determine count of Proth primes until environment n

proth.gp

\\ detemine count of Proth primes until environment n
\\ environment l (loops) allows to tune work distribution
\\
\\ based on Karim's isProth2(): 
\\ https://pari.math.u-bordeaux.fr/archives/pari-users-2403/msg00064.html
\\ and Bill's parallel implementation:
\\ https://pari.math.u-bordeaux.fr/archives/pari-users-2403/msg00067.html
\\
isProth2(p) = !(p >> (valuation(p-1,2)<<1));
export(isProth2);
nbt=default(nbthreads);

doit(n,l)={
  my(c=0,B=n\nbt\if(l,l,1));
  parfor(ii=0,
    (n+B-1)\B,
    my(cc=0,i=ii*B+1);
      forprime(p=i,
        min(i+B-1,n),
        if(isProth2(p),cc++));cc,
    C,
    c+=C
  );
  c
};

n=eval(getenv("n"));
gettime();
t0=getwalltime();
r=doit(n,eval(getenv("l")));
wt=getwalltime()-t0;
tt=gettime();
print(nbt);
printf("%s × %.3f = %s",strtime(wt),tt/wt/1.0,strtime(tt));
print(r);
t0=getwalltime();
c=0;forprime(p=3,n,if(isProth2(p),c++));
wts=getwalltime()-t0;
printf("%s (%.3f×)",strtime(wts),wts/wt/1.0);
print(c);

For 2.5× n even closer to factor 32×, but the other factor 13.046× seems to be maxed out:

hermann@7950x:~$ n=10^11 l=32 gp -q < proth.gp 
32
1min, 10,058 ms × 31.784 = 37min, 6,695 ms
39170
15min, 13,952 ms (13.046×)
39170
hermann@7950x:~$