On the primality of $j(n)=\varphi(p_n+1-n)+1$ when $j(n) \equiv 19 \pmod {100}$

Question

Related to Power of primes.

Let $p_n$ denote n-th prime and $\varphi$ the totient function.

For natural $n$, define $j(n)=\varphi(p_n+1-n)+1$.

For $n$ up to $10^9$ if $j(n) \equiv 19 \pmod {100}$ then $j(n)$ is prime. The computation took about 1 hour and 50 minutes.

Q1 How to explain this numerical result?

Q2 Does this hold for all $n$?

pari/gp code in case someones wants to test it.

{
test3(lim=10^2)=
/*
We avoid `prime(n)` for performance reasons
*/
default(primelimit,2*lim);
cou=0;coup=0;
n=2;
p=prime(n);
while(n<lim,
n += 1;
p=nextprime(p+1);
t=eulerphi(p+1-n)+1;
if(t%100==19,
    if(!isprime(t),print(" bad ",n);cou+=1;
    ,
    coup += 1);
);
);
return([cou,coup]);
}

Answer 1

Using the observations in Max's answer, we can find an explicit counterexample: for $n = 19179214864, p_n = 496527469717$, we have $p_n + 1 - n = 477348254854 = 2 \cdot 6203^3$, and $\varphi(p_n + 1 - n) + 1 = 238635650219 = 23 \cdot 103 \cdot 100732651$. I used this SageMath code.

Answer 2

Since $j(n)\equiv 19\pmod{100}$, we have $\varphi(p_n+1-n)\equiv 18\pmod{100}$ meaning that $\nu_2(\varphi(p_n+1-n))=1$. That is, $p_n+1-n$ is $q^k$ or $2q^k$ for a prime $q\equiv 3\pmod4$.

If $k=1$, then $j(n)=q$ is prime, and this is the case that you observe. A counterexample may exist only among those $n$ for which $p_n+1-n$ is $q^k$ or $2q^k$ with $k>1$ and $(q-1)q^{k-1}\equiv18\pmod{100}$, which tend to be rare (if any exists at all).

ADDED. Impossibility of $(q-1)q\equiv 3\pmod{5}$ rules out the case $k=2$.

In search for a counterexample, let $d:=p_n+1-n$. We want it to be of the form $q^k$ or $2q^k$ with $k>1$, $(q-1)q^{k-1}\equiv18\pmod{100}$, and number $\varphi(d)+1$ being composite. Leaving aside the question whether a suitable $n$ exists for a given value of $d$, the aforementioned conditions restrict the values of $d$ below $10^{10}$ to the set $S\cup 2S$ with $S:=\{103^3, 503^3, 1103^3, 1303^3, 79^5, 2003^3\}$. However, none of them gives a soluble $p_n+1-n=d$, and the chance of solubility is even smaller for larger $d$.