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T. Popoviciu, Sur les indicateurs, Bull. Sc. de l’Ecole Polytechnique de Timişoara, 3 (1930) nos. 1-2, pp. 72-80 (in French).
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- The inverse problem of calculating the indicator consists in solving the equation:
(1)
varphi(N)=A.\varphi(\mathrm{N})=\mathrm{A} .
First, note that it suffices to search for even numbers N satisfying (1). Indeed, from property 30, it follows that:
varphi(N)=varphi(2N)\varphi(\mathrm{N})=\varphi(2 \mathrm{~N})
ifNNis odd. We will constantly use this remark in what follows.
But before addressing the solution of equation (1), we must examine whether it is possible or not. It is obvious that if A is odd and not equal to 1, the equation is impossible, but it is easy to see that it can be impossible even if A is even. For example, the relation:
But before addressing the solution of equation (1), we must examine whether it is possible or not. It is obvious that if A is odd and not equal to 1, the equation is impossible, but it is easy to see that it can be impossible even if A is even. For example, the relation:
varphi(N)=14φ(N) = 14
is impossible.
More generally, one could propose solving the equation:
More generally, one could propose solving the equation:
{:(2)varphi_(i)(N)=A:}\begin{equation*} \varphi_{i}(\mathrm{~N})=\mathrm{A} \tag{2} \end{equation*}
Given A. It is evident that if equation (2) is possible fori=mi=m, all equations (2), wherei=1,2,dots m-1i=1,2, \ldots m-1will be possible and if the equation withi=mi=mis impossible, all equations withi=m+1i=m+1,m+2,dotsm+2, \ldotswill be impossible.
Yes, the equation:
varphi_(i)(N)=A\varphi_{i}(\mathrm{~N})=\mathrm{A}
Given that it is possible, the following:
varphi_(i+1)(N)=A\varphi_{i+1}(\mathrm{~N})=\mathrm{A}
is impossible, we will say that A can be an indicator of order i.
If equation (2) is possible for allii, A can be an indicator of any order.
If equation (2) is possible for allii, A can be an indicator of any order.
The formula:
(3)
(3)
varphi(2^(alpha+1))=2^(alpha)\varphi\left(2^{\alpha+1}\right)=2^{\alpha}
Let's take:
varphi_(1)(N)=varphi(N),quadvarphi_(2)(N)=varphi(varphi_(1)((N))),quadvarphi_(3)(N)=varphi(varphi_(2)((N))),quad" etc."\varphi_{1}(\mathrm{~N})=\varphi(\mathrm{N}), \quad \varphi_{2}(\mathrm{~N})=\varphi\left(\varphi_{1}(\mathrm{~N})\right), \quad \varphi_{3}(\mathrm{~N})=\varphi\left(\varphi_{2}(\mathrm{~N})\right), \quad \text { etc. }
varphi_(i)(N)\varphi_{i}(\mathrm{~N})will be the iterated indicator or simply the indicator of order i of
N.1^(0)1^{0}And2^(0)2^{0}it follows that there exists a numbernndependent on N such that:
N.1^(0)1^{0}And2^(0)2^{0}it follows that there exists a numbernndependent on N such that:
varphi_(n)(N)=1^(1))\varphi_{n}(\mathrm{~N})=1^{1)}
varphi_(n)(N)\varphi_{n}(\mathrm{~N})is then the last indicator of N.
This shows us that powers of 2 can be indicators of any order. Similarly, the equality:
{:(4)varphi(2^(alpha)3^(beta-1-1))=2^(alpha)3^(beta):}\begin{equation*} \varphi\left(2^{\alpha} 3^{\beta-1-1}\right)=2^{\alpha} 3^{\beta} \tag{4} \end{equation*}
proves to us that numbers of the form2^(alpha)3^(beta)(a!=0)2^{\alpha} 3^{\beta}(a \neq 0)can be indicators of any order.
{:(4)varphi(2^(alpha)3^(beta-1-1))=2^(alpha)3^(beta):}\begin{equation*} \varphi\left(2^{\alpha} 3^{\beta-1-1}\right)=2^{\alpha} 3^{\beta} \tag{4} \end{equation*}
proves to us that numbers of the form2^(alpha)3^(beta)(a!=0)2^{\alpha} 3^{\beta}(a \neq 0)can be indicators of any order.
We propose to demonstrate that:
The only numbers that can be indicators of any order are numbers of the form:
The only numbers that can be indicators of any order are numbers of the form:
{:(5)2^(alpha)","quad" ou "2^(alpha)3^(beta)","quad" avec "quad a!=0:}\begin{equation*}
2^{\alpha}, \quad \text { ou } 2^{\alpha} 3^{\beta}, \quad \text { avec } \quad a \neq 0 \tag{5}
\end{equation*}
We need only examine the even numbers that are not of the form (5). Before continuing, note that if:
varphi(N)=2^(alpha)A\varphi(\mathrm{N})=2^{\alpha} \mathrm{A}
A being odd and not a power of 3 (A=1=3^(0)\mathrm{A}=1=3^{0}is a power of 3), the number N cannot be of any of the forms (5). We will always keep this remark in mind.
3. Lemma. Let p be a positive integer, in what follows:
3. Lemma. Let p be a positive integer, in what follows:
p_(1)=2p+1,quadp_(2)=2p_(1)+1,quadp_(3)=2p_(2)+1,dotsp_{1}=2 p+1, \quad p_{2}=2 p_{1}+1, \quad p_{3}=2 p_{2}+1, \ldots
heyyhas at least one non-prime number.
Indeed:
1^(0)1^{0}. If :
Indeed:
1^(0)1^{0}. If :
p-=0quad(mod3)p \equiv 0 \quad(\bmod 3)
we have:
{:[p_(1)-=1,(mod3)],[p_(2)-=0,(mod3)]:}\begin{array}{ll}
p_{1} \equiv 1 & (\bmod 3) \\
p_{2} \equiv 0 & (\bmod 3)
\end{array}
and it is evident thatp_(2) > 3p_{2}>3.
2^(@)*Si:2^{\circ} \cdot \mathrm{Si}:
p-=1p \equiv 1
(mod. 3),
we have:
we have:
{:(mod.3)p_(1)-=0:}\begin{equation*}
p_{1} \equiv 0 \tag{mod.3}
\end{equation*}
unlessppnot equal to 1. In this casep_(1)=3p_{1}=3And:
p_(8)-=0quad(mod3)p_{8} \equiv 0 \quad(\bmod 3)
- If :
{:(mod.3)p-=-1:}\begin{equation*}
p \equiv-1 \tag{mod.3}
\end{equation*}
we can ask:
p=3k-1p=3 k-1
So what?
{:[p_(1)=2*3*k-1],[p_(2)=2^(2)*3*k-1],[p_(3)=2^(3)*3*k-1]:}\begin{aligned}
& p_{1}=2 \cdot 3 \cdot k-1 \\
& p_{2}=2^{2} \cdot 3 \cdot k-1 \\
& p_{3}=2^{3} \cdot 3 \cdot k-1
\end{aligned}
Let's state:
3k-1=2^(m)*k_(1)3 k-1=2^{m} \cdot k_{1}
k_(1)k_{1}being odd, then of congruence:
2^(varphi(k_(1)))-=12^{\varphi\left(k_{1}\right)} \equiv 1
we deduce:
{:(1)2^(varphi(k_(1)))3*h-=2^(m)k_(1)+1-=1:}\begin{equation*}
2^{\varphi\left(k_{1}\right)} 3 \cdot h \equiv 2^{m} k_{1}+1 \equiv 1 \tag{1}
\end{equation*}
SOp_(varphi(k_(1)))p_{\varphi\left(k_{1}\right)}is divisible byk_(1)k_{1}which is certainly smaller than him. The reasoning is flawed ifk_(1)=1k_{1}=1. In this case:
3k-1=2^(m)=2^(2n+1)3 k-1=2^{m}=2^{2 n+1}
and we have:
{:[2^(2n)*3k-=1,(mod5),n" impair "],[2^(2n+5)*3k-=1,(mod5),n" pair "]:}\begin{array}{lll}
2^{2 n} \cdot 3 k \equiv 1 & (\bmod 5) & n \text { impair } \\
2^{2 n+5} \cdot 3 k \equiv 1 & (\bmod 5) & n \text { pair }
\end{array}
The lemma is therefore proven.
4. We will now prove the property, first for the case of a number of the form:
4. We will now prove the property, first for the case of a number of the form:
2 A
(Odd)!=3^(p)\neq 3^{p}).
We divide the demonstration into three parts.
I. A number of the form:
We divide the demonstration into three parts.
I. A number of the form:
2p2 p
p being prime> 3>3cannot be an indicator of just any order.
From the equation:
From the equation:
varphi(N)=2p\varphi(\mathrm{N})=2 p
We deduce:
N=2q^(m)\mathrm{N}=2 q^{m}
(qqfirst)
hence:
hence:
varphi(2q^(m))=q^(m-1)(q-1)=2p\varphi\left(2 q^{m}\right)=q^{m-1}(q-1)=2 p
SO:
1^(0).quad m=2,quad q=p=3quad1^{0} . \quad m=2, \quad q=p=3 \quadimpossible by hypothesis.
2^(@).quad m=1,quad q=2p+1=p_(1)2^{\circ} . \quad m=1, \quad q=2 p+1=p_{1}.
Ifp_(1)p_{1}is not first,2p2 pcannot be an indicator. Otherwise:
1^(0).quad m=2,quad q=p=3quad1^{0} . \quad m=2, \quad q=p=3 \quadimpossible by hypothesis.
2^(@).quad m=1,quad q=2p+1=p_(1)2^{\circ} . \quad m=1, \quad q=2 p+1=p_{1}.
Ifp_(1)p_{1}is not first,2p2 pcannot be an indicator. Otherwise:
varphi(2p_(1))=2p\varphi\left(2 p_{1}\right)=2 p
Ifp_(2)=2p_(1)+1p_{2}=2 p_{1}+1is not first,2p_(1)2 p_{1}therefore cannot be an indicator2p2 pIt cannot be a second-order indicator. Otherwise.
varphi_(2)(2p_(2))=varphi_(1)(2p_(1))=2p\varphi_{2}\left(2 p_{2}\right)=\varphi_{1}\left(2 p_{1}\right)=2 p
We finally see that in order for2p2 peither an order indicatoriiand not of orderi+1i+1The numbers must be:
p_(1)=2p+1,quadp_(2)=2p_(1)+1,dotsquadp_(i)=2p_(i-1)+1p_{1}=2 p+1, \quad p_{2}=2 p_{1}+1, \ldots \quad p_{i}=2 p_{i-1}+1
are first and that2p_(i)+12 p_{i}+1not be prime. The property in question then follows from the proven lemma.
II. A number of the form:
II. A number of the form:
{:(a>1)2p^(ℓ):}\begin{equation*}
2 p^{\ell} \tag{a>1}
\end{equation*}
ppbeing first> 3>3cannot be an indicator of any order.
The equation:
varphi(N)=2p^(alpha)\varphi(\mathrm{N})=2 p^{\alpha}
gives again:
N=2q^(''')\mathrm{N}=2 q^{\prime \prime \prime}
(q first)
and:
and:
q^(m-1)(q-1)=2p^(e)q^{m-1}(q-1)=2 p^{e}
SO.
1^(0)m=a+1,quad q=p=31^{0} m=a+1, \quad q=p=3impossible by hypothesis.
2^(0)quad m=1,quad q=2p^('')+1=p_(1)2^{0} \quad m=1, \quad q=2 p^{\prime \prime}+1=p_{1}.
Ifp_(1)p_{1}is not first,2p^(alpha)2 p^{\alpha}cannot be an indicator. Otherwise:
1^(0)m=a+1,quad q=p=31^{0} m=a+1, \quad q=p=3impossible by hypothesis.
2^(0)quad m=1,quad q=2p^('')+1=p_(1)2^{0} \quad m=1, \quad q=2 p^{\prime \prime}+1=p_{1}.
Ifp_(1)p_{1}is not first,2p^(alpha)2 p^{\alpha}cannot be an indicator. Otherwise:
f(2p_(1))=2p^(a)f\left(2 p_{1}\right)=2 p^{a}
and the2p_(1)2 p_{1}is of a form already studied. Property II then follows from property I.
III. A number of the form:
2p_(1)^(u_(1))-p_(2)^(u_(2))cdotsp_(v)^(u_(v))quad(p_(1) < p_(2) < dots < p_(v),v > 1)2 p_{1}^{u_{1}}-p_{2}^{u_{2}} \cdots p_{v}^{u_{v}} \quad\left(p_{1}<p_{2}<\ldots<p_{v}, v>1\right)
p_(1),p_(2),dotsp_(0)p_{1}, p_{2}, \ldots p_{0}being prime, cannot be an indicator of any order.
We must always have:
N=2q^(m)\mathrm{N}=2 q^{m}
(because if N contained several odd prime factors,varphi(N)\varphi(\mathrm{N})would be divisible at least purely2^(2)2^{2}).
Therefore, we have:
q^(m-1)(q-1)=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))dotsp_(v)^(alpha_(n))q^{m-1}(q-1)=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{v}^{\alpha_{n}}
1^(0)1^{0}It is obvious that one can never haveq=p_(1),p_(2),dotsq=p_{1}, p_{2}, \ldotsonp_(0-1)p_{0-1}.
2^(0)2^{0}.
2^(0)2^{0}.
m=a_(v)+1quad q=p_(v)=2p_(1)^(a_(1))p_(2)^(a_(2))dotsp_(v-1)^(a_(v)-1)+1m=a_{v}+1 \quad q=p_{v}=2 p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{v-1}^{a_{v}-1}+1
So either this equality does not exist, or we have:
varphi(2p_(v)^(alpha_(v))+1)=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))cdotsp_(v)^(alpha_(v))\varphi\left(2 p_{v}^{\alpha_{v}}+1\right)=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{v}^{\alpha_{v}}
and the2p_(v)^(alpha_(v)+1)2 p_{v}^{\alpha_{v}+1}is of a form already studied. Property III follows from property II.
3^(0)3^{0}.
3^(0)3^{0}.
m=1,quad4=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))dotsp_(v)^(alpha_(v))+1=p^(')m=1, \quad 4=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{v}^{\alpha_{v}}+1=p^{\prime}
Ifp^(')p^{\prime}If the product is not prime, equality is impossible. Otherwise:
varphi(2p^('))=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))cdotsp_(v)^(alpha_(v))\varphi\left(2 p^{\prime}\right)=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{v}^{\alpha_{v}}
and property III then follows from property 1.
We can therefore state the following proposition:
We can therefore state the following proposition:
A number of the form:
Since A is an odd number not a power of 3, it cannot be an indicator of any order.
5. We will now show that the numbers:
5. We will now show that the numbers:
2^(a)Aquad(" A impair "!=3^(3))2^{a} \mathrm{~A} \quad\left(\text { A impair } \neq 3^{3}\right)
enjoy the same property. The proposition has been demonstrated fora=1a=1It is therefore sufficient to prove that it remains true fora=ka=kassuming it to be trueu=1,2,dots k-1u=1,2, \ldots k-1We break down the proof as before for the casea=1a=1IV
. A number of the form:
. A number of the form:
2^(k)p2^{k} p
(p prime > 3)
cannot be an indicator of any order.
The equation:
cannot be an indicator of any order.
The equation:
varphi(N)=2^(k)p\varphi(\mathrm{N})=2^{k} p
gives us:
(6)
(6)
N=2^(i)p^(j)q_(1)q_(2)dotsq_(1)\mathrm{N}=2^{i} p^{j} q_{1} q_{2} \ldots q_{1}
(q_(1),q_(2),dotsq_(r)q_{1}, q_{2}, \ldots q_{r}first!=p\neq p) and distinct from others,
hence:
hence:
{:[q(N)=2^(i-1)p^(j-1)(p-1)(q_(1)-1)(q_(2)-1)dots(q_(r)-1),j!=0],[p(N)=2^(i-1)(q_(1)-1)(q_(2)-1)dots(q_(r)-1),j=0]:}\begin{array}{ll}
q(\mathrm{~N})=2^{i-1} p^{j-1}(p-1)\left(q_{1}-1\right)\left(q_{2}-1\right) \ldots\left(q_{r}-1\right) & j \neq 0 \\
p(\mathrm{~N})=2^{i-1}\left(q_{1}-1\right)\left(q_{2}-1\right) \ldots\left(q_{r}-1\right) & j=0
\end{array}
and it is necessary that:
(8)
(8)
{:[p=2^(n)+1],[q_(1)=2^(n_(1))p^(n_(1)^('))+1],[q_(2)=2^(n_(2))p^(n_(2)^('))+1],[***],[q_(r)=2^(n*)p^(n_(1)^('))+1]:}\begin{aligned}
& p=2^{n}+1 \\
& q_{1}=2^{n_{1}} p^{n_{1}^{\prime}}+1 \\
& q_{2}=2^{n_{2}} p^{n_{2}^{\prime}}+1 \\
& \cdot \cdot \cdot \\
& q_{r}=2^{n \cdot} p^{n_{1}^{\prime}}+1
\end{aligned}
1^('')1^{\prime \prime}. Ifj!=0j \neq 0we have:
{:[i-1+n+n_(1)+n_(2)+dots+n_(r)=k],[j-1+n_(1)^(')+n_(2)^(')+dots+n_(r)^(')=1]:}\begin{aligned}
& i-1+n+n_{1}+n_{2}+\ldots+n_{r}=k \\
& j-1+n_{1}^{\prime}+n_{2}^{\prime}+\ldots+n_{r}^{\prime}=1
\end{aligned}
We can't havei > ki>k. Ifi < ki<kthe number N obtained falls into the casea < ka<kfor which the property is true by hypothesis. Ifi < ki<kwe must have:
n=1,quadn_(1)+n_(2)+dots+n_(r)=0quad(n_(1)=n_(2)=dots=n_(r)=0)n=1, \quad n_{1}+n_{2}+\ldots+n_{r}=0 \quad\left(n_{1}=n_{2}=\ldots=n_{r}=0\right)
that's to sayN=2^(k)p^(2)N=2^{k} p^{2}Andp=3p=3which is excluded by hypothesis.
20. Ifj=0j=0we have:
20. Ifj=0j=0we have:
{:[i-1+n_(1)+n_(2)+dots+n_(r)=k],[n_(2)^(')+n_(2)^(')+dots+n_(r)^(')=1]:}\begin{gathered}
i-1+n_{1}+n_{2}+\ldots+n_{r}=k \\
n_{2}^{\prime}+n_{2}^{\prime}+\ldots+n_{r}^{\prime}=1
\end{gathered}
and the only case that needs to be studied isi=ki=k, SO:
{:[n_(1)=1","quad hat(n_(2))+dots+n_(r)=0","quad r=1],[n_(1)^(')=1]:}\begin{gathered}
n_{1}=1, \quad \hat{n_{2}}+\ldots+n_{r}=0, \quad r=1 \\
n_{1}^{\prime}=1
\end{gathered}
And :
varphi(2^(h)(2p+1))=2^(h)p\varphi\left(2^{h}(2 p+1)\right)=2^{h} p
Ifp_(1)=2p+1p_{1}=2 p+1If the product is not prime, equality is impossible. Otherwise2^(k)(2p+1)=2^(k)p_(1)2^{k}(2 p+1)=2^{k} p_{1}is of the same shape as2^(k)p2^{k} pIt is easy to see that everything boils down to showing that the following:
p_(1)=2p-1,quadp_(2)=2p_(1)+1,quadp_(3)=2p_(2)+1,dotsp_{1}=2 p-1, \quad p_{2}=2 p_{1}+1, \quad p_{3}=2 p_{2}+1, \ldots
contains at least one non-prime number, which follows from the proven lemma.
V. A number of the form:
V. A number of the form:
2^(h)p^(beta)quad(p" premier " > 3,beta > 3)2^{h} p^{\beta} \quad(p \text { premier }>3, \beta>3)
cannot be an indicator of just any order.
We always have the form (6) of N withj=0,1,2,dots beta+1j=0,1,2, \ldots \beta+1We will therefore write relations (7) and (8).
1^(@)1^{\circ}. Ifj!=0j \neq 0we have:
We always have the form (6) of N withj=0,1,2,dots beta+1j=0,1,2, \ldots \beta+1We will therefore write relations (7) and (8).
1^(@)1^{\circ}. Ifj!=0j \neq 0we have:
{:[i-1+n+n_(1)+dots+n_(r)=k],[j-1+n_(1)^(')+n_(2)^(')+dots+n_(r)^(')=beta.]:}\begin{aligned}
& i-1+n+n_{1}+\ldots+n_{r}=k \\
& j-1+n_{1}^{\prime}+n_{2}^{\prime}+\ldots+n_{r}^{\prime}=\beta .
\end{aligned}
We can still see that the casei > ki>kis impossible and thati < ki<kreduces to a supposedly proven case. It therefore remainsi=ki=kas the last possible case. But in this casen=1n=1Andp=3p=3, which is by definition impossible.
2^(0)2^{0}. Ifj=0j=0The only case to consider isi=ki=k, SO :
2^(0)2^{0}. Ifj=0j=0The only case to consider isi=ki=k, SO :
{:[n_(1)=1","quadn_(2)+n_(3)+dots+n_(r)=0","quad r=1],[n_(1)^(')=beta]:}\begin{gathered}
n_{1}=1, \quad n_{2}+n_{3}+\ldots+n_{r}=0, \quad r=1 \\
n_{1}^{\prime}=\beta
\end{gathered}
And:
N=2^(k)(2p^(alpha)+1)\mathrm{N}=2^{k}\left(2 p^{\alpha}+1\right)
assuming of course that2p^(c)+12 p^{c}+1is prime. But the number2^(h)(2p^(c)+1)2^{h}\left(2 p^{c}+1\right)taken as an indicator has already been studied and property V results from property IV.
Finally, we must demonstrate that:
VI. A number of the form:
VI. A number of the form:
2^(h)p_(1)^(alpha_(1))p_(2)^(alpha_(2))cdotsp_(v)^(alpha_(v))quad(p_(1),p_(2),dotsp_(v)" premiers, "v > 1)2^{h} p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{v}^{\alpha_{v}} \quad\left(p_{1}, p_{2}, \ldots p_{v} \text { premiers, } v>1\right)
cannot be an indicator of any order.
We must have:
N=2^(i)p_(1)^(j_(1))p_(2)^(j_(2))dotsp_(v)^(j_(v))q_(1)q_(1)dotsq_(,)quad([q_(1)","q_(2)","dotsq_(,)"premiers distincts "],[" et distincts de "p_(1)","p_(2)","dotsp_(v)],[j_(2)=0","1","dotsa_(lambda)+1","lambda=1","2","dots v])\mathrm{N}=2^{i} p_{1}^{j_{1}} p_{2}^{j_{2}} \ldots p_{v}^{j_{v}} q_{1} q_{1} \ldots q_{,} \quad\left(\begin{array}{l}
q_{1}, q_{2}, \ldots q_{,} \text {premiers distincts } \\
\text { et distincts de } p_{1}, p_{2}, \ldots p_{v} \\
j_{2}=0,1, \ldots a_{\lambda}+1, \lambda=1,2, \ldots v
\end{array}\right)
and we obtain:
{:[p(N)=2^(i-1)p_(s)^(j_(s)-1)p_(s+1)^(j_(s+1)-1)dotsp_(v)^(j_(v)-1)(p_(s)-1)(p_(s+1)-1)dots(p_(v)-1)],[(q_(1)-1)(q_(2)-1)dots(q_(v)-1)],[(j_(1)=j_(2)=dots=j_(s-1)=0,j_(s)=j_(s+1),dots j!=0:}],[p(N)=2^(i-1)(q_(1)-1)(q_(2)-1)dots(q_(1)-1)quad(j_(1)=j_(2)=dots=j_(v)=0).]:}\begin{array}{r}
p(\mathrm{~N})=2^{i-1} p_{s}^{j_{s}-1} p_{s+1}^{j_{s+1}-1} \ldots p_{v}^{j_{v}-1}\left(p_{s}-1\right)\left(p_{s+1}-1\right) \ldots\left(p_{v}-1\right) \\
\left(q_{1}-1\right)\left(q_{2}-1\right) \ldots\left(q_{v}-1\right) \\
\left(j_{1}=j_{2}=\ldots=j_{s-1}=0, j_{s}=j_{s+1}, \ldots j \neq 0\right. \\
p(\mathrm{~N})=2^{i-1}\left(q_{1}-1\right)\left(q_{2}-1\right) \ldots\left(q_{1}-1\right) \quad\left(j_{1}=j_{2}=\ldots=j_{v}=0\right) .
\end{array}
To satisfy these equalities, we must assume that:
{:[p_(mu)=2^(m_(mu))B_(mu)+1],[q_(mu)=2^(n_(mu))B_(mu)^(')+1]:}\begin{aligned}
& p_{\mu}=2^{m_{\mu}} \mathrm{B}_{\mu}+1 \\
& q_{\mu}=2^{n_{\mu}} \mathrm{B}_{\mu}^{\prime}+1
\end{aligned}
for allp_(mu)p_{\mu}which intervene in N as factors, and forq_(mu),mu=1q_{\mu}, \mu=1,2,dots r,B_(u),B_(u)^(')2, \ldots r, \mathrm{~B}_{u}, \mathrm{~B}_{u}^{\prime}are numbers of the form:
p_(v_(1))^(n_(1)^('))p_(v_(2))^(n_(2)^('))dotsq_(v_(s))^(n_(s)^('))p_{v_{1}}^{n_{1}^{\prime}} p_{v_{2}}^{n_{2}^{\prime}} \ldots q_{v_{s}}^{n_{s}^{\prime}}
v_(1),v_(2),dotsv_(s)v_{1}, v_{2}, \ldots v_{s}beingssnumbers of the sequence1,2,dots nu1,2, \ldots \nu.
1^('')1^{\prime \prime}. Ifj_(1)=j_(2)=dots=j_(s-1)=0,j_(s)!=0,j_(s+1)!=0,dotsj_(v)!=0j_{1}=j_{2}=\ldots=j_{s-1}=0, j_{s} \neq 0, j_{s+1} \neq 0, \ldots j_{v} \neq 0, We have :
1^('')1^{\prime \prime}. Ifj_(1)=j_(2)=dots=j_(s-1)=0,j_(s)!=0,j_(s+1)!=0,dotsj_(v)!=0j_{1}=j_{2}=\ldots=j_{s-1}=0, j_{s} \neq 0, j_{s+1} \neq 0, \ldots j_{v} \neq 0, We have :
i-1+sum_(mu=s)^(v)m_(mu)+sum_(mu=1)^(y)n_(mu)=ki-1+\sum_{\mu=s}^{v} m_{\mu}+\sum_{\mu=1}^{y} n_{\mu}=k
and the only case that requires study isi=ki=kSo we deduce:
s=v,quadm_(v)=1,quadn_(1)+n_(2)+dots+n_(r)=0s=v, \quad m_{v}=1, \quad n_{1}+n_{2}+\ldots+n_{r}=0
And:
N=2^(k)P_(v)^(a_(v)+1)\mathrm{N}=2^{k} \mathrm{P}_{v}^{a_{v}+1}
with:
p_(v)=2p_(1)^(alpha_(1))p_(2)^(alpha_(3))cdotsp_(v-1)^(alpha_(v-1))+1p_{v}=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{3}} \cdots p_{v-1}^{\alpha_{v-1}}+1
If such an equality is possible (by interchanging the order of the factors if necessary in2^(k)p_(1)^(alpha_(1))p_(2)^(alpha_(2))dotsp_(v)^(alpha_(v))2^{k} p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{v}^{\alpha_{v}}), the number N falls into a category already studied and the propertyv_(1)v_{1}results from property V:
2^(0)2^{0}. Ifj_(1)=j_(2)=dots=j_(v)=0j_{1}=j_{2}=\ldots=j_{v}=0, We have :
2^(0)2^{0}. Ifj_(1)=j_(2)=dots=j_(v)=0j_{1}=j_{2}=\ldots=j_{v}=0, We have :
i-1+sum_(mu=1)^(x)n_(mu)=ki-1+\sum_{\mu=1}^{x} n_{\mu}=k
done ifi=k,r=1i=k, r=1Andn_(1)=1n_{1}=1The number N is of the form:
2^(k)q2^{k} q
with:
q=2p_(1)^(alpha_(1))p_(2)^(alpha_(2))dotsp_(v)^(alpha_(v))+1q=2 p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{v}^{\alpha_{v}}+1
and when this equality is possible we are brought back to property IV.
We can therefore state:
Theorem. A number of the form:
Theorem. A number of the form:
{:((k!=0)")"2^(k)A:}\begin{equation*}
2^{k} \mathrm{~A} \tag{$k\neq0$}
\end{equation*}
Since A is odd and not a power of 3, it cannot be an indicator of any order.
It follows that for such a number one can determine aiisuch that the equation:
varphi_(i)(N)=2^(k)A\varphi_{i}(\mathrm{~N})=2^{k} \mathrm{~A}
either possible and the following:
varphi_(i+1)(N)=2^(k)A\varphi_{i+1}(\mathrm{~N})=2^{k} \mathrm{~A}
impossible.
The proof given for the stated theorem can be used as a calculation method for the numberii.
The proof given for the stated theorem can be used as a calculation method for the numberii.
- We studied the numbernnin a short work which will soon appear in the "Gazeta Matematică".