OFFSET
1,3
COMMENTS
Leading zeros in the substrings are allowed so 103^2 = 10609 is rejected because 1{060}9 contains a palindromic substring.
Probabilistic analysis strongly suggests that this sequence is not finite. - Franklin T. Adams-Watters, Nov 15 2006
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10001
EXAMPLE
118^2 = 13924 -> substrings 13, 39, 92, 24, 139, 392, 924, 1392, 3924 and 13924 are all non-palindromic.
PROG
(PARI) noPalSub(n)={my(d); local(digit); digit=eval(Vec(Str(n))); d = #digit; for(len=2, d, for(i=1, d-len+1, if(isPalSub(i, len), return(0)))); 1};
isPalSub(start, len)={my(b=start-1, e=start+len); for(j=1, len>>1, if(digit[b+j] != digit[e-j], return(0))); 1};
for(n=0, 200, if(noPalSub(n^2), print1(n", ")))
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Jan 15 2000
EXTENSIONS
Program and b-file from Charles R Greathouse IV, Sep 09 2009
STATUS
approved