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\topmatter
 \leftheadtext{\hskip 8pt \smalltt INTEGERS: 
\smallrm Electronic J. of Combinatorial Number Theory \smalltt 5
(2005), \#A12 }
 \rightheadtext{\smalltt INTEGERS: 
\smallrm Electronic J. of Combinatorial Number Theory \smalltt x
(2005), \#A12 \hskip 8pt}
 \endtopmatter

%\title
%On Consecutive Integer Pairs With the Same Sum of Distinct Prime Divisors
%\endtitle
%\rightheadtext{On Consecutive Integer Pairs}
%\leftheadtext{D. Iannucci and A. Mintos}
%\author
%Douglas E. Iannucci\\
%Alexia S. Mintos\\
%\endauthor
%\affil
%University of the Virgin Islands
%\endaffil
%\address
%St. Thomas VI 00802
%USA
%\endaddress
%\email
%diannuc\@uvi.edu
%\endemail
%\keywords
%Ruth--Aaron pairs, cyclotomic polynomials, Pell sequences, primes
%\endkeywords
%\subjclass
%11A25, 11Y55 
%\endsubjclass

\document

\centerline{\smalltt INTEGERS: 
\smallrm Electronic J. of Combinatorial Number Theory \smalltt 5
(2005), \#A12 }
\vskip 20pt


 \centerline{\uppercase{\bf On Consecutive Integer Pairs With the Same
Sum}}
 \centerline{\uppercase{\bf of Distinct Prime Divisors}}
 \vskip 20pt
 \centerline{\smallit Douglas E. Iannucci, University of the Virgin
Islands, St. Thomas VI 00802,
USA}
 \centerline{\tt diannuc\@uvi.edu} 
 \vskip 20pt
 \centerline{\smallit Alexia S. Mintos, University of the Virgin
Islands, St. Thomas VI 00802,
USA}
 \vskip 30pt
 \centerline{\smallit Received: 1/10/05, Accepted: 5/28/05, Published:
6/29/05}
 \vskip 30pt

\centerline{\bf Abstract}

We define the arithmetic function $P$ by $P(1)=0$, 
and $P(n)=p_1+p_2+\cdots+p_k$ if $n$ has the unique prime factorization
given by $n=\prod_{i=1}^{k}p_i^{a_i}$; we also define $\omega(n)=k$ and
$\omega(1)=0$. We study pairs $(n,n+1)$ of consecutive integers such that
$P(n)=P(n+1)$. We prove that $(5,6)$, $(24,25)$, and $(49,50)$ are the
only such pairs $(n,n+1)$ where $\{\omega(n),\omega(n+1)\}=\{1,2\}$. We
also show how to generate certain pairs of the form $(2^{2n}pq,rs)$, with
$p<q$, $r<s$ odd primes, and lend support to a conjecture that infinitely
many such pairs exist.


\noindent
{\bf Keywords:} Ruth--Aaron pairs, cyclotomic polynomials, Pell
sequences, primes

\noindent
{\bf Subject Class:}  11A25, 11Y55 


%
%
%
%
%     Section 1   INTRODUCTION
%
%
\subhead \nofrills{1. Introduction} \endsubhead

For positive integers $n$, we define the arithmetic function $P(n)$ by\break $P(1)=0$, and, for a positive integer $n$ having as its unique prime factorization $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, 
$$P(n)=p_1+p_2+\cdots+p_k.$$
That is, $P(n)$ gives the sum of prime divisors of $n$ without multiplicity taken into account. The function is additive, in that $P(m)+P(n)=P(mn)$ if $(m,n)=1$.

This function compares to the arithmetic function defined for positive integers $n$ by $S(1)=0$ and $S(n)=\sum_{i=1}^{k}a_ip_i$ whenever $n=\prod_{i=1}^{k}p_i^{a_i}$; that is, $S(n)$ gives the sum of primes dividing $n$, taken with multiplicity. Then $S(n)$ is completely additive, in that $S(mn)=S(m)+S(n)$ for any two positive integers $m$ and $n$. A {\it Ruth--Aaron pair\/} is a pair $(n,n+1)$ of consecutive integers such that $S(n)=S(n+1)$. These were first discussed by Pomerance et. al.~[4], and have been the subject of several articles (such as by Pomerance~[6], Drost~[2]) and numerous websites since. 

However, in this article we are interested in finding pairs of consecutive positive integers $(n,n+1)$, such that $P(n)=P(n+1)$. For the sake of easy reference, we may call these {\it Ruth--Aaron pairs of the second kind,\/} or RAP2s for short. Note, however, that a RAP2 is also an ordinary Ruth--Aaron pair if both members $n$ and $n+1$ are square-free. 

Some observations regarding RAP2s are immediate. For example, the members $(n,n+1)$ of a RAP2 are of opposite parity, and are relatively prime. Let $n$ be a positive integer. If $n$ has the unique prime factorization $n=\prod_{i=1}^{k}p_i^{a_i}$, then the prime powers $p_i^{a_i}$, $1\le i\le k$, are called the {\it components\/} of $n$, and we define $\omega(n)=k$, $\omega(1)=0$ (thus $\omega$ counts the components of $n$). For any given RAP2 $(n,n+1)$, since $2$ divides exactly one of the members (all other prime divisors of $n$ and $n+1$ being odd), we see that $\omega(n)$ and $\omega(n+1)$ are of opposite parity. In this article we shall completely determine all RAP2s $(n,n+1)$ whose members have one or two components. We will also investigate RAP2s of the form $(2^{2n}pq,rs)$, with $p<q$, $r<s$ odd primes.  
%
%                               2. PRELIMINARIES
%
%
\vskip 30pt
\subhead \nofrills{2. Preliminaries} \endsubhead

If $p$ is a prime and $a$, $m$, are positive integers we write $p^m\|a$ if $p^m\mid a$ and $p^{m+1}\nmid a$.  In this case we say $p^m$ {\it exactly\/} divides $a$. For distinct primes $p$ and $q$ we write $e_p(q)$ to denote the exponent to which $q$ belongs modulo~$p$.

For positive integers $n$, we denote the $n^{th}$ cyclotomic polynomial evaluated at $x$ by $\Phi_n(x)$. The cyclotomic polynomials (as shown by Niven~[5], Ch. 3) may be defined inductively by
$$x^n-1=\prod_{d\mid n}\Phi_n(x).\tag1$$
By Theorems~94 and~95, Nagell~[3], Ch. 5, we have
\proclaim{Lemma 1}
Let $p$ be and $q$ be odd primes and let $m$ be a positive integer. Let $h=e_p(q)$. Then $p\mid\Phi_m(q)$ if and only if $m=hp^j$ for some integer $j\ge0$. If $j>0$ then $p\|\Phi_{hp^j}(q)$.
\endproclaim
\proclaim{Lemma 2}
Let $q$ be an odd prime and let $m$ be a positive integer. Then $2\mid\Phi_m(q)$ if and only if $m=2^j$ for some integer $j\ge0$. If $j>1$ then $2\|\Phi_{2^j}(q)$.
\endproclaim 
Let $q$ be prime and let $m>0$ be an integer. Since, by definition, 
$$\Phi_m(q)=\prod \Sb k=1\\(k,m)=1 \endSb ^{m-1}(q-e^{2\pi ik/m}),$$
and since $\Phi_m(q)>0$, we have
$$\Phi_m(q)=\prod \Sb k=1\\(k,m)=1 \endSb ^{m-1}\left|q-e^{2\pi ik/m}\right|,$$
and since $|q-e^{2\pi ik/m}|\ge q-1$ for $1\le k\le m-1$, we have
\proclaim{Lemma 3}
For a prime $q$ and an integer $m>0$ we have $\Phi_m(q)\ge(q-1)^{\phi(m)}$.
\endproclaim  
%
%                                3. < 2^a p^b , q^c >
\vskip 30pt
\subhead \nofrills{3. RAP2s of the form $(2^ap^b, q^c)$} \endsubhead

The smallest numbers of components the members of a RAP2 can have are~1 and~2. In this instance, the members of the RAP2 have the form $2^ap^b$ and $q^c$ for positive integers $a$, $b$, and $c$, where $p$ and $q$ are necessarily twin (odd) primes (that is, $p+2=q$). We have two cases arising in this instance, those being $2^ap^b=q^c\pm1$. In this section we consider the easier case of the two,
$$2^ap^b=q^c-1.\tag2$$
Clearly $c>1$, since $2^ap^b\ge2(q-2)=q+(q-4)>q-1$. Thus, since $q=p+2$, (2) factors as
$$2^ap^b=(p+1)(q^{c-1}+q^{c-2}+\cdots+q+1).$$
Since $(p,p+1)=1$, it follows that $p+1=2^t$ for some positive integer $t$. Hence $p=2^t-1$, $q=2^t+1$, which is possible only if $t=2$; that is, $p=3$ and $q=5$. Then~(2) becomes
$$2^a3^b=5^c-1.$$
Since $5^c\equiv1\pmod3$, we have $2\mid c$, so we write $c=2\gamma$ for some positive integer $\gamma$. Thus
$$2^a3^b=(5^{\gamma}-1)(5^{\gamma}+1).$$
Since $2\|5^{\gamma}+1$, we must have $3\mid5^{\gamma}+1$. Since $(5^{\gamma}+1,5^{\gamma}-1)=2$, we have $3\nmid5^{\gamma}-1$. Hence
$$5^{\gamma}-1=2^{a-1},\qquad5^{\gamma}+1=2\cdot3^b.$$
Certainly $\gamma$ is odd (as $3\nmid5^\gamma-1$). Suppose $\gamma>1$. Then
$$5^{\gamma}-1=(5-1)(5^{\gamma-1}+5^{\gamma-2}+\cdots+5+1).$$
But the second factor is odd, and greater than~1; this contradicts $5^{\gamma}-1=2^{a-1}$. Therefore $\gamma=1$, and so $c=2$. Hence~(2) becomes $2^3\cdot3=5^2-1$; that is, $a=3$, $b=1$,  and we have the RAP2 $(24,25)$. Hence the only RAP2 of the form $(2^ap^b, q^c)$ is $(24,25)$.
%
%                                4. < q^c , 2^a p^b >
%
\vskip 30pt
\subhead \nofrills{4. RAP2s of the form $(q^c, 2^ap^b)$} \endsubhead

Suppose now that
$$2^ap^b=q^c+1\tag3$$
for positive integers $a$, $b$, and $c$, where $p$ and $q$ are primes such that $p+2=q$. This case is more difficult than that in Section~3. 

By~(1), we see that~(3) is equivalent to
$$2^ap^b=\prod \Sb d\mid2c\\d\nmid c \endSb \Phi_d(q).\tag4$$
Let $h=e_p(q)$; we observe $h=e_p(2)$ as well (since $q=p+2$). By Lemmas~1 and~2, each divisor $d\mid2c$ such that $d\nmid c$ must either have the form $hp^j$ for some integer $j\ge0$ or the form $2^k$ for some integer $k\ge1$. Writing $c=2^ms$ for some integer $m\ge0$ and odd integer $s$, we see that $2^{m+1}\|d$ for all divisors $d\mid 2c$ such that $d\nmid c$. In particular $2^{m+1}\|h$. 

Suppose $s$ is composite. Then $t\mid s$ for some odd integer $t$ such that $1<t<s$. Then by~(4), $\Phi_s(q)\Phi_t(q)\mid2^ap^b$. This is impossible as $2\nmid\Phi_s(q)\Phi_t(q)$ by Lemma~2, and, as $2\mid h$, we have $h\nmid s$, and so $p\nmid\Phi_s(q)\Phi_t(q)$ by Lemma~1. Hence either $s$ is prime or $s=1$. 

Suppose $s$ is prime. Then $h=2^{m+1}s$. For, if this were not the case then we would have $h=2^{m+1}$; since $2\nmid\Phi_{2c}(q)$ by Lemma~2, it follows that either $p\nmid\Phi_{2c}(q)$ (if $s\ne p$) or $p\|\Phi_{2c}(q)$ (if $s=p$) by Lemma~1. The former possibility clearly contradicts~(4); the latter implies $\Phi_{2c}(q)=p$, which is impossible as $\Phi_{2c}(q)>p$ by Lemma~3.

Therefore, since $h=2^{m+1}s$, we have
$$2^ap^b=\Phi_{2^{m+1}s}(q)\Phi_{2^{m+1}}(q)\tag5$$
by~(4). This implies $m=0$ because otherwise~(5) is impossible since we have $p\nmid\Phi_{2^{m+1}}(q)$, $2\|\Phi_{2^{m+1}}(q)$,  and $\Phi_{2^{m+1}}(q)>2$ by Lemmas~1, 2, and~3 respectively. Therefore $h=2s$ and
$$2^ap^b=\Phi_2(q)\Phi_{2s}(q),$$ 
with $2^a=\Phi_2(q)=q+1$ and $p^b=\Phi_{2s}(q)$. But then $q=2^a-1$, so that $p=2^a-3$. It is clear that $a>2$, hence $p\equiv5\pmod8$. Thus~2 is a quadratic nonresidue of~$p$, and hence $2^{(p-1)/2}\equiv-1\pmod{p}$ by Euler's criterion. Since $2^2\mid p-1=2^a-4$, it follows that $2^2\mid e_p(2)$. But $e_p(2)=h$, and since $h=2s$, we have $2\|h$, a contradiction. 

Therefore $s=1$ and hence $c=2^m$ for some integer $m\ge0$. Thus~(3) becomes
$$2^ap^b=q^{2^m}+1.\tag6$$
First let us suppose that $m>1$. Then $q^{2^m}\equiv1\pmod{4}$ so that $a=1$. Hence
$$2p^b=q^{2^m}+1.\tag7$$
Since $p\mid q^{2^m}+1=\Phi_{2^{m+1}}(q)$, it follows from Lemma~1 that $h=2^{m+1}$; recalling as well $h=e_p(2)$, it follows that $p\equiv1\pmod{2^{m+1}}$. Since $e_p(2)=2^{m+1}$ and $\Phi_{2^{m+1}}(2)=2^{2^m}+1$, it follows from Lemma~1 that 
$$p\mid2^{2^m}+1.\tag{8}$$
Suppose $p=2^{m+1}t+1$ for some odd integer $t$. Then, as $2^{2^m}\equiv-1\pmod{p}$ by~(8), 
$$2^{(p-1)/2}=2^{2^mt}=(2^{2^m})^t\equiv(-1)^t\equiv-1\pmod{p},$$
and hence $\left(\frac2{p}\right)=-1$ by Euler's criterion, where $\left(\frac{\cdot}{\cdot}\right)$ denotes the Legendre symbol. But, $p\equiv1\pmod8$, which implies $\left(\frac2{p}\right)=+1$, a contradiction. Therefore
$$p\equiv1\pmod{2^{m+2}}.\tag9$$
Suppose $b>2^m$. Then from~(7) we have
$$2p^{b-2^m}=\left(1+\frac2{p}\right)^{2^m}+\frac1{p^{2^m}}.$$
By~(9), $p>2^{m+2}$, and so
$$2p^{b-2^m}<\left(1+\frac1{2^m}\right)^{2^m}+1<e+1<4,$$
which implies $2p<4$, a contradiction. On the other hand, suppose $b<2^m$. Then by~(6),
$$2=\left(1+\frac2{p}\right)^b(p+2)^{2^m-b}+\frac1{p^b}>(p+2)^{2^m-b}\ge p+2,$$
a contradiction. Therefore we must have $b=2^m$, so that~(7) becomes
$$2p^{2^m}=q^{2^m}+1.\tag{10}$$
Since $q=p+2$, (10) becomes
$$\align
p^{2^m}&=-p^{2^m}+(p+2)^{2^m}+1\\
&=\sum_{k=1}^{2^m}\binom{2^m}{k}p^{2^m-k}2^k+1.\tag{11}
\endalign$$
Since by~(9) $p>2^{m+2}$, we have for each $k$ such that $1\le k\le2^m$, 
$$\align
\binom{2^m}{k}p^{2^m-k}2^k&=\frac{2^m(2^m-1)(2^m-2)\cdots(2^m-k+1)}{k!}\cdot p^{2^m-k}2^k\\
&<\frac{2^{mk}}{k!}\cdot p^{2^m-k}2^k\\
&=\frac1{2^k}\cdot\frac1{k!}\cdot p^{2^m}.
\endalign$$ 
Hence by~(11),
$$p^{2^m}<\sum_{k=1}^{2^m}\frac1{2^k}\cdot\frac1{k!}\cdot p^{2^m}+1<p^{2^m}\left(\sqrt{e}-1+\frac1{p^{2^m}}\right)<0.8p^{2^m},$$
a contradiction. 

Hence we have~(6) with either $m=0$ or $m=1$. If $m=0$ then~(6) becomes
$$2^ap^b=q+1=p+3,$$
implying $p\mid3$, and hence $p=3$, $q=5$. Therefore $2^a3^b=6$, and so $a=b=1$, and we have the RAP2~$(5,6)$.

If $m=1$ then~(6) becomes
$$2^ap^b=q^2+1,$$
which implies $a=1$ since $q^2\equiv1\pmod4$. Therefore
$$2p^b=q^2+1=(p+2)^2+1=p^2+4p+5,$$
implying $p\mid5$, and hence $p=5$, $q=7$. Therefore $2\cdot5^b=50$, and so $b=2$, and we have the RAP2~$(49,50)$. Therefore the only RAP2s of the form $(q^c, 2^ap^b)$ are $(5,6)$ and $(49,50)$. 

We summarize our results from this and the previous section:
\proclaim{Theorem 1}
The only RAP2s $(n,n+1)$ such that $\{\omega(n),\omega(n+1)\}=\{1,2\}$ are $(5,6)$, $(24,25)$, and $(49,50)$.
\endproclaim
%
%                         5. < 2^{2n} p q , r s >
%
\vskip 30pt
\subhead \nofrills{5. RAP2s of the form $(2^{2n}pq, rs)$} \endsubhead

We now turn our attention to RAP2s $(n,n+1)$ where $\{\omega(n),\omega(n+1)\}=\{2,3\}$. There are 88 such pairs less than $10^9$. Of these, 41 have the form $(4pq,rs)$, six have the form $(16pq,rs)$, and three have the form $(64pq,rs)$, with $p<q$, $r<s$ odd primes. Among the remaining 38 pairs, no discernable patterns emerged. These data led us to narrow our investigation to those pairs of the form $(2^{2n}pq,rs)$, $n\ge1$. 

Given such a pair, we have
$$\align
2+p+q&=r+s,\tag{12}\\
2^{2n}pq+1&=rs.\tag{13}
\endalign$$
By~(12) we have integers $x$, $y$, and $z$ such that
$$\alignat 2
r&=x-y,&\qquad s&=x+y,\tag{14}\\
p&=x-1-z,&\qquad q&=x-1+z.\tag{15}
\endalignat$$
Substituting~(14) and~(15) into~(13), and simplifying, gives us
$$((2^{2n}-1)x-(2^{2n}+1))(x-1)=(2^nz-y)(2^nz+y),$$
which may be expressed as
$$\frac{(2^{2n}-1)x-(2^{2n}+1)}{2^nz-y}=\frac{2^nz+y}{x-1}=\frac{a}{b},\tag{16}$$
where $a/b$ represents the fractions in~(16) in their lowest terms; thus $(a,b)=1$. Separating the variables $x$, $y$, and $z$ in~(16) gives us
$$\align
(2^{2^n}-1)bx+ay-2^naz&=(2^{2^n}+1)b,\\
ax-by-2^nbz&=a,
\endalign$$
which we solve for $x$, $y$, in terms of $z$:
$$\align
(a^2+(2^{2^n}-1)b^2)x&=2^{n+1}abz+a^2+(2^{2^n}+1)b^2,\tag{17}\\
(a^2+(2^{2^n}-1)b^2)y&=2^n(a^2-(2^{2^n}-1)b^2)z+2ab.\tag{18}
\endalign$$
Our data of RAP2s less than $10^9$ revealed to us many different rational numbers for the quotient $a/b$ in~(16), but some persisted more than others, especially $2/1$ and $7/4$ in the cases where $n=1$. Recognizing these values as solutions to the Pell equation $a^2-3b^2=1$, we decided to assume that $a$, $b$, solved the Pell equation 
$$a^2-(2^{2^n}-1)b^2=1\tag{19}$$
in the general case for $n\ge1$. Under this hypothesis, (17) and~(18) simplify to
$$\align
(2a^2-1)x&=2^{n+1}abz+2a^2+2b^2-1,\\
(2a^2-1)y&=2^nz+2ab.\tag{20}
\endalign$$
It is well known (e.g., as shown by Shockley~[7], Ch. 12) that all positive solutions to~(19) are given by 
$$\alignat 2
a_1&=2^n,&\qquad b_1&=1,\\
a_{j+1}&=2^na_j+(2^{2^n}-1)b_j&\qquad (j&\ge1),\\
b_{j+1}&=a_j+2^nb_j&\qquad (j&\ge1).\tag{21}
\endalignat$$
One shows by induction that $2^n\mid a_jb_j$ for all $j\ge1$. Hence we may parametrize $z$ from~(20): since $y$ is an integer it follows that $2a^2-1$ divides $2^nz+2ab$, and since $2a^2-1$ is odd we have 
$$z\equiv-\frac{2ab}{2^n}\pmod{2a^2-1}.$$
Thus $z$ has the form given by
$$z=(2a^2-1)k+2a^2-1-\frac{2ab}{2^n}\tag{22}$$
for integers $k\ge0$. Substituting~(22) into~(17) and~(18) gives us
$$\align
x&=2^{n+1}abk+2^{n+1}ab-2b^2+1,\tag{23}\\
y&=2^nk+2^n.\tag{24}
\endalign$$
Substituting~(22), (23), and~(24) into~(14) and~(15) gives us
\proclaim{Theorem 2}
Let integral $n\ge1$ be given and let $a$, $b$, be solutions to the Pell equation~(19). Then $(2^{2n}pq,rs)$ is a RAP2 if, for an integer $k\ge0$, the following four quantities are all prime:
$$\align
p&=2(2^{n+1}ab-2a^2+1)k+\left(2^{n+1}-2b^2-2a^2+1+\frac{2ab}{2^n}\right),\\
q&=2(2^{n+1}ab+2a^2-1)k+\left(2^{n+1}-2b^2+2a^2-1-\frac{2ab}{2^n}\right),\\
r&=2^{n+1}(2ab-1)k+2^n(2ab-1)-2b^2+1,\\
s&=2^{n+1}(2ab+1)k+2^n(2ab+1)-2b^2+1.
\endalign$$
\endproclaim
Note that we substituted $2k$ instead of $k$ to ensure $p$ and $q$ as given in Theorem~2 are odd. We also kept $2ab$ in the numerators above (rather than reduce to $ab/2^{n-1}$) since the Pell sequences~(21) have the property $b_{2j}=2a_jb_j$ (as well as $a_{2j}=2a_j^2-1$). Moreover, one shows by induction that for all $n$ and $k$, if $a_{3j}$, $b_{3j}$ in~(21) are the solutions used in applying Theorem~2, then at least one of $p$, $q$, $r$, and $s$ is divisible by~3 (hence no RAP2 is produced).

There are 149 RAP2s of the form $(2^{2n}pq,rs)$ whose elements are less than $2^{34}$. Of these, 116 correspond to $n=1$, and 16 of these involve the solutions $a_1=2$, $b_1=1$ of the Pell equation $a^2-3b^2=1$, while an additional 3 involve $a_2=7$, $b_2=4$. Also, 16 such RAP2s correspond to $n=2$, 3 of which involve the solutions $a_1=4$, $b_1=1$ of the equation $a^2-15b^2=1$, and 9 of the RAP2s correspond to $n=3$, 3 of which involve $a_1=8$, $b_1=1$ ($a^2-63b^2=1$). Finally, 3 of the RAP2s involve $n=4$. 

We had found the RAP2s less than $2^{34}$ by a straightforward computer search. Later on, we applied Theorem~2 to search for the RAP2s of the special form described in that theorem.  We found literally thousands of them. We computed them on a PC, using the UBASIC software package. Primality of $p$, $q$, $r$, $s$, were verified by the APR primality test due to Adleman, Pomerance, and Rumely~[1].
%
%                         6. conclude
%
\vskip 30pt
\subhead \nofrills{6. Concluding Remarks} \endsubhead

It is unknown if there are infinitely many RAP2s. The question of infinitude also remains open for ordinary Ruth--Aaron pairs---see Pomerance~[6] for a detailed history. In light of Theorem~2, fixing $n$ at say $n=1$, if one could show that for each solution $a_j$, $b_j$, $3\nmid j$, to~(19), there exists at least one $k$ for which $p$, $q$, $r$, $s$, are all prime, then a proof of infinitely many RAP2s of the form $(4pq,rs)$ would be obtained. We have not been able to produce such a proof, but we conjecture the existence of infinitely many RAP2s nonetheless. 

We have also considered RAP2s $(n,n+1)$ for which $\{\omega(n),\omega(n+1)\}=\{1,4\}$. These would be obtained by finding distinct odd primes $p_1$, $p_2$, $p_3$, $q$, and positive integers $a$, $b_1$, $b_2$, $b_3$, $c$, such that
$$2+p_1+p_2+p_3=q,\tag{25}$$
and such that
$$2^ap_1^{b_1}p_2^{b_2}p_3^{b_3}=q^c\pm1.\tag{26}$$
Let $h=[e_{p_1}(q),e_{p_2}(q),e_{p_3}(q)]$. Then $p_1$, $p_2$, $p_3$, all divide $q^c-1$ only if $h\mid c$, in which case $q^h-1$ divides $2^ap_1^{b_1}p_2^{b_2}p_3^{b_3}$.  Thus if $q^h-1$ is found to contain any prime factors other than 2, $p_1$, $p_2$, $p_3$, then a contradiction is obtained. Using modular arithmetic, we can find $\alpha$, $\beta_1$, $\beta_2$, $\beta_3$, such that $2^{\alpha}\|q^h-1$ and $p_i^{\beta_i}\|q^h-1$ ($1\le i\le 3$). A contradiction is obtained if $2^{\alpha}p_1^{\beta_1}p_2^{\beta_2}p_3^{\beta_3}<q^h-1$.

In the case of $q^c+1$, (26) becomes
$$2^ap_1^{b_1}p_2^{b_2}p_3^{b_3}=\prod \Sb d\mid2c\\d\nmid c \endSb \Phi_d(q)$$
by~(1). By Lemma~1, the primes $p_1$, $p_2$, $p_3$, all divide $q^c+1$ only if $e_{p_1}(q)$, $e_{p_2}(q)$, and $e_{p_3}(q)$ are all even such that each quantity is exactly divisible by the same power of~2. In this case we have $q^{h/2}+1\mid2^ap_1^{b_1}p_2^{b_2}p_3^{b_3}$. Thus a contradiction is obtained if $q^{h/2}+1$ contains any prime factors other than 2, $p_1$, $p_2$, or $p_3$.

For all odd primes $q<20000$, we found all triples of odd primes $p_1<p_2<p_3$ satisfying~(25), and then we disproved the possibility of~(25) and~(26) by computation. We conjecture the nonexistence of RAP2s $(n,n+1)$ for which $\{\omega(n),\omega(n+1)\}=\{1,4\}$, although we have not yet obtained a proof.

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\enddocument
%
%
%                THE END 
%
%The authors should be made aware of the article by
% Pomerance entitled "Ruth-Aaron numbers revisited" in Paul Erdos and his
%mathematics, I, Budapest (1999), 567--579, Bolyai Soc. Math. Stud., 11,
%Janos Bolyai Math. Soc., Budapest, 2002. Paul Erdös and His Mathematics
%Series: Bolyai Society Mathematical Studies, Vol. 11 
%Halasz, G.; Lovasz, L.; Simonovits, M.; Sós, V.T. (Eds.) 
%2002,