1. 1.

   J.A. Blake noted [1] that if we consider all irreducible ordinary fractions, smaller than 1 and whose denominator is$\leq n$, the sum of the denominators is equal to twice the sum of the numerators. This is an immediate consequence of the property that if we consider all ordered pairs of natural numbers ($a,b$) Or$a<b=n$, the numbers$has$And$b$being relatively prime, the sum$S_{2}(n)$secondary components$b$is equal to 2 times the sum$S_{1}(n)$the first components$has$Moreover, we have$S_{2}(n)=n\varphi(n)$And$S_{1}(n)=\frac{1}{2}n\varphi(n)$is equal to the sum of the natural numbers$\leqq n$and first with$n$It is assumed, of course, that$n>1$And$\varphi(n)$is the well-known Euler function! (1' "indicator" of$n$).
   I. Katz [2] and A. Zane [3] revisited the previous result of JA Blake.

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In what follows, we will give a generalization of this property.
2. We will demonstrate the:

Theorem. Let$k,n$two given natural numbers, where$2\leqq k\leqq n$and consider all increasing sequences ($a_{1}, a_{2}, ..., a_{k}$) of$k$natural numbers where:$a).a_{k}=n,b$).$a_{1}, a_{2}, ..., a_{k}$are relatively prime. Let us designate by$S_{\nu}(n)$the sum of$v^{\text{viths}}$terms$a_{v}$of all these sequels for$v=1,2,\ldots,k$.

We then have
(1)$\quad S_{\nu}(n)=\nu S_{1}(n),\nu=1,2,\ldots,k$.

We will deduce the proof of this theorem from a similar property of sequences.$\left(a_{1},a_{2},\ldots,a_{k}\right)$whose terms are not subject to restriction b) of the text. We will therefore demonstrate the:

Lemma. Let$k,n$two given natural numbers, where$2\leqq k\leqq n$and consider all increasing sequences ($a_{1}, a_{2}, ..., a_{k}$) of$k$natural numbers$a_{1}, a_{2}, ..., a_{k}$, the last term of which is equal to$n$Let us designate by$A_{v}(n)$the sum of$v^{\text{ths}}$terms of all these sequences for$v=1,2,\ldots,k$.

We then have

1. 3.

   Let's move on to the proof of the lemma. We can calculate the sums$A_{v}(n)$by induction, but they can also be obtained quickly by using a certain "generating function". Let's consider the generating function of$k$variables$x_{1},x_{2},\ldots,x_{k},F=F\left(x_{1},x_{2},\ldots,x_{k}\right)==\sum x_{1}^{a_{1}}x_{2}^{a_{2}}\ldots x_{k}^{a_{k}}$the summation being extended to all increasing sequences$\left(a_{1},a_{2},\ldots,a_{k}\right)$of natural numbers. We then have

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To simplify the notation, let [f] denote the function of a variable$z$which is obtained from the function$f\left(x_{1},x_{2},\ldots,x_{k}\right)$by asking$x_{1}=x_{2}=\ldots==x_{k-1}=1,x_{k}=z$We then have

For$v=1,2,\ldots,k-1$And

For$n<\nu,\nu=1,2,\ldots,k$the coefficients$A_{\psi}(n)$do not appear in the lemma. They are defined by formulas (4) and (5). We will see, moreover, that they are all zero. To calculate the left-hand sides of equalities (4) and (5), note that

Therefore, if we apply the rule for differentiating a product, we obtain

Formulas (2) follow immediately, and the lemma is thus proven. Moreover, the expression for$A_{v}(n)$in the form

1. 4.

   We will now demonstrate the stated theorem. Let us denote by$\mathfrak{M}(n)$the set of all sequences of$k$natural numbers that appear in the lemma and by$\mathfrak{M}_{a}(n)$the subset formed by the elements of$\mathfrak{M}(n)$whose terms are all divisible by the number$d$It is clear that$d$is a divisor of$n$We deduce the set$\mathfrak{M}_{d}(n)$by multiplying by$d$each term of the elements of$\mathfrak{M}\left(\frac{n}{d}\right)$The result is that the whole$\mathfrak{M}_{d}(n)$enjoys the same ownership as the whole$\mathfrak{M}(n)$The sums$A_{v}(n),\nu=1,2,\ldots,k$corresponding to the set$\mathfrak{M}_{d}(dn)$are equal to$dA_{v}\left(\frac{n}{d}\right),v=1,2,\ldots,k$respectively.

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To obtain the sequences that appear in the theorem, we must remove from$\mathfrak{M}(n)$the elements (sequences) whose terms are not coprime. Let us denote by$p_{1},p_{2},\ldots,p_{r}$(all) the (distinct) prime factors of the number$n$The deleted sequels of$\mathfrak{M}(n)$are those for which the greatest common divisor (GCD) of the terms is divisible by at least one of the prime numbers$p_{1},p_{2},\ldots,p_{r}$By applying the well-known principle of inclusion and exclusion, we obtain

where the summons$\sum$is extended to all combinations$i_{1},i_{2},\ldots,i_{s}$, & has$s$clues$1^{(8)},2,\ldots,r$.

Given (2), the equalities (1) follow. The theorem is therefore proven.
5. From formulas (6) we also deduce:

Or$\binom{n-1}{k-1}$is precisely the number of elements of$\mathfrak{M}(n)$Similarly, if we designate by$N(n)$the number of sequences that appear in the theorem, we have

Finally, from formula (7), for$v=k$, we deduce

For$k=2$We have$N(n) = φ(n)$Or$\varphi(n)$is the Euler function (the indicator). For$k>2$the arithmetic structure of the number$N(n)$Therefore, also numbers$S_{\nu}(n),\nu=1,2,\ldots,k$is more complicated. Let's designate by$C_{k}^{v},v=0,1,\ldots,7-1$the factorial coefficients of rank$h_{1}$defined by polynomial equality

Let us also consider arithmetic functions

THE$p_{i},i=1,2,\ldots,r$being always the distinct primary factors of$n$For$\nu=1$we find the Euler function$\varphi_{1}(n)=\varphi(n)$. In general,$\varphi_{v}(n)$is equal to the number of sequences of v natural numbers whose terms are$\leqq n$and have their greatest common divisor (GCD) prime with$n$.

If we notice that$\binom{n-1}{k-1}=\frac{(n-1)(n-2)\ldots(n-k+1)}{(k-1)!}$From formula (8) we deduce

1. 6.

   We can draw some conclusions from this and we can generalize in several ways what we have demonstrated.

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The property we are examining in this work remains true for the set of increasing sequences of relatively prime natural numbers whose last term is$\leqq n$This results from the fact that the set under consideration is the union of the sets$\mathfrak{M}(v),v=k,k+1,\ldots,n$which are disjoint in pairs. Therefore, in this case$S_{y}^{\prime}(n)$is the sum of the sixth terms and$N^{\prime}(n)$the number of sequences in the set considered, these arithmetic functions are the summation functions of the functions$S_{\mathrm{y}}(n)$And$N(n)$respectively. We have done

where, moreover, the$S_{v}(i),N(i)$are useless for$i<k$7.
The property stated in the theorem is also true for the set of all non-decreasing sequences of$k$coprime, non-negative integers whose last term is equal to$n$If we designate in this case by$S_{v}^{*}(n),v=1,2,\ldots,k$And$N^{*}(n)$the numbers corresponding to$S_{v}(n),v=1,2,\ldots,k$And$N(n)$respectively, we have

And

This is a formula analogous to formula (9).
This result is obtained by establishing a property entirely analogous to that stated in the lemma and relating to the set of all non-decreasing sequences of$k$integers whose last term is equal to$n$(the terms are not necessarily relatively prime). The number of these sequences is equal to$\binom{n+k-1}{k-1}=\frac{(n+1)(n+2)\ldots(n+k-1)}{(k-1)!}$There is no need to detail the proof. I will simply say that this time, instead of (3), we need to rely on the generating function.

We can obviously also give a property analogous to the consequence examined in n ⁰ 6. I propose to the reader to state this property.