{(n)}\right]$generalizes the number$\binom{m}{n}$and reduces$\grave{\mathfrak{m}}^{2}$the latter for$\mathrm{P}^{\prime}{}_{1}=0$.

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$$\text{ Si }\mathrm{P}_{1}^{\prime}\neq 0\text{ la série }$$

$$\sum_{n=1}^{\infty}\left[\mathrm{U}_{0}^{(0)},\mathrm{U}_{0}^{(1)},\ldots\mathrm{U}_{0}^{(n)}\right]$$

converges absolutely regardless of$m$It can indeed be demonstrated without
difficulty that

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$$\left|\left[\mathrm{U}_{0}^{(0)},\mathrm{U}_{0}^{(1)},\ldots\mathrm{U}_{0}^{(n)}\right]\right|<\frac{\binom{n}{n^{\prime}}}{n!\left(\mathrm{P}^{\prime}{}_{1}\right)^{n}}\sum_{i=0}^{n}\left|\mathrm{U}_{0}^{(i)}\right|m$$

$n$being equal to$\frac{n}{2}$Or$\frac{n-1}{2}$following that$n$is even or odd.
4. On some particular sequences. Let us designate by$[a],[b],\ldots$the sequences of numbers, therefore the sequences (1) such that$p(n)\leq 0n=0,1,2,\ldots$A sequence of numbers is always normal. Its order is equal to its class with the sign reversed.

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Number sequences of class 0 form a permutable subgroup of the group of normal sequences of class 0.

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A normal sequence of class 0 that is permutable with a sequence of numbers of class 0 is not necessarily a sequence of numbers. If the sequence of numbers has all its elements non-zero, then any sequence permutable with it is a sequence of numbers.

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Let us consider sequences of the form
(8)

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$$\mathrm{P}_{n}=a_{n}\frac{x^{n}}{n!},\quad n=0,1,2,\ldots$$

These sequences are characterized by the sequence of numbers [a]. We will denote them by$[\mathrm{P};[a]]$For such a sequence to be normal, it must belong to class 0. The conditions for normality are then

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$$\sum_{\begin{subarray}{c}j=0\\ i=0\end{subarray}}^{i}\binom{i}{j}a_{j}\neq 0$$

Normal sequences of the form (8) form a permutable subgroup of the group of normal sequences of class 0.

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The reverse sequence

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$$[\mathrm{P};[a]]^{-1}=\left[{}_{-1}\mathrm{P};[-1a]\right]$$

is determined by the equations

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The product

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$$\begin{gathered}\left|\sum_{i=0}^{n}\binom{n}{i}_{-1}a_{i}\right|\left\{\sum_{i=0}^{n}\binom{n}{i}a_{i}\right\}=1\\ n=0\quad 1,-2,\ldots\end{gathered}$$

$$[Q;[b]]\cdot[\mathrm{P};[a]]=[\mathrm{R};[c]]$$

is determined by the equations

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$$\sum_{i=0}^{n}\binom{n}{i}c_{i}=\left(\sum_{i=0}^{n}\binom{n}{i}a_{i}\right)\left(\sum_{i=0}^{n}\binom{n}{i}b_{i}\right)$$

(9)

The power$[\mathrm{P};[a]]^{m}=\left[{}_{m}\mathrm{P};\left[{}_{m}a\right]\right]$is given by

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$$\begin{gathered}\sum_{i=0}^{n}\binom{n}{i}_{m}a_{i}=\left(\sum_{i=0}^{n}\binom{n}{i}a\right)^{m}\\ n=0,1,2,\ldots\end{gathered}$$

From equations (9) we easily derive

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$$c_{n}=\sum_{i=0}^{n}\binom{n}{i}b_{i}\left(\sum_{r=0}^{i}\binom{i}{r}a_{n-i+r}\right)=\sum_{i=0}^{n}\binom{n}{i}a_{i}\left(\sum_{r=0}^{i}\binom{i}{r}b_{n-i+}\right).$$