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A050647
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a(n+1)^3 is next smallest cube ending with a(n)^3, initial term is 1.
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3
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OFFSET
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1,2
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COMMENTS
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Frank Ellermann conjectures (Jun 11 2001) that if b(n) = 10^(3^n -1), i.e., 1, 100, 100000000, etc. then a(n) = concatenation b( n-2 ) || a( n-1 ) for n > 1.
a(6) is too large to include (contains 122 digits).
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LINKS
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FORMULA
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a(n) = Sum_{i=0..n-1} 10^((3^i-1)/2). - Max Alekseyev, Jan 07 2015
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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More terms from Ulrich Schimke (ulrschimke(AT)aol.com), Nov 06 2001, who remarks that a(6) = 10^121+a(5) = concat (10^80, a(5)) and a(7) = 10^364+a(6) = concat (10^242, a(6)), which supports Ellermann's conjecture
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STATUS
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approved
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