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A078221
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a(1) = 1, a(n+1) > a(n) is the smallest multiple of a(n) using only odd digits.
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14
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OFFSET
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1,2
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LINKS
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FORMULA
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a(n) = 10^(2^(n-3)) - 1 for n >= 3. (Proof by induction. Consider a(n)*f, L = ceiling(log(f)/log(10)), g1 = number formed by the first L digits of a(n)*f, g2 = number formed by the last L digits of a(n)*f => g1 + g2 = number formed by L 9's, if L <= 10^(2^(n-2)) + 1). - Sascha Kurz, Jan 04 2003
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MAPLE
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1, 3, seq(10^(2^(n-3))-1, n=3..11);
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PROG
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(Python)
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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