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%I #12 Feb 12 2015 04:04:20
%S 1,4,13,130,1300,13000,130000,1300000,13000000,130000000,1300000000,
%T 13000000000,130000000000,1300000000000,13000000000000,
%U 130000000000000,1300000000000000,13000000000000000,130000000000000000
%N a(1) = 1; string of digits of a(n)^2 is a substring of the string of digits of a(n+1)^2.
%C Provably infinite.
%C 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
%Y Cf. A014563, A066825.
%K nonn,base
%O 1,2
%A _David W. Wilson_, Feb 05 2002
%E More terms from _David W. Wilson_, Feb 05 2002
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