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Search: id:A035014
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| A035014 |
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a(n) contains n digits (either '3' or '4') and is divisible by 2^n. |
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+0
25
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| 4, 44, 344, 3344, 33344, 433344, 3433344, 33433344, 333433344, 3333433344, 43333433344, 343333433344, 3343333433344, 33343333433344, 433343333433344, 3433343333433344, 43433343333433344, 443433343333433344
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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If (n-1)-th term is divisible by 2^n, then n-th term begins with a 4. If not, then n-th term begins with a 3.
Proof of conjecture that a(n) ends with a(n-1): If a(n) is divisible by 2^n, then a(n) is divisible by 2^(n-1), so a(n)-k*10^(n-1) is divisible by 2^(n-1) for integer k, but if k is first digit of a(n) then a(n)-k*10^(n-1) is an n-1 digit number made up of 3s and 4s and divisible by 2^(n-1) and so must be a(n-1).
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FORMULA
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a(n)=a(n-1)+10^(n-1)*(4-[a(n-1)/2^(n-1) mod 2]) i.e. a(n) ends with a(n-1).
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CROSSREFS
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Cf. A050620, A050621, A050622, A023402.
Sequence in context: A074751 A129551 A081078 this_sequence A030987 A043039 A002754
Adjacent sequences: A035011 A035012 A035013 this_sequence A035015 A035016 A035017
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KEYWORD
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nonn,base
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AUTHOR
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J. Lowell (jhbubby(AT)avana.net)
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EXTENSIONS
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Corrected and extended by Patrick De Geest (pdg(AT)worldofnumbers.com), Jun 15 1999.
Formula, proof of conjecture and more terms from Henry Bottomley (se16(AT)btinternet.com), Feb 14 2000
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