OFFSET
0,2
COMMENTS
Note that 48384 (k=12) is a 'Droll' number: see A019507.
There are 3 more known terms: a(15)=405504, a(16)=40955904, a(20)=677707776. Any other terms would have at least 18 decimal digits. Conjecture: The sequence is finite and has no other terms than those shown here. - Hugo Pfoertner, Aug 13 2019
EXAMPLE
a(7) = 82728 because it is the smallest palindrome with 7 palindromic and no other prime factors: 82728 = 2^3 * 3^3 * 383. If other prime factors are not excluded, A309565(7) = 29792 = 2^5 * 7^2 * 19 also has exactly 7 palindromic factors and the additional factor 19.
PROG
arepalf(nf, x)={forstep(j=nf, 1, -1, if(is_A002113(x[j, 1]), , return(0))); return(1)};
md=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]; \\ Middle digits in odd length palindromes
a=vector(64); a[1]=2; a[2]=4; a[3]=8;
for(d=2, 11, print("Digits: ", d); if(d%2==0, for(k=10^((d-2)/2), 10*10^((d-2)/2)-1, my(dv=digits(k)); P=fromdigits(concat(dv, Vecrev(dv))); x=factor(P); bigom=vecsum(x[, 2]); nf=#x[, 2]; for(j=1, #a, if(a[j], , if(j==bigom, if(arepalf(nf, x), print("a(", j, ")=", a[j]=P)))))), for(k=10^((d-3)/2), 10*10^((d-3)/2)-1, my(dv=digits(k)); for(m=1, 10, P=fromdigits(concat(concat(dv, md[m]), Vecrev(dv))); x=factor(P); bigom=vecsum(x[, 2]); nf=#x[, 2]; for(j=1, #a, if(a[j], , if(j==bigom, if(arepalf(nf, x), print("a(", j, ")=", a[j]=P))))))))); a \\ Hugo Pfoertner, Aug 13 2019
CROSSREFS
KEYWORD
nonn,base,more,hard
AUTHOR
Patrick De Geest, Jun 15 1998
EXTENSIONS
Definition clarified by Hugo Pfoertner, Aug 08 2019
STATUS
approved