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Search: id:A067634
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| A067634 |
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a(1) = 1; string of digits of a(n)^2 is a substring of the string of digits of a(n+1)^2. |
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+0
3
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| 1, 4, 13, 130, 1300, 13000, 130000, 1300000, 13000000, 130000000, 1300000000, 13000000000, 130000000000, 1300000000000, 13000000000000, 130000000000000, 1300000000000000, 13000000000000000, 130000000000000000
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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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.
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CROSSREFS
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Cf. A014563, A066825.
Sequence in context: A015460 A121813 A006104 this_sequence A042537 A132512 A050629
Adjacent sequences: A067631 A067632 A067633 this_sequence A067635 A067636 A067637
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KEYWORD
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nonn,base
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AUTHOR
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David W. Wilson (davidwwilson(AT)comcast.net), Feb 05 2002
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EXTENSIONS
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More terms from David W. Wilson (davidwwilson(AT)comcast.net), Feb 05 2002
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