OFFSET
1,3
COMMENTS
Leading zeros in substring allowed so 103^2 = 10609 is rejected because 1{060}9 contains a palindromic substring.
A comment in A052061 suggests that this sequence is infinite.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10001
FORMULA
a(n) = A052061(n)^2. - Andrew Howroyd, Aug 11 2024
EXAMPLE
2304 (= 48^2) -> substrings 23, 30, 04, 230, 304 and 2304 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, 100, if(noPalSub(n^2), print1(n^2", ")))
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Jan 15 2000
EXTENSIONS
Program and b-file from Charles R Greathouse IV, Sep 09 2009
Offset changed by Andrew Howroyd, Aug 11 2024
STATUS
approved
