\\========================================================================
\\ Search for Palindromic SUBSTRINGS - Patrick De Geest - Pari/GP - Version [2, 11, 4] - 8/12/2021
\\========================================================================
\p100;
f=1040; inp=fibonacci(f);
inpd=digits(inp);
linp=#(inpd);
spp=linp;
{
print();
cnt=0;
for(i=1,spp-1,
for(n=1,linp-spp+1,
extract=fromdigits(inpd[n..spp+n-1]);
if(inpd[n]!=[0],
print1(extract); d=digits(extract); lext=#(d);
if(d==Vecrev(d) && (extract>9),
cnt+=1;
prf=isprime(extract); if(prf==1, color="\e[38;5;93m",color="");
print(" is palindromic ! "color"fib="f"\e[38;5;37m linp=["linp"] lext=[\e[38;5;92m"lext"\e[38;5;37m] at=["n"]");
lenpp1=#(inpd[1..n-1]);
lenpp2=#(inpd[n..spp+n-1]);
lenpp3=#(inpd[spp+n..linp]);
pp3=fromdigits(inpd[spp+n..linp]);
if(lenpp1!=0, pp1=fromdigits(inpd[1..n-1]); print1(pp1) );
pp2=fromdigits(inpd[n..spp+n-1]); print1("\e[38;5;92m"pp2"\e[38;5;37m");
cntzero=lenpp3-#(digits(pp3)); if(cntzero, for(k=1,cntzero, print1("0");) );
if(lenpp3!=0, print(pp3));
\\ break(2);
z=1;while(z<20000000, z++); , print();
);
);
);
print();
spp-=1;
);
det=0;
for(k=1,linp-1,
if(inpd[k]==0 && inpd[k+1]==0, det+=1););
if(cnt==0 && det, print("Fib("f") "inp" Strings of zero's detected!"));
if(cnt==0 && det==0, print("Fib("f") "inp" is palfree !"));
}