OFFSET
1,2
COMMENTS
Provably infinite.
The obvious pattern continues. Proof: By induction, assume that 13*10^k is the (k+2)nd element in the sequence for some k >= 1. Clearly 13*10^{k+1} satisfies the required condition; we need to show that no other number works. Equivalently, we need to show that 169*10^{2k+2} is the smallest square of one of the forms: 169*10^{2k+1}+a, a*10^{2k+3}+169*10^{2k}, 169*10^{2k+2}+a*10+b, a*10^{2k+4}+169*10^{2k+1}+b, a*10^{2k+4}+b*10^{2k+3}+169*10^{2k}, where 0 <= a,b <= 9. Insisting that the number be less than 169*10^{2k+2} and checking that it is a 2-adic, 3-adic and 5-adic square eliminates all but 169*10^{2k+1}+9 and 1169*10^{2k+1}+1. To eliminate these, reduce modulo the primes 101, 137=(10^4+1)/173 and 5882353=(10^8+1)/17; these all divide 10^16+1, so it suffices to check k=0,1,2,3,4,5,6,7. QED. - Eric Rains, Jan 29 2002
MATHEMATICA
Join[{1, 4}, NestList[10#&, 13, 20]] (* Harvey P. Dale, Jul 25 2024 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
David W. Wilson, Feb 05 2002
EXTENSIONS
More terms from David W. Wilson, Feb 05 2002
STATUS
approved